I really love this problem. I solved it using only Pythagoras' theorem (multiple times) to find the equations for the coordinates (x, y) of the third vertex (using expressions of r). I got 3 equations. After eliminating x and y I got the same equation that was shown in the video for r and solved it by substituting (r-2) = z. The only valid solution for z is 2*cos(20°) which I think is quite a tricky calculation - but I really enjoyed solving it (30+ years after finishing school).
It's possible to figure out the sides a, b and c even if you forgot about (or never knew) the chord theorem. For instance in the case for a) you get a right angled triangle with sides a/2, r - 2 and r. Thus we get (a/2)^2 + (r - 2)^2 = r^2 => a ^2 = 16(r - 1).
I did it, my calculations were long but is easy to understand compared to this. I got the same equation r3-6r2+9r-3=0 but yeah I admit I used calculator to find r here. Even 3 year olds can solve it in seconds
It is possible to solve this without any special knowledge about chords. Start with image at 1:20 and mark point above number 2 D (intersection of chord and circle). Than there are two possible ways to express length |CD|. When you do this for all three sides, you get this equations: 𝛼+β+ɤ=180; 1=R(1-cos𝛼); 2=R(1-cosβ);3=R(1-cosɤ). After some algebra and elimination impossible answer you get R and rest is simple.
I got as far as the intersecting chords, but 3:30 "... after we do quite a bit of algebra ..." 4:25 "... I myself wouldn't be able to solve this cubic equation ..." I don't feel _that_ bad about giving up.
I'm so glad I watched the video instead of trying to solve the problem. I knew I had to use the relationship between the angles - and that looked wicked.
Fundamentally this is a cubic eqn solving problem. Before coming to that point, my way is... Let the intersection of the 3 lines be O, which is the center of the circle. From any vertex, say, B, draw diameter BOD, thus ΔABD and ΔCBD must be rt Δ. So，from midlines, know, AD=2(r-1), CD=2(r-2), from Gougu theorem, can solve AB and BC, and likewise, for AC. Note ABCD 4pt concyclic, so, as circumferential angles, ∠A=∠BDC, ∠C=∠ADB, equate their sines or cosines, got same cubic eqn... The rest is the same.
I was thinking... We know the chord is twice the angle. There is a relationship between the chord's arc length and its height above the line connecting the endpoints. I don't know what that is off hand, but it exists and will be a trig function of some kind. We know the relative sizes of the heights, and we know that the sum of the chords must be 2pi. So take one of them, plus twice that, plus triple that, to get 6 of them equals 2pi. From there solve for the original relationship formula's values.
The area is 17.18. The corresponding central angles to the cords with segment heights 1,2,3 are 84.5°, 122° and 153.5° respectively (rounded). They add up to 360. Cord lengths c1,c2,c3 are 5.19, 6.79, 7.59, respectively (rounded). The radius of the outer circle is approx. 3.9. Equations used: h segment height (sagitta) R radius c cord length (c1,c2,c3) ß central angle (ß1,ß2,ß3) s semi-circumference inscribed triangle A area irregular triangle h=c/2 * tan(ß/4) ==> c=2h * cot(ß/4); ß=arctan(2h/c) * 4; R=h/2 + c^2/(8h); A=Sqrt[s*(s-c1)*(s-c2)*(s-c3)] (Heron); c2^2/c3^2 = 12/15, c1^2/c2^2 = 7/12, c1^2/c3^2 = 7/15 (home-cooked assumption) ;-)
I tried this problem again, but using a different approach. As it turns out, you can use analytic geometry. First, place the circle and the triangle on a coordinate plane. You can then label the points of the triangle as: (x1, y1), (-x0, y0), and (x0, y0). Next, label the mid-points of each of the legs in terms of x0, x1, y0, and y1. Once this is done, you can set up 5 equations in 5 unknowns (r, x0, x1, y0, y1), using distance formulas and the coordinates of the points and the mid-points of the triangle. If you solve for r, you end up with the cubic equation in terms of r - the same one shown in the video. This approach is conceptually much easier than the approach shown in the video. It involves a quite a bit of tedious algebra, although not much worse than the one shown in the video.
hey presh. im a 7th grader in india and i found an approach without trigonometry but didnt solve it. heres my approach lets first do some preliminaries 1. how to calculate the height of a triangle when you are given it has side lengths a, b and c and c is the base and the angles adjacent to c are less than 90°: first draw an altidude with length h to the side c such that it divides c into lengths of x and c - x. you now have 2 right triangles - one with legs h and x and hypotenuse a; and the other with legs c - x and h and hypotenuse b. you can use the gougu theorem and get two equations in the variables x and h. solve for h in both equations then set them equal to each other. then solve for x and substitute into one of the two original equations and solve for h. you will get the formula for the height of the triangle 2. to prove that when you connect the midpoints of any triangle they will bound a similar triangle that is 2x smaller than the original triangle: when you connect the midpoints of the sides consider one of the outer triangles. two of its sides are half the sides of the original triangle and it shares an angle with the original triangle therefore it is congruent to a triangle that is a 2x scaled down version of the original triangle. therefore the side of the triangle that connects two midpoints will be parallel to the other side of the original triangle. the same applies for the other sides of the middle triangle and the original triangle. since the sides of the middle triangle are parallel to the sides of the original triangle the middle triangle is similar to the original triangle and is a 2x scaled down version of the original triangle since it shares sides with the outer triangles now lets solve the problem: first extend the perpendicular bisector of the sides of triangle ABC. let the centre of the circle be M. the perpendicular bisectors of triangle ABC will meet a point M. let X, Y and Z be the midpoints of AB, BC and AC respectively. Mark points P, Q and R along the circumference of the circle such that respectively such that PX perpendicularly bisects AB and PX = 1, YQ perpendicularly bisects BC and YQ = 2 and ZR perpendicularly bisects AC and ZR = 3. if r is the radius of the circle then XM = r - 1, YM = r - 2 and ZM = r - 3. now draw AM. AM = r and XM = r - 1 therefore you can calculate XA using the gougu theorem XA = √(AM² - XM²) = √(r² - (r - 1)²) = √(2r - 1) AB = 2(XA) = 2√(2r - 1) similarly you can calculate BC and AC in terms of r. now consider triangle XYZ. since X, Y and Z are the midpoints of AB, BC and AC respectively that means triangle XYZ is a 2x scaled down version if triangle ABC. since XM = r - 1, YM = r - 2 and ZM = r - 3 and triangle XYZ is a 2x scaled down version of triangle ABC then there must be a point, say point K such that CK = 2(r - 1), AK = 2(r - 2) and BK = 2(r - 3). since triangle XYZ is similar to triangle ABC that means BK and ZM are parallel. since ZM is perpendicular to AC that means BK is perpendicular to AC when extended. extend BK until it intersects AC at point W. KW is perpendilcular to AC therefore BW is the height of triangle ABC and KW is the height of triangle AKC. since you know the sides of triangle ABC and triangle AKC in terms of r that means you can calculate BW and KW in terms of r. BK + KW = BW since you know BK, BW and KW in terms of r you can substitute and solve for r. you can now substitute the value of r in the formula for BW and AC. now the area of triangle ABC = ½(BW)(AC) and you can substitute the values of BW and AC in the formula for the area of triangle ABC and solve for the area of triangle ABC. this is under the assumption that angle BCA and angle CAB are less than 90° and that point M is in the interior of triangle XYZ
You can, but you need to add infinitely many terms. Using the perimeter formula from calculus and expanding the function using Taylor series, you can integrate term by term and get an infinite series.
Only in the same sense that we can't work out the value of Pi. We can get as arbitrarily close as we want, so it's solved for all intents and purposes, there's just no analytical solution.
I hate not being able to solve a problem and when my answer was 17.2 a got frustrated and spent around 2 hours trying to figure out what’s wrong until I finally decided to watch your answer . Please don’t do me like that again .
I did not use the Chord Theorem to get the lengths of the sides of ABC in terms of r.I used simple triangle manipulation. I got a=2sqrt(4r-4),b=2sqrt(6r-9) and c=2sqrt(2r-1) I plugged these values into Wolfram Alpha and got the area ABC=4sqrt(r(2r-3)). This means r>1.5 I then equated area BOC+ area AOC +area BOA= area ABC (r-2)sqrt(4r-4)+(r-3) sqrt(6r-9)+(r-1)sqrt(2r-1)= 4sqrt(r(2r-3)) (r-2)sqrt(4r-4)+(r-3) sqrt(6r-9)+(r-1)sqrt(2r-1)-4sqrt(r(2r-3))=0.................(1) I plugged equation (1) into Wolfram Alpha for a numerical answer and got r=3.8794 approx. another root was r=1.5 but this was less than r>1.5 therefore I plugged r>3 into Area ABC=4sqrt(r(2r-3) Area ABC was 4sqrt(3.8794(4.7588))=17.18664073 Area ABC=17.18664073 sq. units.This was a transcendental equation because it contained a circle and therefore pi. Check answer; For r=3.8794, (r-2)sqrt(4r-4)=6.37823 (r-3)sqrt(6r-9)=3.32274 (r-1)sqrt(2r-1)=7.48578 Total =17.18675 The slight difference at the fourth decimal place is caused by rounding errors.If r was taken to ten decimal places the 'error' would be insignificant. Try r=3.8793852415718, (r-2)sqrt(4r-4)=6.378164190722 (r-3)sqrt(6r-9)=3.322669765738 (r-1)sqrt(2r-1)=7.485720779302 Area large tri=17.186554735762 4sqrt(r(2r-3) =17.186554735762 The area of triangle ABC=17.186554735762 square units to 13 decimal places. Presh equation r^3-6r^2+9r-3=0 This equation had three real roots according to Wolfram Alpha:- r=0.46791,r=1.6527 and r=3.8794. The sum of the roots=6. The sum of the roots of the cubic ax^3+bx^2+cx+d=0 is -b/a.
Al-Tulsi's law of sines is not well explained at 4:07. Go back to 1:03 and it can be figured out by pointing out that the arc BC subtends twice the angle A. Then draw a radius from B to the center. The vertex angle at the center equals angle A. Then apply the law of sines to that sub triangle.
It took me several hours to solve this interesting problem; thank you Presh. My answer was Area = 17.798. I didn't use trigonometry - just geometry and algebra.
In my earlier explanation, I have related the values of a, b and c in terms of radius of the circle, r. Such that a^2 = 16*(r -1), b^2 = 4 * (2 *r - 1) and c^2 = 12 * (2* r - 3). Also the area of the circumscribed triangle is known as, Area of the tringle = (a * b* c) / (4* r ). Using these two relations, we can also derive the area of the triangle directly in terms of r as follows: Area of the triangle = sqrt (48 * (4 * r - 12 +(11/r) - (3/r^2)). Therefore the radius of the circle, r, alone determines the area of the triangle inside it.
Thank you Presh sir for making these videos, you have helped me alot. I found your channel about 6 months ago and have watched almost all of your videos you have proved that Math is awesome keep making such videos thanks once again
I used intersecting chords to get expressions for a,b,c, sine law to get radius. But when I got cubic equation, I thought I went in the wrong direction. Just in case I checked the roots in program, but they're not simple. With no desire to look for mistakes I gave up and watched the video. And what do I see?! Presh does the same and solves cubic equation, that leads to that weird expression for area!!! Possibly there was a typo in the task, in other case this problem is a beast!
There is a different (and i think easier) way to solve this problem: Start with the formula Alpha = 2 arccos (1-h/r) where h are the lengths 1, 2 and 3. This results in 3 equations with 4 variables. Ad the equation that the sum of the three angles is equal to 360 and you have your 4th equation, so you can calculate the value of r. Then just use the formula for a cut circle piece: A=r²/2*(alpha-sin(alpha) Total up the aera of the segments and subtract it from the aera of the circle. Done This also needs a lot of algebra but is much easier, as you dont have to "guess" the right way to solve the problem.
Another approach: Recall Heron's Theorem and the formula for the area of a triangle: 4R/abc. As we can express a,b,c with R we cn get a polynomial of a high degree, that we can solve for R and then substitute back into one of the formulas for the answer. Ideed, no trig is needed.
Great conclusion at the end of the video!! You can't make a RIGHT choice if you have the WRONG options!! Exactly as in real life... btw: you forgot to say this is a question for kindergarten kids age 3.5 in Japan...
Ok, I admit, I didn't solve it. However I have a couple of comments: 1) At 1:08 if you draw lines from the center of the circle to the triangle's vertices you divide the main triangle into 6 right triangles which are equal 2 by 2. Those 6 triangles have hypotenuse=r and the other side that connects to the circle's center is (r-1), (r-2) or (r-3). Using Goku's theorem 😁 you can find your a/2, b/2 and c/2 in terms of 'r'. From here, to express the area of the big triangle in terms of 'r' it's trivial (= sum of the 6 right triangle's area) 2) Also from the picture at 1:08 it is obvious that r > 3 since 3 is the distance from the circle to AC. Could the circle's center be outside of the triangle? Is that why you use the chord theorem?
Good life advice at the end. I thought to myself if the answers were that clean, there has to be some not so crazy way that involve pi or trigonometry to get it. So I tried to look for a simple solution with no luck. In life people tell you things are easy, but lead you on a wrong path.
5:57 Shouldn't r >3 be obvious? I haven't tried to solve this, but I'm not particularly happy with the method. I'll give it a go (if I have time) after the IGCSE finals. ~ Nov-Dec. If I remember.
Interesting problem. The method presented in the video gives a third degree polynomial equation and the exact value of the radius r : r^3 - 6r^2 + 9r - 3 = 0 ==> r = 2(1 + cos(pi/9)) Different methods give more complex equations verified by the radius r, for example : Area = (a/2)(r - 2) + (b/2)(r - 3) + (c/2)(r - 1) = (abc)/(4r) 2(r - 2)sqrt(r - 1) + (r - 3)sqrt(6r - 9) + (r - 1)sqrt(2r - 1) - 4sqrt((r - 1)(6r - 9)(2r - 1))/r = 0 The numerical solution obtained with a computer is obviously the same : r = 3.879385 Area = 17.186555
5:30 I think we should discuss the case: 3t =-π/3 + 2nπ, although we immediately know it was another exclusion because x = 2cos(t) is negative in this case.
A neat solution! I could get as far as the cubic equation for R: 4*R^3-4*R^2*(a+b+c)+R*(a+b+c)^2-2*a*b*c=0 But once there, I had no idea what to do further.
This video does not only demonstrate a cool solution to an elegant trig question but also how westernized the education system is. Everyone is taught about the Pythagorean theorem and the laws of sines and cosines in geometry classes. But, I never knew about Al-Kashi or Al-Tusi until now. I am pretty sure the majority of the audience didn't know either. We could have called the Pythagorean theorem as the law of hypotenuse but instead, we are honoring the great mathematician behind it which is the way it should be. But when it comes to the scientists from the eastern civilization, we don't feel the need to honor their names even if their discoveries are extensively used. Am I the only one who thinks that this is big hypocrisy?
I think there should be two valid solutions, i. e. including the value 1.65 of r that was ruled out. When drawing the bisectors, it wasn't mentioned that they should connect to the *closest* point on the boundary of the circle, the opposite side should also be valid, even if that wasn't visualized.
I find the solution of 17.1865..cm² via the general determination of tangents to a function. I start with the calculation of a tangent through the point (r | 0) to the radius of the inner circles of the triangle shortened by 1 or 2 cm and get the other two points of intersection with the circumference of the triangle as a function of r. The midpoint of the line between the two other points must lie on the innermost circle (with a 3 cm shortened radius). The result is a complicated root equation, that only depends on r. The result of r was approximately 3,87939..Now it is possible to calculate the area.
I still have a doubt while solving the problem in this way. Given that the center of the circle lies inside the triangle and non of the side of the triangle passes through the radius, we can have innumerable solutions to the problem. Example: Given that a, b and c are the lengths of the sides of the triangle and r is the length of radius of the circle inside the triangle, a little effort will prove that each side of the triangle is a function of the radius of the circle, such that a^2 = 16*(r-1); b^2 = 4*(2*r-1) and c^2 = 12*(2*r-3). These relations can be established by joining points from the center of the circle to the points A, B and C, the apex points of the triangle and by using Pythagoras theorem. (The values of a, b and c can be interchanged depending on how you draw the triangle ABC). Now for any value of radius (at least greater than 3 as the center lies inside the circle and therefore will be more than 3), we will have a set of values for a, b and c using the above relations. And the area of the triangle will thus be calculated using the well-known relation, Area = sqrt((s)*(s-a)*(s-b)*(s-c)) where s is 0.5*(a + b + c). For example if we take r = 5, then we get a =8, b = 6 and c =9.165 using the above relations, so that area of the triangle becomes 23.664 sq. units. Again if change the value of r, say r = 8, we have new sets of values of a, b and c such that a =10.583, b = 7.746 and c =12.49, thus measuring the area of the triangle as 40.792 sq units. So the results (area of the triangle) all depends on the value of radius of the circle chosen. Interestingly, one of the answers to the problem as Presh has explained above can be 'E' which is 14.5 when we chose r = 3.42. In this case our a = 6.223, b = 4.833 and c = 6.788, the sides of the triangle.
Tried to us e Heron's formula after solving for the sides in terms of r. The algebra didn't play nice though. I was hoping the r would drop out. It seems there should be an easier way than what he presents though.
no need to do sin/cos law 2:24 Draw 3 line from the center O to the the angle of A, B, C, form ABO, ACO, BCO triangle, just simply calculate the area of three by base*height/2 and add them all.
@@JohnDlugosz M is the mid - point of AB, MO is the Perpendicular bisector of AB, also a part of radius of circle height of ABO is MO, in this case MO=r-1 Base of ABO is AB, In this case AB=2sqrt(2r-1) Area= (r-1)*sqrt(2r-1) do the same thing obtain can obtain the area of ACO and BCO
I notice that the areas of the segments sum to 2x (r squared) so the area of the triangle is (pi - 2) x r squared. I used Desmos to graph a sum of angle plus 2 arccos expressions to find the angle that satisfies summing to 2pi.
I never remember all those formulas, unfortunately, so I first sketched it and realized the radius had to be near 4 (and then a little under) At that, I measured the drawing and got a quick estimate of 16+ for the area and knew all the answers were wrong. Then because I'm me, I did a recursive look for the radius in a Libreoffice Calc spreadsheet and found R = ~ 3.87938, ran formulas for the angles to get angle BAC, trig for the HT, and voila area was 17.19 Very gratifying to see the answer after that.
In CAD drafting software it's a matter of trial and error to get the circle the right size to exactly touch the points of the triangle, when you don't even have the lengths of the sides of the triangle yet. In 5 iterations, I got the last corner of the triangle too small by 1/256", and the area of the triangle came to 17.165 square inches. That tiny amount of error in my drawing made a big difference in the area because all three sides are too close together by a hair each.
Al Whoosey? Sorry Presh this is just confusing. The sine rule has been attributed to Al-Tusi among others but this isn't it. What you give is a formula for a chord, a=2rsinA where A is the angle subtended at the circumference. This result was well known to the Greeks, you don't need the sine rule to get it (as A is the same anywhere on the circumference and you can take the symmetrical case). FWIW I checked your solution value, it's fine, but I used a more direct method i.e. use 1=r(1-cos A/2), 2=r(1-cos B/2), 3=r(1-cos C/2); eliminate r to solve for B and C in terms of A; vary A till A+B+C=2pi; solve for r; area=0.5r^2(sinA+sinB+sinC)
Sir you have chosen most sophisticated way to eliminate two options of radius but we can simply eliminate that two options because the one of the perpendicular bisector is 3 so obviously radius is more than 3
I knew something was up when Presh said "when I was shown this problem, these were the answer choices presented to me". It just seemed an odd way of phrasing it. So when I got 17.2 I wasn't too surprised.
Thank you for giving credit to AlKashi and AlTusi. So many forget the contributions of Arab and Persian mathematicians. What would algebra, trigonometry, and geometry be with out them?
Hey I'm here to shorten the solution: Circumradius = r = (abc)/4delta Delta means the area of involved triangle = (abc)/4r and we have a,b and c in terms of r .. now we have to find out r as already shown in the video... And that's it
Your last statement was awesome Presh. 😁 You cant make the right choice if you have the wrong options. But if you mind your decisions, you can solve anything.🙌🙌🙌
This is one of the most crazy problems so far. Whenever Al-Kashi and his law of co-sins enters the scene, I’m usually out, because we didn‘t cover this in school. But this time, Al-Kashi also invited his friend Al-Tushi or what his name was, and all the crazy cubic substitutions showed me, where my limits are. Impressive! Incomprehensible, but impressive!
The solution to this viral problem results in 2 valid answers for the radius of the incircle circle. The areas of the triangle are: 11.826sq units and 13.9876sq units So, in approximation to the nearest whole number or to the nearest 0.5, the answers are 12sq units and 14sq units.
OMG, I actually got it right and I thought it was wrong cause my answer wasn't one of the options. I didn't get a nice algebraic expression like Presh; I had to resort to a numerical solution, but I did get the right answer. I did find the lengths of the three sides of the triangle in terms of r and got the same result as Presh. Then I looked at the angles formed at the center of the circle by the altitudes and segments drawn from the center of the circle to the vertices of the triangle. (Basically they form 6 right triangles in three pairs of congruent right triangles.) I knew these angles had to add up to 360 degrees, and that allowed me to solve numerically for r. I then solved for the area in two ways to check my result: first I summed the areas of the 6 right triangles using 1/2 * base * height and added them. Then I used Heron's formula using my expressions for the sides of the triangle. They results agreed (and furthermore, they didn't agree if I plugged in random values for r, suggesting my calculation for r was correct.) So, I did get the right answer, albeit it was a numerical approximation. I didn't get an exact expression, so Presh had the better answer.
If you look at the cubic equation in terms of x, 2^3-3(2)-1>0, and since the term x^3 increases at a faster rate than 3x, any x>=2 cannot be a possible solution. Also both wolfram alpha and symbolab offer approximate solutions to the original cubic equation, and in all cases r
Not exactly. If you found all of 3 roots by this method it's correct. It would be wrong if you couldn't find an answer and said the problem had no solutions.
@@abysslight2490 Yes. It happened to be correct but mathematically not acceptable. Suppose we double the given lengths on the chord distances then r will be 8 and in that case the method fails. Hence for this particular case this might be right but mathematically wrong approach.
@@HemantPandey123 The method would give an cossine bigger than 1. It isn't incorrect because of that, it just says that there is no solution in that range.
i used another approach. i taked 3 angles(ai) from the centre of the circle and wrote the lenghts of the triangle(Li) as Li/2=R*sin(ai) and the angles as a1=arccos(1-2/R) a2=arccos(1-3/R) a3=arccos(1-1/R) then I changed the radius value to obtain a1+a2+a3=180° once i found R, a1, a2, a3 i've calculated the lenghts and used Heron's formula for the area. when i found a completely different answer from the possible choises i tought i was wrong XD
I tried calculating everything with the only variable r, the radius. The formula for the ABC area has square roots of (r-1/2), (r-1) and (r-3/2), but the offered solutions are all rational. No value of r can allow all 3 parenthesis to be squares. I guess there is a theorem saying that do not exist 3 squares in the form a^2, a^2-b, a^2+b... Sorry is a little messy but I think it works.
Hint for those scrolling down: - Forget about and just watch the video! - OK, if you insist it involves combining law of cosines with law of sines... and a depressed cubic! (not convinced yet?)
This is a hard puzzle which I still have not figured out. I eventually convinced myself that the measurements of 1, 2 and 3 as shown do not uniquely define the radius of the circle, nor the measurements of the triangle as I first thought. This means the area of the triangle isn't defined either. I think you need more information to solve it. Still trying to figure this one out.
When you got to the point where you had three possible choices for "r," why didn't you go back to the original drawing to see that "r" had to be greater than 3? PS given the solutions offered did you check the possibility of the value of "r" that didn't fit the drawing, but was an otherwise valid value of "r?"
This isn't an impossible problem... just a non-deterministic problem (i.e. infinitely many solutions). So long as the radius R of the circle is strictly greater than 3, there exists some (unique in this case) solution set (a,b,c) for each of those chord lengths, because each chord length is solely a function of R. And for every value R > 3 there is exactly ONE value for the area of the triangle. Presh derived a formulation of the area as a function of the Radius R in his video. I had a different one only because I didn't simply all the way. Didn't see the point. EDIT: Ahhh, apologies. I see others have mentioned this before in older comments.
I got the same cubic equation, but since the question was multiple choice I did not go through the splendid method you used. I said the value of r had to be >3 since r-3 was one of our lengths, and a little iteration gave me 3.88. It was horrible enough from there to get 17.19 as the area. It would have been quite enough as a question to just ask for the radius!
yea i solved it all the way until i got stuck at the cubic equation. i tried using the rational root theorem, but there was no rational roots :( . so i continued from there, and I was not so sad, because all those substititions were knowledge in which i never knew existed, i had never learned it.
Being an actual living fossil (i.e. a guy old enough to remember when computers were used to actually compute novel things…), I decided to do a simpler iterative solution. Vary 𝒓 from 2.5 to 10, using 𝒅𝒓 = 0.1 … and calculate the sum of the 6 interior angles created by the sub-triangles going between △ABO, △BCO and △CAO, where O is the center of the enclosing circle. I suppose, it is important to stipulate that since the chords AB, BC and CA are each exactly divided in 2, then the 1, 2, 3 length outer bits are normal to the circle, and thus MUST each hit the centerpoint O. No matter. The algorithm is simple enough… #!/usr/bin/perl -w use Math::Trig; use strict; my $r; my $dr = 0.2; my $ae; my $bf; my $cg; my $s; my $t; my $theta; my $phi; my $gamma; my $sum; my $x; my $bd; my $area; my $ab; my $bc; my $ca; for( $r = 2.5; $r < 15; $r += $dr ) { $s = $r - 1; $ae = sqrt( $r ** 2 - $s ** 2 ); $ab = 2 * $ae; $s = $r - 2; $bf = sqrt( $r ** 2 - $s ** 2 ); $bc = 2 * $bf; $s = $r - 3; $cg = sqrt( $r ** 2 - $s ** 2 ); $ca = 2 * $cg; $theta = asin( $ae / $r ); $phi = asin( $bf / $r ); $gamma = asin( $cg / $r ); $sum = 2 * $theta + 2 * $phi + 2 * $gamma; next if $sum > 2 * pi; $r -= $dr; $dr /= 2.718281828; last if $dr < 1e-11; printf "r %.4f 2r %.4f th %.4f ph %.4f ga %.4f sum %.4f ", $r, 2 * $r, rad2deg ( $theta ), rad2deg ( $phi ), rad2deg ( $gamma ), rad2deg ( $sum ); $x = ($bc ** 2 - $ab ** 2 + $ca ** 2 ) / ( 2 * $ca ); $bd = sqrt( $bc ** 2 - $x ** 2 ); $area = 1/2 * $bd * $ca; printf "ab %.4f bc %.4f ca %.4f ", $ab, $bc, $ca; printf "x %.4f bd %.4f area %.4f ", $x, $bd, $area; printf " "; } which produces: r 3.7000 2r 7.4000 th 41.9619 ph 60.8446 ga 76.6576 sum 358.9283 ab 5.2154 bc 6.8118 ca 7.5895 x 5.0596 bd 4.5607 area 17.3066 r 3.8472 2r 7.6943 th 41.8456 ph 60.6666 ga 76.4180 sum 357.8605 ab 5.2312 bc 6.8361 ca 7.6222 x 5.0814 bd 4.5728 area 17.4273 r 3.8742 2r 7.7484 th 41.9546 ph 60.8335 ga 76.6427 sum 358.8618 ab 5.2163 bc 6.8133 ca 7.5915 x 5.0610 bd 4.5615 area 17.3141 r 3.8742 2r 7.7484 th 42.0513 ph 60.9816 ga 76.8421 sum 359.7501 ab 5.2032 bc 6.7931 ca 7.5644 x 5.0429 bd 4.5514 area 17.2145 r 3.8779 2r 7.7558 th 42.0662 ph 61.0045 ga 76.8729 sum 359.8873 ab 5.2012 bc 6.7900 ca 7.5602 x 5.0402 bd 4.5499 area 17.1991 r 3.8792 2r 7.7585 th 42.0717 ph 61.0129 ga 76.8842 sum 359.9378 ab 5.2004 bc 6.7889 ca 7.5587 x 5.0391 bd 4.5493 area 17.1935 r 3.8792 2r 7.7585 th 42.0766 ph 61.0203 ga 76.8942 sum 359.9822 ab 5.1998 bc 6.7879 ca 7.5573 x 5.0382 bd 4.5488 area 17.1885 r 3.8792 2r 7.7585 th 42.0784 ph 61.0230 ga 76.8979 sum 359.9986 ab 5.1995 bc 6.7875 ca 7.5568 x 5.0379 bd 4.5486 area 17.1867 r 3.8794 2r 7.7587 th 42.0783 ph 61.0229 ga 76.8977 sum 359.9976 ab 5.1996 bc 6.7876 ca 7.5569 x 5.0379 bd 4.5487 area 17.1868 r 3.8794 2r 7.7587 th 42.0785 ph 61.0232 ga 76.8982 sum 359.9998 ab 5.1995 bc 6.7875 ca 7.5568 x 5.0379 bd 4.5486 area 17.1866 r 3.8794 2r 7.7588 th 42.0785 ph 61.0232 ga 76.8982 sum 359.9997 ab 5.1995 bc 6.7875 ca 7.5568 x 5.0379 bd 4.5486 area 17.1866 r 3.8794 2r 7.7588 th 42.0785 ph 61.0232 ga 76.8982 sum 359.9998 ab 5.1995 bc 6.7875 ca 7.5568 x 5.0379 bd 4.5486 area 17.1866 r 3.8794 2r 7.7588 th 42.0785 ph 61.0233 ga 76.8982 sum 359.9999 ab 5.1995 bc 6.7875 ca 7.5568 x 5.0379 bd 4.5486 area 17.1866 r 3.8794 2r 7.7588 th 42.0785 ph 61.0233 ga 76.8982 sum 360.0000 ab 5.1995 bc 6.7875 ca 7.5568 x 5.0379 bd 4.5486 area 17.1866 Thus the radius is 3.879385… and the various interior angles shown. I guess that's the solution! (area = 17.1866 ✓) ⋅-=≡ GoatGuy ✓ ≡=-⋅
Can someone help me..... When I have many ways to solve a question I often get confused so much that I waste much time in thinking which method shld I go with..... Do someone has a way to guess which method would be best(less messy/ time consuming) in such cases ?
Experience with your subject and lots of practice can help build speed and intuition. Another useful method is to learn how to make estimates that eliminate unreasonable possibilities.
The formulas for a, b, c are as shown by Presh. Are A=0.5[a(r-2)+b(r-3)+c(r-1)]. Use Trial and Error Method. Assume r=4; calculate a, b, c, A. Check r using the circumradius formula r=abc/4A. Repeat the calculation for r=3.9, r=3.8. Rapidly value of r approaches to 3.88. Not a very difficult problem after all!
@@rohangeorge712 we engineers are have high regards to mathematicians. Without you could not figure out solutions. Isn't solution the ultimate goal. Give us some credit. Please.
This solution turned me into a depressed cubic equation
you made my day bro xD
😂😂😂
A depressed cubic!? Is there anything we can do to cheer it up??
🤣
**sarcastic claps**
tickle?
@@zlodevil426
You meant sarcastic slaps ??? 😂
Nope. You will become depressed trying to cheer it up. Such is the misery :)
I like the way this guy speaks.
He speaks using logical methods!
That's true, really make the puzzle looks more interesting, in the way he speaks and solve it :)
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I understood nothing except he said me to mind my decision.
😂😂
🤣🤣🤣🤣
Haha
Why is this relatable?
Me too
I loved the last words you said sir.
We always need to mind our decisions.& skipping this question on a test could've been a wiser one.😅
I really love this problem. I solved it using only Pythagoras' theorem (multiple times) to find the equations for the coordinates (x, y) of the third vertex (using expressions of r). I got 3 equations. After eliminating x and y I got the same equation that was shown in the video for r and solved it by substituting (r-2) = z. The only valid solution for z is 2*cos(20°) which I think is quite a tricky calculation - but I really enjoyed solving it (30+ years after finishing school).
Can u plz provide ur solution
7:45
'...we solve the world's problems one video at a time.'
I liked that
Ah, I see you're a man of culture as well
"Chugging the world one bottle at a time"
I think you heard this
Maxamed Axmedn What??
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Syllabus: 2²=4
The test:
And the teacher won't explain the reason why 2²=4 you learn it like a fact...
Infinity 9151 It’s kinda hard to explain basic equations, you can use Peano arithmetic but that’s seen as overkill.
@@integralboi2900 well yeah, but I was talking about concepts in general...
Good
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It's possible to figure out the sides a, b and c even if you forgot about (or never knew) the chord theorem. For instance in the case for a) you get a right angled triangle with sides a/2, r - 2 and r. Thus we get
(a/2)^2 + (r - 2)^2 = r^2 => a ^2 = 16(r - 1).
I thought of that as well. I'm betting Presh used the chords theorem to avoid mentioning the Gougu/Pythagorean Theoem, ho ho...
Ooo, even 3 year olds can solve it. How old are you? Your observation was nice.
GoodBye Gougu. Hello Al Kashi and Al Tusi.
I am not surprised if he uses tri-sides instead triangle.
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So, Gougu is taking a break....
haha!
Al-Kashi didn't *sigh*
He avoided it using the intersecting chords. I got the values of sides using Pythagoras, I mean, Gougu. XD
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Impossible turns to possible haha LOVE trig functions for sure!
Love it dear
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Civil Engineering Q&A Thanks haha
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Turns back to impossible due to incorrect multiple choice answers.
I did it, my calculations were long but is easy to understand compared to this. I got the same equation r3-6r2+9r-3=0 but yeah I admit I used calculator to find r here. Even 3 year olds can solve it in seconds
It is possible to solve this without any special knowledge about chords. Start with image at 1:20 and mark point above number 2 D (intersection of chord and circle). Than there are two possible ways to express length |CD|. When you do this for all three sides, you get this equations: 𝛼+β+ɤ=180; 1=R(1-cos𝛼); 2=R(1-cosβ);3=R(1-cosɤ). After some algebra and elimination impossible answer you get R and rest is simple.
I got as far as the intersecting chords, but
3:30 "... after we do quite a bit of algebra ..."
4:25 "... I myself wouldn't be able to solve this cubic equation ..."
I don't feel _that_ bad about giving up.
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Haha I know that feeling. I once solved a "VERY HARD GEOMETRIC" problem. At least, that was how he described it in the title. I felt so proud haha
I'm so glad I watched the video instead of trying to solve the problem. I knew I had to use the relationship between the angles - and that looked wicked.
Good for u
Fundamentally this is a cubic eqn solving problem. Before coming to that point, my way is...
Let the intersection of the 3 lines be O, which is the center of the circle. From any vertex, say, B, draw diameter BOD, thus ΔABD and ΔCBD must be rt Δ. So，from midlines, know, AD=2(r-1), CD=2(r-2), from Gougu theorem, can solve AB and BC, and likewise, for AC. Note ABCD 4pt concyclic, so, as circumferential angles, ∠A=∠BDC, ∠C=∠ADB, equate their sines or cosines, got same cubic eqn... The rest is the same.
"Multiple choices format" is an *absolute evil* in math exams.
Wouldn't it be slightly faster to use Heron's formula at 2:54 instead of calculating cos(A)?
Thought so. Im trying it now
oh.. never try it. Total mess
You in 9th grade?
How about abc/4R = ∆
@@virajagr Yes but what is R?
I was thinking...
We know the chord is twice the angle. There is a relationship between the chord's arc length and its height above the line connecting the endpoints. I don't know what that is off hand, but it exists and will be a trig function of some kind.
We know the relative sizes of the heights, and we know that the sum of the chords must be 2pi. So take one of them, plus twice that, plus triple that, to get 6 of them equals 2pi. From there solve for the original relationship formula's values.
Yeah!! You are right.Its coming!!
I'm not going to read what you have written. With 25 likes, it sounds like you've solved it.
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The area is 17.18. The corresponding central angles to the cords with segment heights 1,2,3 are 84.5°, 122° and 153.5° respectively (rounded). They add up to 360. Cord lengths c1,c2,c3 are 5.19, 6.79, 7.59, respectively (rounded).
The radius of the outer circle is approx. 3.9.
Equations used:
h segment height (sagitta)
R radius
c cord length (c1,c2,c3)
ß central angle (ß1,ß2,ß3)
s semi-circumference inscribed triangle
A area irregular triangle
h=c/2 * tan(ß/4) ==> c=2h * cot(ß/4);
ß=arctan(2h/c) * 4;
R=h/2 + c^2/(8h);
A=Sqrt[s*(s-c1)*(s-c2)*(s-c3)] (Heron);
c2^2/c3^2 = 12/15, c1^2/c2^2 = 7/12, c1^2/c3^2 = 7/15 (home-cooked assumption) ;-)
I tried this problem again, but using a different approach.
As it turns out, you can use analytic geometry. First, place the circle and the triangle on a coordinate plane. You can then label the points of the triangle as: (x1, y1), (-x0, y0), and (x0, y0). Next, label the mid-points of each of the legs in terms of x0, x1, y0, and y1. Once this is done, you can set up 5 equations in 5 unknowns (r, x0, x1, y0, y1), using distance formulas and the coordinates of the points and the mid-points of the triangle.
If you solve for r, you end up with the cubic equation in terms of r - the same one shown in the video.
This approach is conceptually much easier than the approach shown in the video. It involves a quite a bit of tedious algebra, although not much worse than the one shown in the video.
hey presh. im a 7th grader in india and i found an approach without trigonometry but didnt solve it. heres my approach
lets first do some preliminaries
1. how to calculate the height of a triangle when you are given it has side lengths a, b and c and c is the base and the angles adjacent to c are less than 90°:
first draw an altidude with length h to the side c such that it divides c into lengths of x and c - x. you now have 2 right triangles - one with legs h and x and hypotenuse a; and the other with legs c - x and h and hypotenuse b. you can use the gougu theorem and get two equations in the variables x and h. solve for h in both equations then set them equal to each other. then solve for x and substitute into one of the two original equations and solve for h. you will get the formula for the height of the triangle
2. to prove that when you connect the midpoints of any triangle they will bound a similar triangle that is 2x smaller than the original triangle:
when you connect the midpoints of the sides consider one of the outer triangles. two of its sides are half the sides of the original triangle and it shares an angle with the original triangle therefore it is congruent to a triangle that is a 2x scaled down version of the original triangle. therefore the side of the triangle that connects two midpoints will be parallel to the other side of the original triangle. the same applies for the other sides of the middle triangle and the original triangle. since the sides of the middle triangle are parallel to the sides of the original triangle the middle triangle is similar to the original triangle and is a 2x scaled down version of the original triangle since it shares sides with the outer triangles
now lets solve the problem:
first extend the perpendicular bisector of the sides of triangle ABC. let the centre of the circle be M. the perpendicular bisectors of triangle ABC will meet a point M. let X, Y and Z be the midpoints of AB, BC and AC respectively. Mark points P, Q and R along the circumference of the circle such that respectively such that PX perpendicularly bisects AB and PX = 1, YQ perpendicularly bisects BC and YQ = 2 and ZR perpendicularly bisects AC and ZR = 3. if r is the radius of the circle then XM = r - 1, YM = r - 2 and ZM = r - 3. now draw AM. AM = r and XM = r - 1 therefore you can calculate XA using the gougu theorem
XA = √(AM² - XM²)
= √(r² - (r - 1)²)
= √(2r - 1)
AB = 2(XA) = 2√(2r - 1)
similarly you can calculate BC and AC in terms of r.
now consider triangle XYZ. since X, Y and Z are the midpoints of AB, BC and AC respectively that means triangle XYZ is a 2x scaled down version if triangle ABC. since XM = r - 1, YM = r - 2 and ZM = r - 3 and triangle XYZ is a 2x scaled down version of triangle ABC then there must be a point, say point K such that CK = 2(r - 1), AK = 2(r - 2) and BK = 2(r - 3). since triangle XYZ is similar to triangle ABC that means BK and ZM are parallel. since ZM is perpendicular to AC that means BK is perpendicular to AC when extended. extend BK until it intersects AC at point W. KW is perpendilcular to AC therefore BW is the height of triangle ABC and KW is the height of triangle AKC. since you know the sides of triangle ABC and triangle AKC in terms of r that means you can calculate BW and KW in terms of r.
BK + KW = BW
since you know BK, BW and KW in terms of r you can substitute and solve for r. you can now substitute the value of r in the formula for BW and AC. now the area of triangle ABC = ½(BW)(AC) and you can substitute the values of BW and AC in the formula for the area of triangle ABC and solve for the area of triangle ABC.
this is under the assumption that angle BCA and angle CAB are less than 90° and that point M is in the interior of triangle XYZ
..:..but man still can't work out the circumference of a ellipse.
TOM SCOTT TOLD ME
Matt Parker told me.
You can, but you need to add infinitely many terms. Using the perimeter formula from calculus and expanding the function using Taylor series, you can integrate term by term and get an infinite series.
@@integralboi2900 that is why man (or woman) cannot work it out. We can't do infinite series. Although, that is true with circles too 😣
Must be an elliptic function...
Only in the same sense that we can't work out the value of Pi. We can get as arbitrarily close as we want, so it's solved for all intents and purposes, there's just no analytical solution.
I hate not being able to solve a problem and when my answer was 17.2 a got frustrated and spent around 2 hours trying to figure out what’s wrong until I finally decided to watch your answer . Please don’t do me like that again .
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I thought he made it extremely obvious at the start that the correct answer was not one of the multiple choice options. Go back and listen again.
I tried solving from the thumbnail
@@thenegotiator5471 That's fine, but don't then complain about how you've been wronged.
For a circumcircle, radius= abc/6A. Area A can be found from herons formula.a, b,c is already found out from chords theorem in terms of r.
I did not use the Chord Theorem to get the lengths of the sides of ABC in terms of r.I used simple triangle manipulation.
I got a=2sqrt(4r-4),b=2sqrt(6r-9) and c=2sqrt(2r-1)
I plugged these values into Wolfram Alpha and got the area ABC=4sqrt(r(2r-3)).
This means r>1.5
I then equated area BOC+ area AOC +area BOA= area ABC
(r-2)sqrt(4r-4)+(r-3) sqrt(6r-9)+(r-1)sqrt(2r-1)= 4sqrt(r(2r-3))
(r-2)sqrt(4r-4)+(r-3) sqrt(6r-9)+(r-1)sqrt(2r-1)-4sqrt(r(2r-3))=0.................(1)
I plugged equation (1) into Wolfram Alpha for a numerical answer and got r=3.8794 approx.
another root was r=1.5 but this was less than r>1.5 therefore I plugged r>3 into
Area ABC=4sqrt(r(2r-3)
Area ABC was 4sqrt(3.8794(4.7588))=17.18664073
Area ABC=17.18664073 sq. units.This was a transcendental equation because it contained a circle and therefore pi.
Check answer;
For r=3.8794,
(r-2)sqrt(4r-4)=6.37823
(r-3)sqrt(6r-9)=3.32274
(r-1)sqrt(2r-1)=7.48578
Total =17.18675
The slight difference at the fourth decimal place is caused by rounding errors.If r was taken to ten decimal places the 'error' would be insignificant.
Try r=3.8793852415718,
(r-2)sqrt(4r-4)=6.378164190722
(r-3)sqrt(6r-9)=3.322669765738
(r-1)sqrt(2r-1)=7.485720779302
Area large tri=17.186554735762
4sqrt(r(2r-3) =17.186554735762
The area of triangle ABC=17.186554735762 square units to 13 decimal places.
Presh equation r^3-6r^2+9r-3=0
This equation had three real roots according to Wolfram Alpha:-
r=0.46791,r=1.6527 and r=3.8794.
The sum of the roots=6.
The sum of the roots of the cubic ax^3+bx^2+cx+d=0 is -b/a.
thanks for the information about wolfram alpha
Gosh, I never felt so excited for the end of a video
Love what you do Presh! Thanks for all the exam help! 😂
Bless u
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Al-Tulsi's law of sines is not well explained at 4:07. Go back to 1:03 and it can be figured out by pointing out that the arc BC subtends twice the angle A. Then draw a radius from B to the center. The vertex angle at the center equals angle A. Then apply the law of sines to that sub triangle.
It took me several hours to solve this interesting problem; thank you Presh. My answer was Area = 17.798. I didn't use trigonometry - just geometry and algebra.
How???
yea how??
and its not exact answer
In my earlier explanation, I have related the values of a, b and c in terms of radius of the circle, r. Such that a^2 = 16*(r -1), b^2 = 4 * (2 *r - 1) and c^2 = 12 * (2* r - 3). Also the area of the circumscribed triangle is known as, Area of the tringle = (a * b* c) / (4* r ). Using these two relations, we can also derive the area of the triangle directly in terms of r as follows: Area of the triangle = sqrt (48 * (4 * r - 12 +(11/r) - (3/r^2)). Therefore the radius of the circle, r, alone determines the area of the triangle inside it.
No maths teacher in my school can explain as good as you do even the toughest problems
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Thank you Presh sir for making these videos, you have helped me alot. I found your channel about 6 months ago and have watched almost all of your videos you have proved that Math is awesome keep making such videos thanks once again
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I used intersecting chords to get expressions for a,b,c, sine law to get radius. But when I got cubic equation, I thought I went in the wrong direction. Just in case I checked the roots in program, but they're not simple. With no desire to look for mistakes I gave up and watched the video. And what do I see?! Presh does the same and solves cubic equation, that leads to that weird expression for area!!! Possibly there was a typo in the task, in other case this problem is a beast!
There is a different (and i think easier) way to solve this problem:
Start with the formula
Alpha = 2 arccos (1-h/r) where h are the lengths 1, 2 and 3. This results in 3 equations with 4 variables.
Ad the equation that the sum of the three angles is equal to 360 and you have your 4th equation, so you can calculate the value of r.
Then just use the formula for a cut circle piece: A=r²/2*(alpha-sin(alpha)
Total up the aera of the segments and subtract it from the aera of the circle. Done
This also needs a lot of algebra but is much easier, as you dont have to "guess" the right way to solve the problem.
Another approach: Recall Heron's Theorem and the formula for the area of a triangle: 4R/abc. As we can express a,b,c with R we cn get a polynomial of a high degree, that we can solve for R and then substitute back into one of the formulas for the answer. Ideed, no trig is needed.
Just think how much he had thick to solve impossible problem really you are the most intelligent person I had on the TH-cam (legendary presh)
This was the most logically, conceptually& morally one of the best video in whole videos u made up till👌👌👌👌
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Great conclusion at the end of the video!!
You can't make a RIGHT choice if you have the WRONG options!! Exactly as in real life...
btw: you forgot to say this is a question for kindergarten kids age 3.5 in Japan...
Well I solved it!! It's comparatively easy to solve. Maths is Love!
How did you solve it?
How?
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This year i graduated from high school, now im studying medicine. I have no maths lesson but maths is my passion and i keep studying math on my own
Riola same here🥺 when i see an equations in my books I immediately turned on😅
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Ok, I admit, I didn't solve it. However I have a couple of comments:
1) At 1:08 if you draw lines from the center of the circle to the triangle's vertices you divide the main triangle into 6 right triangles which are equal 2 by 2.
Those 6 triangles have hypotenuse=r and the other side that connects to the circle's center is (r-1), (r-2) or (r-3). Using Goku's theorem 😁 you can find your a/2, b/2 and c/2 in terms of 'r'.
From here, to express the area of the big triangle in terms of 'r' it's trivial (= sum of the 6 right triangle's area)
2) Also from the picture at 1:08 it is obvious that r > 3 since 3 is the distance from the circle to AC.
Could the circle's center be outside of the triangle? Is that why you use the chord theorem?
Fantastic conclusion. Amazing problem.
Good life advice at the end. I thought to myself if the answers were that clean, there has to be some not so crazy way that involve pi or trigonometry to get it. So I tried to look for a simple solution with no luck. In life people tell you things are easy, but lead you on a wrong path.
5:57 Shouldn't r >3 be obvious? I haven't tried to solve this, but I'm not particularly happy with the method. I'll give it a go (if I have time) after the IGCSE finals. ~ Nov-Dec. If I remember.
I paused and searched for that comment
Interesting problem.
The method presented in the video gives a third degree polynomial equation and the exact value of the radius r :
r^3 - 6r^2 + 9r - 3 = 0 ==> r = 2(1 + cos(pi/9))
Different methods give more complex equations verified by the radius r, for example :
Area = (a/2)(r - 2) + (b/2)(r - 3) + (c/2)(r - 1) = (abc)/(4r)
2(r - 2)sqrt(r - 1) + (r - 3)sqrt(6r - 9) + (r - 1)sqrt(2r - 1) - 4sqrt((r - 1)(6r - 9)(2r - 1))/r = 0
The numerical solution obtained with a computer is obviously the same :
r = 3.879385
Area = 17.186555
dude you make such amazing content, you inspired me to create a youtube channel! keep up the AMAZING work :)
As it said "Fusion 360" (17,187)
Circumference of the triangle (19,544)
Circumference of the circle 24.375
Radius of the circle 3.879
The hardest from all of what you've done!!
Presh: Mind my decision
My mind: He is like Google maps. Just go with him.;)
5:30
I think we should discuss the case: 3t =-π/3 + 2nπ,
although we immediately know it was another exclusion
because x = 2cos(t) is negative in this case.
A neat solution! I could get as far as the cubic equation for R:
4*R^3-4*R^2*(a+b+c)+R*(a+b+c)^2-2*a*b*c=0
But once there, I had no idea what to do further.
I wasn't expecting the end result
This video does not only demonstrate a cool solution to an elegant trig question but also how westernized the education system is. Everyone is taught about the Pythagorean theorem and the laws of sines and cosines in geometry classes. But, I never knew about Al-Kashi or Al-Tusi until now. I am pretty sure the majority of the audience didn't know either. We could have called the Pythagorean theorem as the law of hypotenuse but instead, we are honoring the great mathematician behind it which is the way it should be. But when it comes to the scientists from the eastern civilization, we don't feel the need to honor their names even if their discoveries are extensively used. Am I the only one who thinks that this is big hypocrisy?
I think there should be two valid solutions, i. e. including the value 1.65 of r that was ruled out.
When drawing the bisectors, it wasn't mentioned that they should connect to the *closest* point on the boundary of the circle, the opposite side should also be valid, even if that wasn't visualized.
true dat true dat, but i think they wanted u go also think about the diagram given.
How can we substitute x = 2 cos(t) (at 4:56)? That means we already know x can take values only between [-2,2]. Didn't get that part
Same. Didn't understand the depressed cubic part and now I'm a depressed cubic😢
I find the solution of 17.1865..cm² via the general determination of tangents to a function. I start with the calculation of a tangent through the point (r | 0) to the radius of the inner circles of the triangle shortened by 1 or 2 cm and get the other two points of intersection with the circumference of the triangle as a function of r. The midpoint of the line between the two other points must lie on the innermost circle (with a 3 cm shortened radius).
The result is a complicated root equation, that only depends on r. The result of r was approximately 3,87939..Now it is possible to calculate the area.
I still have a doubt while solving the problem in this way. Given that the center of the circle lies inside the triangle and non of the side of the triangle passes through the radius, we can have innumerable solutions to the problem. Example: Given that a, b and c are the lengths of the sides of the triangle and r is the length of radius of the circle inside the triangle, a little effort will prove that each side of the triangle is a function of the radius of the circle, such that a^2 = 16*(r-1); b^2 = 4*(2*r-1) and c^2 = 12*(2*r-3). These relations can be established by joining points from the center of the circle to the points A, B and C, the apex points of the triangle and by using Pythagoras theorem. (The values of a, b and c can be interchanged depending on how you draw the triangle ABC). Now for any value of radius (at least greater than 3 as the center lies inside the circle and therefore will be more than 3), we will have a set of values for a, b and c using the above relations. And the area of the triangle will thus be calculated using the well-known relation, Area = sqrt((s)*(s-a)*(s-b)*(s-c)) where s is 0.5*(a + b + c). For example if we take r = 5, then we get a =8, b = 6 and c =9.165 using the above relations, so that area of the triangle becomes 23.664 sq. units. Again if change the value of r, say r = 8, we have new sets of values of a, b and c such that a =10.583, b = 7.746 and c =12.49, thus measuring the area of the triangle as 40.792 sq units. So the results (area of the triangle) all depends on the value of radius of the circle chosen. Interestingly, one of the answers to the problem as Presh has explained above can be 'E' which is 14.5 when we chose r = 3.42. In this case our a = 6.223, b = 4.833 and c = 6.788, the sides of the triangle.
Tried to us e Heron's formula after solving for the sides in terms of r. The algebra didn't play nice though. I was hoping the r would drop out. It seems there should be an easier way than what he presents though.
no need to do sin/cos law
2:24
Draw 3 line from the center O to the the angle of A, B, C, form ABO, ACO, BCO triangle, just simply calculate the area of three by base*height/2 and add them all.
I don't get it: what is the height and base of ABO?
@@JohnDlugosz
M is the mid - point of AB,
MO is the Perpendicular bisector of AB, also a part of radius of circle
height of ABO is MO, in this case MO=r-1
Base of ABO is AB, In this case AB=2sqrt(2r-1)
Area= (r-1)*sqrt(2r-1)
do the same thing obtain can obtain the area of ACO and BCO
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I notice that the areas of the segments sum to 2x (r squared) so the area of the triangle is (pi - 2) x r squared. I used Desmos to graph a sum of angle plus 2 arccos expressions to find the angle that satisfies summing to 2pi.
I never remember all those formulas, unfortunately, so I first sketched it and realized the radius had to be near 4 (and then a little under) At that, I measured the drawing and got a quick estimate of 16+ for the area and knew all the answers were wrong. Then because I'm me, I did a recursive look for the radius in a Libreoffice Calc spreadsheet and found R = ~ 3.87938, ran formulas for the angles to get angle BAC, trig for the HT, and voila area was 17.19 Very gratifying to see the answer after that.
Sir, Great solution. It is Really really very very tough for me.
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And the title of "seedhi baat,no bakwas" goes to Presh Talwalkar 😂😂😂
Wahi to bakwaas karne ka poora theka to Politicians ne pehle hi le rakha h poori duniya me...
Ara bhai Bhai Bhai 😂
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Acha bacha
I am getting a feeling that I have seen this question earlier also. 🤔🤔🤔
However, dont remember where ...
But great question nevertheless ,👍👍👍👌👌👌
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What song he plays at the end?
@@arnoahmed9269 sorry buddy, that I don't know
@@MaxMathGames Oh man, I am searching this song for ages. If you find please send me the name. Thank you.
@@arnoahmed9269 have you tried using apps like Shazam ?
In CAD drafting software it's a matter of trial and error to get the circle the right size to exactly touch the points of the triangle, when you don't even have the lengths of the sides of the triangle yet.
In 5 iterations, I got the last corner of the triangle too small by 1/256", and the area of the triangle came to 17.165 square inches.
That tiny amount of error in my drawing made a big difference in the area because all three sides are too close together by a hair each.
man started with 8th grade math and ended up with college math
Al Whoosey? Sorry Presh this is just confusing. The sine rule has been attributed to Al-Tusi among others but this isn't it. What you give is a formula for a chord, a=2rsinA where A is the angle subtended at the circumference. This result was well known to the Greeks, you don't need the sine rule to get it (as A is the same anywhere on the circumference and you can take the symmetrical case). FWIW I checked your solution value, it's fine, but I used a more direct method i.e. use 1=r(1-cos A/2), 2=r(1-cos B/2), 3=r(1-cos C/2); eliminate r to solve for B and C in terms of A; vary A till A+B+C=2pi; solve for r; area=0.5r^2(sinA+sinB+sinC)
Whenever I solve MCQ questions and my answer doesn't match with options I always think that all options are wrong🤣🤣
Hehe it happens. So me too look out for none of these option.
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Isn't the Q redundant? MCQ questions = Multiple Choice Question questions.
Sir you have chosen most sophisticated way to eliminate two options of radius but we can simply eliminate that two options because the one of the perpendicular bisector is 3 so obviously radius is more than 3
The error is in the question - it the perimeter you need to calculate - try out and see.
Really? How exactly did you come to know that, could you plz explain...
The perimeter is ≈19.54, which is still not an answer choice.
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I knew something was up when Presh said "when I was shown this problem, these were the answer choices presented to me". It just seemed an odd way of phrasing it. So when I got 17.2 I wasn't too surprised.
wait . ..you got the answer ? omg impressive
Thank you for giving credit to AlKashi and AlTusi. So many forget the contributions of Arab and Persian mathematicians. What would algebra, trigonometry, and geometry be with out them?
Came for the math, stayed for the life lesson
Hey I'm here to shorten the solution:
Circumradius = r = (abc)/4delta
Delta means the area of involved triangle = (abc)/4r and we have a,b and c in terms of r .. now we have to find out r as already shown in the video... And that's it
I was thinking exactly that!
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Your last statement was awesome Presh. 😁
You cant make the right choice if you have the wrong options.
But if you mind your decisions, you can solve anything.🙌🙌🙌
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bro has phd in marketing
This is one of the most crazy problems so far. Whenever Al-Kashi and his law of co-sins enters the scene, I’m usually out, because we didn‘t cover this in school. But this time, Al-Kashi also invited his friend Al-Tushi or what his name was, and all the crazy cubic substitutions showed me, where my limits are. Impressive! Incomprehensible, but impressive!
Those are standard substitutions actually...
What is the law around 1:40 about the bisector?!
I was watching your 5 year old video and just saw this...
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The solution to this viral problem results in 2 valid answers for the radius of the incircle circle. The areas of the triangle are: 11.826sq units and 13.9876sq units So, in approximation to the nearest whole number or to the nearest 0.5, the answers are 12sq units and 14sq units.
OMG, I actually got it right and I thought it was wrong cause my answer wasn't one of the options. I didn't get a nice algebraic expression like Presh; I had to resort to a numerical solution, but I did get the right answer. I did find the lengths of the three sides of the triangle in terms of r and got the same result as Presh. Then I looked at the angles formed at the center of the circle by the altitudes and segments drawn from the center of the circle to the vertices of the triangle. (Basically they form 6 right triangles in three pairs of congruent right triangles.) I knew these angles had to add up to 360 degrees, and that allowed me to solve numerically for r. I then solved for the area in two ways to check my result: first I summed the areas of the 6 right triangles using 1/2 * base * height and added them. Then I used Heron's formula using my expressions for the sides of the triangle. They results agreed (and furthermore, they didn't agree if I plugged in random values for r, suggesting my calculation for r was correct.) So, I did get the right answer, albeit it was a numerical approximation. I didn't get an exact expression, so Presh had the better answer.
may be solving the area as the addition of tree inner triangles of the blue triangle would be much easier.
x = 2cost implies we assumed that x
If you look at the cubic equation in terms of x, 2^3-3(2)-1>0, and since the term x^3 increases at a faster rate than 3x, any x>=2 cannot be a possible solution. Also both wolfram alpha and symbolab offer approximate solutions to the original cubic equation, and in all cases r
Not exactly. If you found all of 3 roots by this method it's correct. It would be wrong if you couldn't find an answer and said the problem had no solutions.
@@abysslight2490 Yes. It happened to be correct but mathematically not acceptable. Suppose we double the given lengths on the chord distances then r will be 8 and in that case the method fails. Hence for this particular case this might be right but mathematically wrong approach.
@@pedronunes3063 All three roots assume x
@@HemantPandey123 The method would give an cossine bigger than 1. It isn't incorrect because of that, it just says that there is no solution in that range.
i used another approach.
i taked 3 angles(ai) from the centre of the circle and wrote the lenghts of the triangle(Li) as
Li/2=R*sin(ai)
and the angles as
a1=arccos(1-2/R)
a2=arccos(1-3/R)
a3=arccos(1-1/R)
then I changed the radius value to obtain a1+a2+a3=180°
once i found R, a1, a2, a3 i've calculated the lenghts and used Heron's formula for the area.
when i found a completely different answer from the possible choises i tought i was wrong XD
Your approach IS right. I find the root of your eqaution a1+a2+a3=180 using Mathematica. r=3.87939, same as MYD.
yes, i found the same area of MYD of 17.19, but not seeing it in the possible choises confused me, i saw MYD answer only after i tried solving it
I tried calculating everything with the only variable r, the radius. The formula for the ABC area has square roots of (r-1/2), (r-1) and (r-3/2), but the offered solutions are all rational. No value of r can allow all 3 parenthesis to be squares. I guess there is a theorem saying that do not exist 3 squares in the form a^2, a^2-b, a^2+b... Sorry is a little messy but I think it works.
Hint for those scrolling down:
- Forget about and just watch the video!
- OK, if you insist it involves combining law of cosines with law of sines... and a depressed cubic!
(not convinced yet?)
This is a hard puzzle which I still have not figured out. I eventually convinced myself that the measurements of 1, 2 and 3 as shown do not uniquely define the radius of the circle, nor the measurements of the triangle as I first thought. This means the area of the triangle isn't defined either. I think you need more information to solve it.
Still trying to figure this one out.
There is a mistake on 3:30, the risult after a little algebra Is sqrt(4r/6r-3), I think, because I have tried a lot of time.
When you got to the point where you had three possible choices for "r," why didn't you go back to the original drawing to see that "r" had to be greater than 3?
PS given the solutions offered did you check the possibility of the value of "r" that didn't fit the drawing, but was an otherwise valid value of "r?"
This isn't an impossible problem... just a non-deterministic problem (i.e. infinitely many solutions). So long as the radius R of the circle is strictly greater than 3, there exists some (unique in this case) solution set (a,b,c) for each of those chord lengths, because each chord length is solely a function of R. And for every value R > 3 there is exactly ONE value for the area of the triangle. Presh derived a formulation of the area as a function of the Radius R in his video. I had a different one only because I didn't simply all the way. Didn't see the point.
EDIT: Ahhh, apologies. I see others have mentioned this before in older comments.
My brain began turning to putty around the 3 minutes mark. It was completely melted within 5 minutes.
07:29 "You can't make the right choice
if you have the wrong options"
I got the same cubic equation, but since the question was multiple choice I did not go through the splendid method you used. I said the value of r had to be >3 since r-3 was one of our lengths, and a little iteration gave me 3.88. It was horrible enough from there to get 17.19 as the area. It would have been quite enough as a question to just ask for the radius!
yea i solved it all the way until i got stuck at the cubic equation. i tried using the rational root theorem, but there was no rational roots :( . so i continued from there, and I was not so sad, because all those substititions were knowledge in which i never knew existed, i had never learned it.
Being an actual living fossil (i.e. a guy old enough to remember when computers were used to actually compute novel things…), I decided to do a simpler iterative solution.
Vary 𝒓 from 2.5 to 10, using 𝒅𝒓 = 0.1 … and calculate the sum of the 6 interior angles created by the sub-triangles going between △ABO, △BCO and △CAO, where O is the center of the enclosing circle.
I suppose, it is important to stipulate that since the chords AB, BC and CA are each exactly divided in 2, then the 1, 2, 3 length outer bits are normal to the circle, and thus MUST each hit the centerpoint O. No matter.
The algorithm is simple enough…
#!/usr/bin/perl -w
use Math::Trig;
use strict;
my $r;
my $dr = 0.2;
my $ae;
my $bf;
my $cg;
my $s;
my $t;
my $theta;
my $phi;
my $gamma;
my $sum;
my $x;
my $bd;
my $area;
my $ab;
my $bc;
my $ca;
for( $r = 2.5; $r < 15; $r += $dr )
{
$s = $r - 1; $ae = sqrt( $r ** 2 - $s ** 2 ); $ab = 2 * $ae;
$s = $r - 2; $bf = sqrt( $r ** 2 - $s ** 2 ); $bc = 2 * $bf;
$s = $r - 3; $cg = sqrt( $r ** 2 - $s ** 2 ); $ca = 2 * $cg;
$theta = asin( $ae / $r );
$phi = asin( $bf / $r );
$gamma = asin( $cg / $r );
$sum = 2 * $theta + 2 * $phi + 2 * $gamma;
next if $sum > 2 * pi;
$r -= $dr;
$dr /= 2.718281828;
last if $dr < 1e-11;
printf "r %.4f 2r %.4f th %.4f ph %.4f ga %.4f sum %.4f ", $r, 2 * $r, rad2deg ( $theta ), rad2deg ( $phi ), rad2deg ( $gamma ), rad2deg ( $sum );
$x = ($bc ** 2 - $ab ** 2 + $ca ** 2 ) / ( 2 * $ca );
$bd = sqrt( $bc ** 2 - $x ** 2 );
$area = 1/2 * $bd * $ca;
printf "ab %.4f bc %.4f ca %.4f ", $ab, $bc, $ca;
printf "x %.4f bd %.4f area %.4f ", $x, $bd, $area;
printf "
";
}
which produces:
r 3.7000 2r 7.4000 th 41.9619 ph 60.8446 ga 76.6576 sum 358.9283
ab 5.2154 bc 6.8118 ca 7.5895 x 5.0596 bd 4.5607 area 17.3066
r 3.8472 2r 7.6943 th 41.8456 ph 60.6666 ga 76.4180 sum 357.8605
ab 5.2312 bc 6.8361 ca 7.6222 x 5.0814 bd 4.5728 area 17.4273
r 3.8742 2r 7.7484 th 41.9546 ph 60.8335 ga 76.6427 sum 358.8618
ab 5.2163 bc 6.8133 ca 7.5915 x 5.0610 bd 4.5615 area 17.3141
r 3.8742 2r 7.7484 th 42.0513 ph 60.9816 ga 76.8421 sum 359.7501
ab 5.2032 bc 6.7931 ca 7.5644 x 5.0429 bd 4.5514 area 17.2145
r 3.8779 2r 7.7558 th 42.0662 ph 61.0045 ga 76.8729 sum 359.8873
ab 5.2012 bc 6.7900 ca 7.5602 x 5.0402 bd 4.5499 area 17.1991
r 3.8792 2r 7.7585 th 42.0717 ph 61.0129 ga 76.8842 sum 359.9378
ab 5.2004 bc 6.7889 ca 7.5587 x 5.0391 bd 4.5493 area 17.1935
r 3.8792 2r 7.7585 th 42.0766 ph 61.0203 ga 76.8942 sum 359.9822
ab 5.1998 bc 6.7879 ca 7.5573 x 5.0382 bd 4.5488 area 17.1885
r 3.8792 2r 7.7585 th 42.0784 ph 61.0230 ga 76.8979 sum 359.9986
ab 5.1995 bc 6.7875 ca 7.5568 x 5.0379 bd 4.5486 area 17.1867
r 3.8794 2r 7.7587 th 42.0783 ph 61.0229 ga 76.8977 sum 359.9976
ab 5.1996 bc 6.7876 ca 7.5569 x 5.0379 bd 4.5487 area 17.1868
r 3.8794 2r 7.7587 th 42.0785 ph 61.0232 ga 76.8982 sum 359.9998
ab 5.1995 bc 6.7875 ca 7.5568 x 5.0379 bd 4.5486 area 17.1866
r 3.8794 2r 7.7588 th 42.0785 ph 61.0232 ga 76.8982 sum 359.9997
ab 5.1995 bc 6.7875 ca 7.5568 x 5.0379 bd 4.5486 area 17.1866
r 3.8794 2r 7.7588 th 42.0785 ph 61.0232 ga 76.8982 sum 359.9998
ab 5.1995 bc 6.7875 ca 7.5568 x 5.0379 bd 4.5486 area 17.1866
r 3.8794 2r 7.7588 th 42.0785 ph 61.0233 ga 76.8982 sum 359.9999
ab 5.1995 bc 6.7875 ca 7.5568 x 5.0379 bd 4.5486 area 17.1866
r 3.8794 2r 7.7588 th 42.0785 ph 61.0233 ga 76.8982 sum 360.0000
ab 5.1995 bc 6.7875 ca 7.5568 x 5.0379 bd 4.5486 area 17.1866
Thus the radius is 3.879385… and the various interior angles shown.
I guess that's the solution! (area = 17.1866 ✓)
⋅-=≡ GoatGuy ✓ ≡=-⋅
Can someone help me.....
When I have many ways to solve a question I often get confused so much that I waste much time in thinking which method shld I go with.....
Do someone has a way to guess which method would be best(less messy/ time consuming) in such cases ?
Experience with your subject and lots of practice can help build speed and intuition. Another useful method is to learn how to make estimates that eliminate unreasonable possibilities.
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The formulas for a, b, c are as shown by Presh. Are A=0.5[a(r-2)+b(r-3)+c(r-1)].
Use Trial and Error Method.
Assume r=4; calculate a, b, c, A. Check r using the circumradius formula r=abc/4A.
Repeat the calculation for r=3.9, r=3.8.
Rapidly value of r approaches to 3.88.
Not a very difficult problem after all!
we want exact answer
we are mathmaticians not engineers my dude.
@@rohangeorge712 we engineers are have high regards to mathematicians. Without you could not figure out solutions. Isn't solution the ultimate goal. Give us some credit. Please.
6:09 is wrong; (2r - 3) has to be positive to get a *real* area, otherwise it’ll be a square root of a negative value and will end up being complex
He said that (2r - 3) had to be positive to get a positive area. It’s a minor mistake; he made the right selection but justified it wrong
The substitution of x=2cost limits the value of x to
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