I was intrigued at this answer, so I tried to work this method out myself! Perhaps there is a better way, but here is what I thought. At 3:26 we have a right triangle with legs of 1 and 2 + sqrt(3). The angle at point A thus has tan(A) = 2 + sqrt(3). I didn't recognize it at first, but this is actually a special value for A = 75°. Then the other angle at D is 15°, so we want to find 1/sin(15°) = 4/(sqrt(6) - sqrt(2)) = sqrt(2) + sqrt(6). Pretty neat way to solve using trigonometry! Exact value of tangent of 75 degrees (uses sum of angles and values of 30° and 45°) www.mathway.com/popular-problems/Trigonometry/308167 Exact value of sine of 15 degrees (uses sum of angles and values of 30° and 45°) www.mathway.com/popular-problems/Trigonometry/301128
@@MindYourDecisions impressive! :) thank you very much! Do you consider a follow up video? Seems to be a pretty interesting thing that those angles turn out to be so nice.
@@manamimnm I mean, I know of problems for the same purpose that are much harder than this. Not saying it wouldn’t be very challenging for someone that age, but it does seem too easy for an Olympiad Qualifier.
I got as far as √(8+4√3) in my head but had no clue how to resolve that expression further. The progression of second half of the video was really pleasing to watch. Thanks.
Olympiad questions are special. They never require some advanced mathematics, but they are still able to let you think in a creative way, because every problem is different.
When solving the two equations at the end, it is easier to take advantage of the fact that x and y are positive integers. The equations are xy = 12 and x + y = 8 There are only 3 possibilities for the first equation, 1-12, 2-6, 3-4. The only pair that sums to 8 is 2-6. Hence, x =2 and y = 6.
When I was in middle school, we students weren’t allowed to draw anymore lines into the drawings because it would make it easier to solve it and it wouldn’t require specific algorithms anymore but now that I know this is how math problems should be solved, I’m now going practice doing this. This channel has opened my mind in math and showed me how I should approach math. +1 subscriber
Consider the intersection pt. as 'O'. The arc BO subtends angle 30° (sin(1/2)) at the centre of the left circle. Therefore, angle BAO=15° Now, AD.sin (15°)= 1 AD= √6 + √2
Few suggestions for simplification : 1. By symmetry it is quite obvious that both the line joining both centers, and AD are going through the tangent point which is center of symmetry ; from there we have 2 right triangles to calculate. 2. Once you reach xy=12, there are not so many possibilities with x,y integers - checking for x=1,2,3 is sufficient to find the single solution that also satisfies x+y=8 where x
5:45 it’s a small refinement and saving, but we’re already done right there. The system of equations is symmetrical to swapping x and y, so the quadratic will necessarily have as solution the values of x and y.
In the process of factoring x^2 - 8x + 12, you had to answer the question "what are two numbers such that their sum is 8 and their product is 12", which is just the question you started with in "solve 8=x+y and 12=xy". Rewriting it as a quadratic does allow you to solve using the quadratic formula, in case you can't find it in your head.
Exactly what I'm thinking. I feel it is similar to when we are kids we have 18-9 where you need to regroup since 8 is bigger than 9. And so we ended up with the same exact problem. Then when in high school we have that. :)
Yeah - you're correct - it is an unnecessary step - that being said it might be useful to know how to solve this type of questions (this channel wants to educate people) and we don't know if for example you were not given points for some proper step by step explanation - depends on a test I suppose. This was quite easy question for this channel (so something I could have resolved on my own - without much trouble) and people going for Olympiad often trip on improper explanation of the answer. Depending on what level you are obviously.
I solved it as follows: I drew a line from D to the intersection point (called M), then the perpendicular from M to the lower line and called their intersection point K. Then I considered the two right triangles CMK and DMK, and exploited two facts: 1. CK + DK = CD = 2; 2. DK : KM = KM : CK (by the proportion of right triangles with one side in common). From 1) and 2) one can find CK, DK and, by Pythagora's theorem, DM. The result we're looking for is 2DM - the double root trick will give the desired form. The solution as Presh presented is more elegant, but this is another way nonetheless!
Having established the length of the first triangle’s long leg (square root of 3 or 1,7320) we have for the second triangle: Short leg = 1 Long leg = 1 + 1,7320 + 1 = 3,7320 Therefore the distance A to D (the hypotenuse of this right triangle) = 3,86 Simpler
I agree. The distance between the semi-circles’ centres is 2x(sqrt(1-0.5^2))=sqrt(3). Hence horizontal distance AD is 2+sqrt(3) and vertical distance is 1. Much easier than the supplied method (for a change).
Obviously, the 2 semicircles are the same so applying the Pythagorean theorem, horizontal distance pt. 2 to 2 is square root of 3, now assuming that the radius equals to 1, therefore the horizontal distance A-D equals square root of 3 + 2 = 3.732
Like most others, getting to the √(8+4√3) solution was fairly simple. I agree that putting it in the required form for the answer was a neat trick. What I'd like to understand is whether there is a good reason for employing such a trick in other circumstances. I suspect the answer is yes, but can't think of any myself.
That´s analytical solutions versus numerical solutions. It looks a bit of a contradition but you can use analytical geometry to compute the numerical solution for most of these kind of problems.
At 3:57 , we can also directly get √2 + √6 as the ans , we can eliminate the bigger square root if we get a square term in the inside , thus we can try to look for (a+b)^2 identity directly in 8 + 4√3 , we notice a^2 + b^2 = 8 , 2ab = 4√3 , ab = 2√3 = √ 12 . and 12 = 3x4 = 2x6 , but we know 3+4 = 7 , which isn't equal to 8 , so we know a= 2 and b=6 , therefore we get 8 + 4√3 = ( √2 + √6 ) ^2 . Thus we get our ans . This method is just to simplify things and get a different approach too . Thanks Thanks...
Presh is making some progress. He's still a pompous ass that fails miserably in his goal of making math approachable and understandable with his resistance to referring to Pythagoras' Theorum the way everyone knows it it as, but I guess he's OK being a modern mathematical laughing stock. Let's face it Presh, half of your views are by folks wanting to see how you contort yourself to avoid saying Pythagoras. People go to 3B1B, Mathologer, and the like to learn something, we watch you for a chuckle.
at 5:14 you dont need to put it in quadratic form, because the problem of finding two numbers with known sum and product is what is always required when solving a rational quadratic. just look at x+y=8, xy=12. we know the factors of 12 so the pair 6,2 comes out immediately.
@@Fidder492 was that a comment to me David Seed. I didn’t say “obviously”. I said that the problem could be solved without forming a quadratic. note that any quadratic can be represented in this way. if the roots if a quadratic in z are x and y then that quadratic may be written as (z-x)(z-y) = z^2 -(x+y)z +xy. . so if you can solve x+y=8, x+y=12 then there is no need to write out the quadratic.
Nice one. But I always find the last step for this a little funny. Find two numbers that adds to 8 and are factors of 12. And we use substitution to make a quadratic equation then use factoring (which is basically finding two numbers that add to -8 and are factors of 12) :)
Hor length of small triangle = 1.732 (sq rt 3) Hor length of large triangle = 3.732 Thus AD = sq rt ((3.732 x 3.732) + (1 x 1)) Thus AD = sq rt 14.92782 = 3.86
It seems easier to just work out the values when you get to the smaller triangle at 3.05 and then use actual values to get the longer triangle. I was fine with the diagram bit but lost on the x and y bit!
I did using trig. 4cos(a) is the total length AD, where 'a' is the angle between AD and one of the base line of diameter. Vertical length between two parallel lines turn out to be 2sin(2a), which is 1. Thus angle a is pi/12. Next step is to express 4cos(pi/12) in the required form, which simplifies to sqrt(2) +sqrt(6).
5:10 "So we have a system of two equations: 8 = x + y; 12 = xy". Quadratic equation?! Come on, just look at it! The variables x and y have to be integers. 12 = 1*12 = 2*6 = 3*4. Which pair add up to 8?
@@joginderpaldua351 Symmetry? Since both semicircles are the same size, the tangent point has to be half the vertical distance between the centers, and halfway between them horizontally, which means it's halfway between A and D.
@@bandar1606 No? AB is a diameter length of the semicircle, with length 2. The distance from A to the tangent point is necessarily less than 2, because it's not a diameter. The distance from the tangent point to D is the same distance, so the length of AD is less than 4. See diagram at 3:21
This was so unbelievably more difficult than getting the literal answer, needing it in a certain form. sin-1(.5) = 30 degrees (because they must intersect halfway the vertical height) cos(30) = x/.5 ==> x = .86 (this is the horizontal position of intersect from one circle center) add 1 for the other radius half we skipped, 1.86 is the horizontal distance from one circle outer edge to intersect double for 3.73, total horizontal distance (not surprisingly slightly under 4) 3.73^2 + 1^2 = h^2 h = 3.86
i had the distance figured out correctly, but that neat bit of shuffling the numbers around to make a nice simply sum of tow square roots is far to advanced for me...
@@bjorntorlarsson The problem has rotational symmetry about the point where the two circles meet. Let’s say the radius is 1, this won’t change any angle. The meeting point will have a height of 1/2. This corresponds with a 30 degree angle from the horizontal (meeting point -> center of circle -> B or C). Cut that in half for the angle with the same endpoints but with the middle one on point A or D, because we’re moving from the center of the circle to the circle itself. 30/2=15
@@jakobr_ I still don't see why the angle must necessarily halve from 30 degrees when rotating the diagonal from the midpoints of the diameters, to their far endpoints. I don't see that the length of the secant from A to the tangent point has any very obvious relationship to the length of the radius. So why does its angle from the horizontal happens to be half? But thank you anyway!
@@jakobr_ Oh, I see now! The base is the tangent point to B, and the angle point is the diameter's mid point, and the same circle's periferi point respectively. The second one with an acute angle that is half of the first. I just didn't come to think of those two triangles having the same chord as base. This is useful thinking!. I will remember this now. Thanks! Any three points in the plane have all kinds of relationships to each other, and sometimes to a fourth point, and more.
Guys actually there is a quick way for ending.Squareroot 8 plus squareroot 3 times 4 we have.Write the 4 times squareroot part as 2 times square root 12 .And look, 12 can be written as 2.6=12 in situations like this,if the total of the multipliers eaquals to the number next to it .For this question 2+6=8 so it fits our rule so you can write it as squareroot 6 + squareroot 2 it is a fast way to solve it.If you guys couldnt get what I am saying you can ask your questions because I couldnt explain it well I guess.Sorry for any misunderstoodments.Just ask me if you have any questions.Have a nice day.
The whole bit from 5:12 to 6:03 could be replaced with this: "Two positive integers with a sum of 8 and a product of 12 can only be 2 and 6." Why make it any more complicated than that. And why does Presh refuse to utter the word "Pythagorean"?
lovely to see there are so many different approaches! I used the cosine rule to find the distance between D and the point where the semicircles meet. (by subtracting (from pi) the bottom right angle of the central triangle which you also constructed, which ends up nicely being pi/6)
Can't you just connect their centers, this line will surely pass through their tangent point hence it's length will be 2 and the height being 1, using Pythagoras to get horizontal distance as √3. Adding two radius front and back that is 2, we get the length of AD as 2+√3
Still even if we need to find the linear line connecting A and D, we know that it's nothing but a hypothenuse to 1 being height and √3+1 being the base. The need for angles is not relevant that is what I meant
Interesting to see your solution! I started out drawing a similar central triangle to you, but then I deviated in my method. Figured I'd share my solution too as I rather enjoyed it! Instead of working out the missing length of the 2-1-√3 triangle, I worked out that the angle opposite the side of length 1 is 30° (arcsin(1/2)). Based on this, I was able to draw a new triangle between A, the center of circle AB, and the tangent point at which the two circles meet. This is an isosceles triangle with angles 150° (180 - 30), 15° and 15°. The two equal sides have length 1 (as they are both radii of circle AB) and the side opposite the 150° angle is half the length of AD (since you could draw an identical triangle in circle CD and the two line segments passing through the tangent point of the two circles would form the length AD). Therefore, by the sin rule, ½AD / sin 150° = 1 / sin 15°. We know that sin 150° = sin 30° = ½. So ½AD / ½ = 1 / sin 15° => AD = 1 / sin 15°. From here, we just need to work out the value of sin 15°, which is sin(45 - 30) = sin 45 cos 30 - cos 45 sin 30 = (1/√2)(√3/2) - (1/√2)(1/2) = (1/(2√2))(√3 - 1). Multiplying both the numerator and the denominator by √2 gives ¼(√6 - √2). Therefore AD, which = 1 / sin 15° = 4/(√6 - √2). Multiplying both the numerator and the denominator by (√6 + √2) gives 4(√6 + √2)/(6 - 2) = √2 + √6 (i.e. x = 2, y = 6).
This is pretty much how I got it, and I thought it seemed more straightforward than that squaring both sides and expanding stuff. But maybe it would have been harder if the 4 didn't cancel itself out at the end...
I'm 38 and I've never been good at math, and thanks to your channel I've learned that you can put triangles and other shapes into shapes to solve your problems :P
@@lime-limelight this could be part of an hexagonal, circle sorting, in wich we can daw a hypotenuse between the two center of the circles, with distance 2, making pithagoras I got a different result of aproximately 3.73
An interesting insight into this problem is to realise that, since AB and CD are diameters, then (just focusing on the semicircle AB), the triangle formed by A, B and the tangent point between the two circle is a right triangle. The height of this triangle is 0.5 (halfway between the parallel horizontal lines), and its base is 2. Thus, its area is 0.5. You can ‘stack’ the identical triangle formed by C, D and the same tangent point, on top of the first triangle to get a rectangle. You will observe that the area of this rectangle is, therefore, equal to 1. If you let the longer side of the rectangle = x, the shorter side = 1/x, and we write x^2 + (1/x)^2 = 4 [the square of the diameter]. Solving from there is straight forward, but the manipulation of the surd (radical) expression in the second half of the video was a delight to watch! Thanks Presh.
@@apostolosspanelis5686. Certainly, AB is the hypotenuse of a right triangle, but the area of any triangle is a base (choose any side of a triangle - in this case, I’m using the hypotenuse of the right triangle) x height of the triangle (measured perpendicular to the chosen base) x 1/2. Given that the base that I have chosen runs along the top horizontal line, its perpendicular height measured vertically down is 0.5 (the tangent point is halfway between the two parallel horizontal lines = 0.5. Therefore the area of the right triangle = 1/2( 2 (base) x 0.5 (height) ) = 0.5. Do you follow my reasoning?
Damn, it was pretty easy to get the 8+4sqrt(3) but after that I had no idea how to make it into sqrt(x) + sqrt(y). Some nice math shenanigans right there.
Before you constructure a line through the two centers of the two circle, you need the "symmetry" argument. Without the symmetry argument you can't be sure whether you can draw a line through the two centers AND the tangent point of the two circles.
Easy solution that came to my mind: Take centre of left circle as origin Take centre of right circle as (x, -1) Take the point D as (x + 1, -1) Take the point A as (-1, 0) Distance of thier centres is 2 By distance formula x = √3 Then by distance formula, AD = 2√3 + 4 AD in terms of √x + √y is, √12 + √16
I solved this on my own, i tried my best :) The distance between both center of the circles is 2, since it's tangent to each other. Draw one line from the radius of the top circle, and i got a right triangle. I used phytagorean theorem (or whatever you called it 😅), i got √3 for the other unknown side of the triangle. I added up 2 radii of the circles and i got 2+√3. And yes, it makes another right triangle, so AD=√[1²+(2+√3)²] and i got AD=√[8+2√12]. And then i use the formula my teacher ever gave, which is > , so since 2+6 is 8, and 2*6 is 12, the final result i got is √2 + √6 I hope that it was right 😅
There is another less complex way to solve the last part. When grouping alike factors xy = 12, with x, y element of N leads to x, y = 1, 12 or 2, 6 or 3, 4. Then from x+y=8 we find it must be x=2 and y=6. The problem and its solution are well presented in this video.
I got to 2(sqrt(2) + sqrt(3)) = 3.863 and then promptly guessed integers (starting at 1 and 2) until I found a match with sqrt(2) + sqrt(6). Not as pretty but it sure was faster than doing the second half the full way 😅
Finding the length of AD is easy: it's +√13. Putting the answer in your required format is harder. Here is how I did it. Once I have the format shown at 6:52 in the video, I can also write two dependent equations: x + y = 8, and (4)(√ 3) = (2)(√x)(√y). Since you have stipulated that x and y cannot be equal but must be positive integers, that leaves only three pairs of addends that satisfy your requirements and the equation x + y = 8, namely: 1 + 7 , 2 + 6, or 3 + 5. A quick test gives the correct answer as 2 and 6.
We have significant fact that circles are equal, meaning their touch point stays exactly at half of radius = 0.5; Knowing that and radius we can know horizontal distance from center of circle till touch point. Two of such distance + diameter = leg of triangle where we should find hypotenuse. Easy as 1-2-3!
If you define point A at coordinates (0,0), then the coordinates of center of the first semi-circle are (1,-1) and the point tangent to the line above the center are (1,0). A line from this point to the center of the second semi-circle is 2 units long and distance between the two centers is 3^0.5. So the coordinates of the second semi-circle center are (1+3^0.5,-1) and the coordinates of point D are (2+3^0.5,-1). Therefore the line AD is ((2+3^0.5)^2 + (1)^2)^0.5 = (8+4*(3)^0.5)^0.5 = 3.8637...
You actually had the answer at 3:04 which is how I solved it. your AD y is the radius 1 and AD in x is 1 (radius) + sqrt of 3 + 1 (radius) again. pythagorean theorem to get 3.86. No need to go around your elbow to get to your thumb.
Nice problem. On the "Pythagoras or not" question - the attribution is to the *Pythagoreans*, if anyone. Standard references on the preSocratic philosophers go through the very tricky question of attributions to the master, if indeed there was one.
You could use the Cosine rule to find the length from D to the tangent point: C²= A² + B² - 2ABCosC°, where A & B are the radius 1 and C° is 150°. This gives C² =(2 + √3). Times by 2 to get the entire length 2√(2 + √3), or √(8 + √12), which equals 3.86.
Made a meal of that one! Mid point between A & B to mid point between C & D has a length of 2. Therefore base of triangle is 1.732. Larger right angle triangle has a base of 3.732 & a height of 1. Therefore its hypotenuse is 3.8654. Done!
A much easier solution can be found as follows. Drop a perpendicular from A and connect the centres through the point of contact of the two circles. Also drop a perpendicular from the centre of the circle. Using Pythagoras, the horizontal line connecting the centre of the first circle to the point of contact to the other circle is sqrt 3. The horizontal distance between the extreme edes of the circles is 1 + sqrt 3 +1 or 2 + sqrt 3. Then using Pythagoras AD is the square root of ( 1 + ( 2 + sqrt 3 ) squared) . Or the sqrt of (8 + 4 times sqrt 3).
I’m happy I’ve made a different solution from you, for the first time. Most of the time I can’t even solve it. I made a right triangle inside a semicircle, with its right angle at the tangent point, and hypotenuse at the semicircle diameter. This triangle has an angle of 15 degrees, and I used its cos, (sqrt(6)+sqrt(2))/4. Great video. As always, mind your decisions!
Same as mine. But in this case, firstly you must prove that the 3 points [A - D - the tangent point] are collinear. It's not too hard but I think that is the most tricky part, many people forget that.
distance between the centre of the 2 semicircle will be 1+1 = 2 so the horizontal distance between the 2 centres will be sqrt(2^2 - 1^2) then find horizontal distance of AD is above answer + 1 + 1 because of 2 radiuses then final answer is just sqrt(1^2 + horizontal distance) = 3.86
The circles are tangent at the point where sin = 0.5 as radius = 1. Take radians for sin(X)=0.5 and apply to cosine. Cos(30) = 0.86 roughly. Take length of circles 4 - (1- cos(30)) = 3.72 √1+(3.72)^2 = 3.86
0:31 I don't have read all those things according to my age but here's how I find the length between a and d = 1) breadth is 4 unit (2+2) 2) length is 1 unit (2/2) 3) It forms a rectriangle, now we can put Pythagoras theorem. Ans is 4.12 unit 😅.
Calculating the distance AD is pretty straightforward. Connect the centers of the two semicircles. By symmetry it will pass through the tangent point and create right triangle with hypotenuse 2 and height 1. This is special 30-60-90 triangle so the horizontal displacement between the two centers is sqrt(3). Therefore the horizontal displacement between A and D is sqrt(3)+2. You can construct a right triangle with height 1 and base (sqrt(3)+2). AD^2 = 1^2+ ((sqrt(3)+2)^2 = 1+3+4sqrt(3)+4 = 4sqrt(3) + 8. Now we would take the square root to get the length of AD. However, we are asked to put it in the form of sqrt(x)+sqrt(y). So AD^2 = x+2sqrt(xy)+y. Therefore x+y = 8 and 2sqrt(xy) = 4sqrt(3) = 2sqrt(12). Therefore xy=12. We could do the algebra but since we're given that x and y are integers we can quickly get x = 2 and y=6 by trial and error. Therefore the answer is sqrt(2)+sqrt(6).
After the basic premise, I guesstimated 3.80. The explanation was more than I could handle but when it came up 3.86, I knew my old guesstimator was still in operable condition. And as we used to say "Good enough for government work".
Thanks for the nice solution. You could also use the power of the point approach. First calculate the power for the point of the left circle center, thus (1+y)(1+y)=1·3, where y is the distance between point B and right circle tangent (T), then using this value (y=√3-1) you can solve the power for the point A, (AD·AD/2 = T·T) i.e. ½x·x = (2+y)(2+y) getting the value for x as √2+√6.
The point where the 2 circles touch is exactly halfway the two lines. Make a triangle there from the center of the circle, the point in the middle of the 2 and to a line and solve for the length of the line. (sqrt(1^2-0.5^2)) Then make a triangle from point A to the middle and to the same point straight above from the middle and solve the hypotenuse. (sqrt(1.866^2+0.5^2)). Then just times that by 2 and you get 3.86 and change.
Thats a good question though. If you are talking about the total white area between the semi circles then it'll be (√3-π/2) and if we consider any one part among those two white areas between the semicircles then it will be the half of the earlier answer i.e. (√3-π/2) /2 ... [ as both the empty areas are identical]
The rectangle formed by the horizontal lines and the verticals from each of the centre points has an area of 2. Within the rectangle you have 2 quarter circles = 2(pi/4) = pi/2. So the area between the semicircles = (2 - pi/2)
Draw to scale and measure. Simple. No complicated math needed. I can measure it right on my screen and get very close to this length, probably within about 2%. From there it is easy to find the soution of x=2 and y=6.
@@batchrocketproject4720 It doesn't have to be. Take a big enough circle (such as a hoola hoop), and some string, then measure the circumference and the diameter, and you can probably get pi correct to 2 or even 3 decimal places which is good enough for many calculations.
@@batchrocketproject4720 Easy, just measure first, then try all possible reasonable tuples of (x,y) such that they sum very close to that measurement, and choose the best tuple. Just try reasonable combinations for x and y such as (1,6), (1,7), (1,8), (1,9), (2,4), (2,5), (2,6). Then we are done since (2,6) will be very close, although (1,8) for (x,y) is also close at about 3.83, so measure accurately. The only other candidate solution would be (3,4) but that evaluates to 3.73, so if measured accurately, the only choices it could be are (1,8) = 3.83 or (2,6) = 3.86. We don't have to check any tuples where x=4 because Sqrt(x) is already 2 so Sqrt(y) has to be greater than 2 and 4 is already too big (more than our measurement). Absolutely nothing wrong with this solution since the geometric details are given as part of the problem, which means it can be drawn to scale and measured accurately. Zero points (my ass). A correct solution (exactly right, not an approximation). Bonus points for finding the most simple solution is more like it. I win, you lose (this is too easy).
I solved it differently: the semi-circle is given by the equation y = sqrt(1-x^2), so for symmetric reasons we are looking for a positive solution y = 1/2 which leads to x = sqrt(3/4) and the total length (not the diagonal) is 2 + 2*sqrt(3/4). = 2 + sqrt(3). The Pythagoras leads to the result AD = sqrt((2 + sqrt(3))^2 + 1^2) = sqrt(8+4sqrt(3)).
this is just a simple right angle triangle, the base is equal to Diameter of circle A plus Diameter of Circle B, the hypotenuse is AD and the height is the radius which is 1, so to find AD we can just do 1^2 + 4^2 = squareroot(AD) = squareroot(1) + squareroot(16) = squareroot(x) + squareroot(y) right?
That 30 degree angle is obvious, because the radius is 1 and the distance between the point of the tangency of both semicircles and CD is 1/2. This allows you to calculate the half of the horizontal leg of the right triangle which has hypotenuse AD. Multiply the leg (sqrt(3)/2 + 1) by 2, square it, add 1 (vertical leg squared), get the same answer.
I actually solved this problem using the trigonometric identity. At y = 1/2, x = +/- 1/2 * sqrt(3). You can use this to calculate the horizontal length, whereas the vertical length is of course 1 (the radius of the circle).
I tried doing a different method in the first part to find the horizontal length between A and D. The tangent point between circles is denoted E and the tangent point between the right circle and top line is F. The angle EBF is 90. By symmetry of the pseudo-triangle EBF that was created, angle BEF must be twice that of BFE. There values are then 60 and 30. The length from B to F is then cos(60) = sqrt(3)/2. The entire horizontal length, is then 2*(sqrt(3)/2)+2*radius = sqrt(3)+2.
from circle centre to circle centre it is exactly 2 units. From line to line it is 1 unit: therfor the centrese are SQRT(3) apart from each other in the x axis. drop a straight line down from A; A' is the crosspomit on the lowest parallel line. so A- A' = 1 and A'-D = 1+ SQRT(3) + 1. this means that A-D^2 = (AA' )^2 + (A'D )^2 = 1^2 + (2+sqrt(3))^2 = 1 + SQRT(3) ^2 + 2*2* SQRT(3) + 2^2 = 1+ 3 + 4SQRT(3) + 4 = 8 + 4SQRT(3) = 4 (2+SQRT(3)) so AD = SQRT( 4 (2+SQRT(3))) = 2 * SQRT(2+sqrt(3)) so thats the answer 2 * SQRT(2+SQRT(3))
√(8+4√3) can be simplified to √2+√6 by using a neat trick. First rewrite √(8+4√3) as √(8+2√12) and ask yourself what two numbers add to 8 and multiply to 12
So the diameters of the circles are equal to 2 units, therefore the radii are equal to 1 unit If you construct a radius from both semi circles to the point where they touch, it will form a straight line of which the total length would be equal to 2 units If you then construct a line from the midpoint of one of the circles, to the base, so that it is perpendicular, you form a right angle triangle, You can then use pythagoras to calculate the horizontal distance between the two centres of the semi circles Then you can use that distance to calculate the total horizontal distance from A to D if A is projected perpendicular down to the base level of D , You can then use pythagoras to calculate the distance AD I just can't use the notation that is required in the question (AD = square root of X + square root of Y) :) Edit: I apologise if my terminology is not very accurate, I don't do maths using english terms very often
The middle length is unknown, but we can form a triangle of hypotense 2 and and a leg of length 1. That means the middle length is sqrt(3). So the entire bottom is 2 + sqrt 3. Now we can just pythagoras. 1² + (2 + sqrt(3)² = 8 + 4sqrt(3) AD = sqrt(8+ 4sqrt(3)) = sqrt((sqrt(2) + sqrt(6))²) = sqrt(2) + sqrt(6)
This was my way of doing it: The point of intersection is at the half way point for height meaning that it is 30 degrees because sin(30)=1/2, then cos(30) is (sqrt3)/2, then subtract this from 1 to get half the distance and then multiply by 2. Then take the diameters and add them to get 4 and subtract the value we got before. The height is one since it is a semi circle Then use the Pythagorean theorem to solve for the distance. Didn't know how to do it for x and y
There is another way to find AD. Let's label the centers of semicircles as O1 and O2, tangent point as T. From right triangle with sides 1, sqrt(3), 2 we obtain that angle O1O2C is 30°. Therefore angle O1O2D = 150°. Then we can calculate length of segment TD by law of cosines and multiply it by 2. Answer will be the same.
Ez. Two circles intersect in the middle, so we know that sinθ=1/2, Thus, θ=30° and cosθ=√3/2, that's the distance from center of a circle to intersection. Add +1 and multiply by x2, so we get AB+CD horizontal distance = √3+2. Use √((√3+2)^2+1) to find AD, which is √(8+4√3), then rewrite as sum √(a+b)^2, which is just (a+b). Then do a little guess work so that a^2+2ab+b^2 = 8+4√3, which is easy using whole numbers.
I solved it in a similar way, but a bit more trig heavy. Both circles are unit circles. The distance between the two lines (top and bottom) is 1, so the height of the tangent point is 1/2 if we consider the height of the bottom line to be 0. Therefore, the horizontal distance from either of the circle centers to the tangent point is cos(30) or sqrt(3)/2, via the unit cirlce. Therefore, the base of the tirangle is sqrt(3)+2. Pythagorean theorem for the rest, and we arrive in the exact same place. Overall, very similar solution, just skipping some geometry and circle properties in favor of a little extra trig.
Simpler solution : The figure is antisymmetrical - the two circles are thus tangent at Y = 0.5. The angle between the X axis and the line leading from the centre of a circle to that point is 30°. Half of AD is the base of the isosceles triangle with sides 1 and a vertex angle of 150°. Therefore AD = 2 x 2 x sin(75°) = 4 sin (30 +45°) = sqrt(2) + sqrt(6).
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√2+√6=4 sin 75°
I was intrigued at this answer, so I tried to work this method out myself! Perhaps there is a better way, but here is what I thought. At 3:26 we have a right triangle with legs of 1 and 2 + sqrt(3). The angle at point A thus has tan(A) = 2 + sqrt(3). I didn't recognize it at first, but this is actually a special value for A = 75°. Then the other angle at D is 15°, so we want to find 1/sin(15°) = 4/(sqrt(6) - sqrt(2)) = sqrt(2) + sqrt(6). Pretty neat way to solve using trigonometry!
Exact value of tangent of 75 degrees (uses sum of angles and values of 30° and 45°)
www.mathway.com/popular-problems/Trigonometry/308167
Exact value of sine of 15 degrees (uses sum of angles and values of 30° and 45°)
www.mathway.com/popular-problems/Trigonometry/301128
@@MindYourDecisions 75° = 30°+45°. Then you can evaluate sin(75°) with the formula sin(a+b)=cos(a)sin(b)+sin(a)cos(b).
That's a great method of solving it!
@@MindYourDecisions impressive! :) thank you very much!
Do you consider a follow up video? Seems to be a pretty interesting thing that those angles turn out to be so nice.
@@camembertdalembert6323 you meant cos(a)sin(b)+sin(a)cos(b)
Honestly surprised by how relatively simple this is.
Same
remember that it's for 15 years old kids to solve
@@manamimnm ok but 15-year old Olympiad-level mathletes should probably also be able to do this in their heads.
@@manamimnm I mean, I know of problems for the same purpose that are much harder than this. Not saying it wouldn’t be very challenging for someone that age, but it does seem too easy for an Olympiad Qualifier.
@@Superman37891 yes
I got as far as √(8+4√3) in my head but had no clue how to resolve that expression further. The progression of second half of the video was really pleasing to watch. Thanks.
Me too
Same. Got stuck at that point.
Me too
th-cam.com/video/lPYxY90nW8M/w-d-xo.html
8th standard mathematics
Olympiad questions are special. They never require some advanced mathematics, but they are still able to let you think in a creative way, because every problem is different.
Damn after being a subscriber for two years I finally managed to solve a problem by my self; im proud of myself
👍
Bro its ez
th-cam.com/video/lPYxY90nW8M/w-d-xo.html
Love that first feeling of satisfaction!
Same
When solving the two equations at the end, it is easier to take advantage of the fact that x and y are positive integers. The equations are
xy = 12 and
x + y = 8
There are only 3 possibilities for the first equation, 1-12, 2-6, 3-4. The only pair that sums to 8 is 2-6. Hence, x =2 and y = 6.
ikr but same thing practically
...but it's a lot more fun to use complex math!
When I was in middle school, we students weren’t allowed to draw anymore lines into the drawings because it would make it easier to solve it and it wouldn’t require specific algorithms anymore but now that I know this is how math problems should be solved, I’m now going practice doing this. This channel has opened my mind in math and showed me how I should approach math. +1 subscriber
How did your teacher think those algorithms were discovered? That's some low iq reasoning.
This is why my kid will not be taught math in public school lol
@@Propane_Acccessories Doesn't mean that private is better
@@aloosh1375 Not always of course. It very much depends on the parents as well.
Consider the intersection pt. as 'O'.
The arc BO subtends angle 30° (sin(1/2)) at the centre of the left circle.
Therefore, angle BAO=15°
Now, AD.sin (15°)= 1
AD= √6 + √2
Wow! Presh's accent makes it legit easier to understand. Wish he taught me maths. Didn't know Olympiad had easy questions too.
Yeah
The easy ones remain hidden like treasures
th-cam.com/video/lPYxY90nW8M/w-d-xo.html
Yup, not difficult. Just more steps and ppl give up ☺️
Few suggestions for simplification :
1. By symmetry it is quite obvious that both the line joining both centers, and AD are going through the tangent point which is center of symmetry ; from there we have 2 right triangles to calculate.
2. Once you reach xy=12, there are not so many possibilities with x,y integers - checking for x=1,2,3 is sufficient to find the single solution that also satisfies x+y=8 where x
5:45 it’s a small refinement and saving, but we’re already done right there. The system of equations is symmetrical to swapping x and y, so the quadratic will necessarily have as solution the values of x and y.
I love the way Presh explains things, especially when he avoids those "weird" names for standard formulae. Great video.
th-cam.com/video/lPYxY90nW8M/w-d-xo.html
@@akashraj6391 Damn, it's truly saddening to see what lack of toilets does to a man. I'm truly sorry for you and your horrible accent.
Gougou theorem avoided! Near miss.
Except that he should just say Pythagoras because that's what's EVERYONE knows it as, and he just obfuscates the problem.
@@JamesWylde - just be thankful for small mercies. LOL
In the process of factoring x^2 - 8x + 12, you had to answer the question "what are two numbers such that their sum is 8 and their product is 12", which is just the question you started with in "solve 8=x+y and 12=xy".
Rewriting it as a quadratic does allow you to solve using the quadratic formula, in case you can't find it in your head.
Exactly what I'm thinking. I feel it is similar to when we are kids we have 18-9 where you need to regroup since 8 is bigger than 9. And so we ended up with the same exact problem. Then when in high school we have that. :)
Yeah - you're correct - it is an unnecessary step - that being said it might be useful to know how to solve this type of questions (this channel wants to educate people) and we don't know if for example you were not given points for some proper step by step explanation - depends on a test I suppose. This was quite easy question for this channel (so something I could have resolved on my own - without much trouble) and people going for Olympiad often trip on improper explanation of the answer. Depending on what level you are obviously.
Well, you could use Discriminant=64-48=16 x=(8(+ or -)4)/2=6 or 2
I solved it as follows:
I drew a line from D to the intersection point (called M), then the perpendicular from M to the lower line and called their intersection point K. Then I considered the two right triangles CMK and DMK, and exploited two facts:
1. CK + DK = CD = 2;
2. DK : KM = KM : CK (by the proportion of right triangles with one side in common).
From 1) and 2) one can find CK, DK and, by Pythagora's theorem, DM. The result we're looking for is 2DM - the double root trick will give the desired form.
The solution as Presh presented is more elegant, but this is another way nonetheless!
In the last part, there is an easier way. (x-y)^2 = (x+y)^2 - 4xy so that, you just need to solve simultaneous equations
x+y= 8, x-y=4
That would give you x = 6, y = 2, but the answer you want is the other way around.
I didn't get the radical conversion before. I tried it again months later and I solved it.
Thank you!
Having established the length of the first triangle’s long leg (square root of 3 or 1,7320) we have for the second triangle:
Short leg = 1
Long leg = 1 + 1,7320 + 1 = 3,7320
Therefore the distance A to D (the hypotenuse of this right triangle) = 3,86
Simpler
I agree. The distance between the semi-circles’ centres is 2x(sqrt(1-0.5^2))=sqrt(3). Hence horizontal distance AD is 2+sqrt(3) and vertical distance is 1. Much easier than the supplied method (for a change).
Seriously. What was the need to solve for x and y? The question was to solve for the distance of AD.
Obviously, the 2 semicircles are the same so applying the Pythagorean theorem, horizontal distance pt. 2 to 2 is square root of 3, now assuming that the radius equals to 1, therefore the horizontal distance A-D equals square root of 3 + 2 = 3.732
Horizontal distance, yes, but the question asks for the total distance. And that it be expressed as √x+√y with integers x, y.
Like most others, getting to the √(8+4√3) solution was fairly simple. I agree that putting it in the required form for the answer was a neat trick. What I'd like to understand is whether there is a good reason for employing such a trick in other circumstances. I suspect the answer is yes, but can't think of any myself.
What trick?
That´s analytical solutions versus numerical solutions. It looks a bit of a contradition but you can use analytical geometry to compute the numerical solution for most of these kind of problems.
At 3:57 , we can also directly get √2 + √6 as the ans , we can eliminate the bigger square root if we get a square term in the inside , thus we can try to look for (a+b)^2 identity directly in 8 + 4√3 , we notice a^2 + b^2 = 8 , 2ab = 4√3 , ab = 2√3 = √ 12 . and 12 = 3x4 = 2x6 , but we know 3+4 = 7 , which isn't equal to 8 , so we know a= 2 and b=6 , therefore we get 8 + 4√3 = ( √2 + √6 ) ^2 .
Thus we get our ans .
This method is just to simplify things and get a different approach too .
Thanks
Thanks...
"...our favorite right-triangle theorem..."
So no specific attribution as opposed to "not Pythagoras." I guess that's progress of a sort :-)
Well "Gougu" was not an attribution, it's the historic Chinese name for the theorem. Not a person's name.
Gougu is now persona non grata
"the not pythagoras" 😂😂😂 is better
Presh is making some progress. He's still a pompous ass that fails miserably in his goal of making math approachable and understandable with his resistance to referring to Pythagoras' Theorum the way everyone knows it it as, but I guess he's OK being a modern mathematical laughing stock. Let's face it Presh, half of your views are by folks wanting to see how you contort yourself to avoid saying Pythagoras. People go to 3B1B, Mathologer, and the like to learn something, we watch you for a chuckle.
@@JamesWylde Damn, things really heating up in the TH-cam math community
at 5:14 you dont need to put it in quadratic form, because the problem of finding two numbers with known sum and product is what is always required when solving a rational quadratic. just look at x+y=8, xy=12. we know the factors of 12 so the pair 6,2 comes out immediately.
It is quite obvious, but you can’t just say “obviously 6,2 comes out immediately” in a proper solution.
@@Fidder492 was that a comment to me David Seed. I didn’t say “obviously”. I said that the problem could be solved without forming a quadratic. note that any quadratic can be represented in this way.
if the roots if a quadratic in z are x and y then that quadratic may be written as (z-x)(z-y) = z^2 -(x+y)z +xy. . so if you can solve x+y=8, x+y=12 then there is no need to write out the quadratic.
Nice one.
But I always find the last step for this a little funny. Find two numbers that adds to 8 and are factors of 12. And we use substitution to make a quadratic equation then use factoring (which is basically finding two numbers that add to -8 and are factors of 12) :)
Haha, good point!
Hor length of small triangle = 1.732 (sq rt 3)
Hor length of large triangle = 3.732
Thus AD = sq rt ((3.732 x 3.732) + (1 x 1))
Thus AD = sq rt 14.92782 = 3.86
It seems easier to just work out the values when you get to the smaller triangle at 3.05 and then use actual values to get the longer triangle. I was fine with the diagram bit but lost on the x and y bit!
I did using trig. 4cos(a) is the total length AD, where 'a' is the angle between AD and one of the base line of diameter. Vertical length between two parallel lines turn out to be 2sin(2a), which is 1. Thus angle a is pi/12. Next step is to express 4cos(pi/12) in the required form, which simplifies to sqrt(2) +sqrt(6).
5:10 "So we have a system of two equations: 8 = x + y; 12 = xy".
Quadratic equation?! Come on, just look at it!
The variables x and y have to be integers.
12 = 1*12 = 2*6 = 3*4. Which pair add up to 8?
How can you confirm AD is a straight line passing through the contact of the circles
@@joginderpaldua351 Symmetry? Since both semicircles are the same size, the tangent point has to be half the vertical distance between the centers, and halfway between them horizontally, which means it's halfway between A and D.
@@bandar1606 No? AB is a diameter length of the semicircle, with length 2. The distance from A to the tangent point is necessarily less than 2, because it's not a diameter. The distance from the tangent point to D is the same distance, so the length of AD is less than 4. See diagram at 3:21
I didn't even care to read the title until I started watching the video and shocked to see it was from the Philippines. Amazing!
This was so unbelievably more difficult than getting the literal answer, needing it in a certain form.
sin-1(.5) = 30 degrees (because they must intersect halfway the vertical height)
cos(30) = x/.5 ==> x = .86 (this is the horizontal position of intersect from one circle center)
add 1 for the other radius half we skipped, 1.86 is the horizontal distance from one circle outer edge to intersect
double for 3.73, total horizontal distance (not surprisingly slightly under 4)
3.73^2 + 1^2 = h^2
h = 3.86
Excellent solution!
I solved many problems of your channel. But I still feel so proud. Thank you.
i had the distance figured out correctly, but that neat bit of shuffling the numbers around to make a nice simply sum of tow square roots is far to advanced for me...
I tried solving just from the thumbnail, and got as far as “AD = csc(15deg)”. I consider that a win. Even though I didn’t touch any square roots.
How did you get the angle BAD(!)=15 degrees?
@@bjorntorlarsson The problem has rotational symmetry about the point where the two circles meet. Let’s say the radius is 1, this won’t change any angle. The meeting point will have a height of 1/2. This corresponds with a 30 degree angle from the horizontal (meeting point -> center of circle -> B or C). Cut that in half for the angle with the same endpoints but with the middle one on point A or D, because we’re moving from the center of the circle to the circle itself. 30/2=15
@@jakobr_ I still don't see why the angle must necessarily halve from 30 degrees when rotating the diagonal from the midpoints of the diameters, to their far endpoints. I don't see that the length of the secant from A to the tangent point has any very obvious relationship to the length of the radius. So why does its angle from the horizontal happens to be half?
But thank you anyway!
@@bjorntorlarsson en.wikipedia.org/wiki/Inscribed_angle
@@jakobr_ Oh, I see now! The base is the tangent point to B, and the angle point is the diameter's mid point, and the same circle's periferi point respectively. The second one with an acute angle that is half of the first. I just didn't come to think of those two triangles having the same chord as base.
This is useful thinking!. I will remember this now. Thanks!
Any three points in the plane have all kinds of relationships to each other, and sometimes to a fourth point, and more.
Guys actually there is a quick way for ending.Squareroot 8 plus squareroot 3 times 4 we have.Write the 4 times squareroot part as 2 times square root 12 .And look, 12 can be written as 2.6=12 in situations like this,if the total of the multipliers eaquals to the number next to it .For this question 2+6=8 so it fits our rule so you can write it as squareroot 6 + squareroot 2 it is a fast way to solve it.If you guys couldnt get what I am saying you can ask your questions because I couldnt explain it well I guess.Sorry for any misunderstoodments.Just ask me if you have any questions.Have a nice day.
I used another way to solve the problem using similar triangles. It was fun.
The whole bit from 5:12 to 6:03 could be replaced with this: "Two positive integers with a sum of 8 and a product of 12 can only be 2 and 6." Why make it any more complicated than that. And why does Presh refuse to utter the word "Pythagorean"?
lovely to see there are so many different approaches!
I used the cosine rule to find the distance between D and the point where the semicircles meet.
(by subtracting (from pi) the bottom right angle of the central triangle which you also constructed, which ends up nicely being pi/6)
It's faster.
I'm so pleased that I was actually able to do this one myself before the video
Can't you just connect their centers, this line will surely pass through their tangent point hence it's length will be 2 and the height being 1, using Pythagoras to get horizontal distance as √3. Adding two radius front and back that is 2, we get the length of AD as 2+√3
yes but we aren't measuring the zig zag distance or the distance along one of the horizontal lines
Still even if we need to find the linear line connecting A and D, we know that it's nothing but a hypothenuse to 1 being height and √3+1 being the base. The need for angles is not relevant that is what I meant
I love that you can tell that math makes him so excited in every video
Interesting to see your solution! I started out drawing a similar central triangle to you, but then I deviated in my method. Figured I'd share my solution too as I rather enjoyed it! Instead of working out the missing length of the 2-1-√3 triangle, I worked out that the angle opposite the side of length 1 is 30° (arcsin(1/2)). Based on this, I was able to draw a new triangle between A, the center of circle AB, and the tangent point at which the two circles meet. This is an isosceles triangle with angles 150° (180 - 30), 15° and 15°. The two equal sides have length 1 (as they are both radii of circle AB) and the side opposite the 150° angle is half the length of AD (since you could draw an identical triangle in circle CD and the two line segments passing through the tangent point of the two circles would form the length AD). Therefore, by the sin rule, ½AD / sin 150° = 1 / sin 15°. We know that sin 150° = sin 30° = ½. So ½AD / ½ = 1 / sin 15° => AD = 1 / sin 15°. From here, we just need to work out the value of sin 15°, which is sin(45 - 30) = sin 45 cos 30 - cos 45 sin 30 = (1/√2)(√3/2) - (1/√2)(1/2) = (1/(2√2))(√3 - 1). Multiplying both the numerator and the denominator by √2 gives ¼(√6 - √2). Therefore AD, which = 1 / sin 15° = 4/(√6 - √2). Multiplying both the numerator and the denominator by (√6 + √2) gives 4(√6 + √2)/(6 - 2) = √2 + √6 (i.e. x = 2, y = 6).
This is pretty much how I got it, and I thought it seemed more straightforward than that squaring both sides and expanding stuff. But maybe it would have been harder if the 4 didn't cancel itself out at the end...
I'm 38 and I've never been good at math, and thanks to your channel I've learned that you can put triangles and other shapes into shapes to solve your problems :P
Easiest way : Use distance between centers and Pythagoras (Gougu) Theorem. 😊
Yeah, did same
@@lime-limelight this could be part of an hexagonal, circle sorting, in wich we can daw a hypotenuse between the two center of the circles, with distance 2, making pithagoras I got a different result of aproximately 3.73
An interesting insight into this problem is to realise that, since AB and CD are diameters, then (just focusing on the semicircle AB), the triangle formed by A, B and the tangent point between the two circle is a right triangle. The height of this triangle is 0.5 (halfway between the parallel horizontal lines), and its base is 2. Thus, its area is 0.5. You can ‘stack’ the identical triangle formed by C, D and the same tangent point, on top of the first triangle to get a rectangle. You will observe that the area of this rectangle is, therefore, equal to 1. If you let the longer side of the rectangle = x, the shorter side = 1/x, and we write x^2 + (1/x)^2 = 4 [the square of the diameter]. Solving from there is straight forward, but the manipulation of the surd (radical) expression in the second half of the video was a delight to watch! Thanks Presh.
While thy are right triangles, the height is not 0.5 because AB is the hypotenuse and not the base..
You really need to apply some trigonometry to find the actual height
@@apostolosspanelis5686. Certainly, AB is the hypotenuse of a right triangle, but the area of any triangle is a base (choose any side of a triangle - in this case, I’m using the hypotenuse of the right triangle) x height of the triangle (measured perpendicular to the chosen base) x 1/2. Given that the base that I have chosen runs along the top horizontal line, its perpendicular height measured vertically down is 0.5 (the tangent point is halfway between the two parallel horizontal lines = 0.5. Therefore the area of the right triangle = 1/2( 2 (base) x 0.5 (height) ) = 0.5. Do you follow my reasoning?
Yes got it now!
Damn, it was pretty easy to get the 8+4sqrt(3) but after that I had no idea how to make it into sqrt(x) + sqrt(y).
Some nice math shenanigans right there.
Base is 2 diameters = 4 , Height is 1 radius =1. AD = right triangle hypotenuse SQRT(4^2 + 1^2) = SQRT(17) = 4.123...
Finally! One I could solve in my head before watching the video :) (Well, for a numeric answer, not in the form requested).
Before you constructure a line through the two centers of the two circle, you need the "symmetry" argument. Without the symmetry argument you can't be sure whether you can draw a line through the two centers AND the tangent point of the two circles.
now I know why I became a PLUMBER..
Ha ha ha
Easy solution that came to my mind:
Take centre of left circle as origin
Take centre of right circle as (x, -1)
Take the point D as (x + 1, -1)
Take the point A as (-1, 0)
Distance of thier centres is 2
By distance formula x = √3
Then by distance formula,
AD = 2√3 + 4
AD in terms of √x + √y is,
√12 + √16
are u jee 2023 aspirant?
@@rahulsupari8640 yes
I solved this on my own, i tried my best :)
The distance between both center of the circles is 2, since it's tangent to each other. Draw one line from the radius of the top circle, and i got a right triangle. I used phytagorean theorem (or whatever you called it 😅), i got √3 for the other unknown side of the triangle. I added up 2 radii of the circles and i got 2+√3. And yes, it makes another right triangle, so AD=√[1²+(2+√3)²] and i got AD=√[8+2√12].
And then i use the formula my teacher ever gave, which is > , so since 2+6 is 8, and 2*6 is 12, the final result i got is √2 + √6
I hope that it was right 😅
There is another less complex way to solve the last part. When grouping alike factors xy = 12, with x, y element of N leads to x, y = 1, 12 or 2, 6 or 3, 4. Then from x+y=8 we find it must be x=2 and y=6. The problem and its solution are well presented in this video.
I got to 2(sqrt(2) + sqrt(3)) = 3.863 and then promptly guessed integers (starting at 1 and 2) until I found a match with sqrt(2) + sqrt(6).
Not as pretty but it sure was faster than doing the second half the full way 😅
Finding the length of AD is easy: it's +√13. Putting the answer in your required format is harder. Here is how I did it. Once I have the format shown at 6:52 in the video, I can also write two dependent equations: x + y = 8, and (4)(√ 3) = (2)(√x)(√y). Since you have stipulated that x and y cannot be equal but must be positive integers, that leaves only three pairs of addends that satisfy your requirements and the equation x + y = 8, namely: 1 + 7 , 2 + 6, or 3 + 5. A quick test gives the correct answer as 2 and 6.
Finally, a question that I was able to solve!!
th-cam.com/video/lPYxY90nW8M/w-d-xo.html
We have significant fact that circles are equal, meaning their touch point stays exactly at half of radius = 0.5; Knowing that and radius we can know horizontal distance from center of circle till touch point. Two of such distance + diameter = leg of triangle where we should find hypotenuse. Easy as 1-2-3!
Can't believe I did this myself. :D
Also seeing the "nice" in the title made me feel like the answer was 69 when it obviously wasn't 😆
🤣
If you define point A at coordinates (0,0), then the coordinates of center of the first semi-circle are (1,-1) and the point tangent to the line above the center are (1,0). A line from this point to the center of the second semi-circle is 2 units long and distance between the two centers is 3^0.5. So the coordinates of the second semi-circle center are (1+3^0.5,-1) and the coordinates of point D are (2+3^0.5,-1). Therefore the line AD is ((2+3^0.5)^2 + (1)^2)^0.5 = (8+4*(3)^0.5)^0.5 = 3.8637...
You actually had the answer at 3:04 which is how I solved it. your AD y is the radius 1 and AD in x is 1 (radius) + sqrt of 3 + 1 (radius) again. pythagorean theorem to get 3.86. No need to go around your elbow to get to your thumb.
Nice problem. On the "Pythagoras or not" question - the attribution is to the *Pythagoreans*, if anyone. Standard references on the preSocratic philosophers go through the very tricky question of attributions to the master, if indeed there was one.
You could use the Cosine rule to find the length from D to the tangent point:
C²= A² + B² - 2ABCosC°, where A & B are the radius 1 and C° is 150°.
This gives C² =(2 + √3). Times by 2 to get the entire length 2√(2 + √3), or √(8 + √12), which equals 3.86.
Made a meal of that one! Mid point between A & B to mid point between C & D has a length of 2. Therefore base of triangle is 1.732. Larger right angle triangle has a base of 3.732 & a height of 1. Therefore its hypotenuse is 3.8654. Done!
A much easier solution can be found as follows. Drop a perpendicular from A and connect the centres through the point of contact of the two circles. Also drop a perpendicular from the centre of the circle. Using Pythagoras, the horizontal line connecting the centre of the first circle to the point of contact to the other circle is sqrt 3. The horizontal distance between the extreme edes of the circles is 1 + sqrt 3 +1 or 2 + sqrt 3. Then using Pythagoras AD is the square root of ( 1 + ( 2 + sqrt 3 ) squared) . Or the sqrt of (8 + 4 times sqrt 3).
I’m happy I’ve made a different solution from you, for the first time. Most of the time I can’t even solve it. I made a right triangle inside a semicircle, with its right angle at the tangent point, and hypotenuse at the semicircle diameter. This triangle has an angle of 15 degrees, and I used its cos, (sqrt(6)+sqrt(2))/4. Great video. As always, mind your decisions!
Same as mine. But in this case, firstly you must prove that the 3 points [A - D - the tangent point] are collinear. It's not too hard but I think that is the most tricky part, many people forget that.
distance between the centre of the 2 semicircle will be 1+1 = 2
so the horizontal distance between the 2 centres will be sqrt(2^2 - 1^2)
then find horizontal distance of AD is above answer + 1 + 1 because of 2 radiuses
then final answer is just sqrt(1^2 + horizontal distance) = 3.86
The circles are tangent at the point where sin = 0.5 as radius = 1. Take radians for sin(X)=0.5 and apply to cosine.
Cos(30) = 0.86 roughly.
Take length of circles 4 - (1- cos(30)) = 3.72
√1+(3.72)^2 = 3.86
0:31 I don't have read all those things according to my age but here's how I find the length between a and d =
1) breadth is 4 unit (2+2)
2) length is 1 unit (2/2)
3) It forms a rectriangle, now we can put Pythagoras theorem.
Ans is 4.12 unit 😅.
Breadth is not 4 units
Look at the fig. of semis carefully
Calculating the distance AD is pretty straightforward. Connect the centers of the two semicircles. By symmetry it will pass through the tangent point and create right triangle with hypotenuse 2 and height 1. This is special 30-60-90 triangle so the horizontal displacement between the two centers is sqrt(3). Therefore the horizontal displacement between A and D is sqrt(3)+2. You can construct a right triangle with height 1 and base (sqrt(3)+2). AD^2 = 1^2+ ((sqrt(3)+2)^2 = 1+3+4sqrt(3)+4 = 4sqrt(3) + 8. Now we would take the square root to get the length of AD. However, we are asked to put it in the form of sqrt(x)+sqrt(y). So AD^2 = x+2sqrt(xy)+y. Therefore x+y = 8 and 2sqrt(xy) = 4sqrt(3) = 2sqrt(12). Therefore xy=12. We could do the algebra but since we're given that x and y are integers we can quickly get x = 2 and y=6 by trial and error. Therefore the answer is sqrt(2)+sqrt(6).
After the basic premise, I guesstimated 3.80. The explanation was more than I could handle but when it came up 3.86, I knew my old guesstimator was still in operable condition.
And as we used to say "Good enough for government work".
AD = 2√(2+√3) = √2 + √6,
Where x = 2, y = 6
And x < y
Easy one without touching pen
U deserve 50 million subscriber .Love from🇧🇩🇧🇩🇧🇩
Thanks for the nice solution. You could also use the power of the point approach. First calculate the power for the point of the left circle center, thus (1+y)(1+y)=1·3, where y is the distance between point B and right circle tangent (T), then using this value (y=√3-1) you can solve the power for the point A, (AD·AD/2 = T·T) i.e. ½x·x = (2+y)(2+y) getting the value for x as √2+√6.
I accidentally solve like you and get the correct answer for the first time! Thank god!
this trick is formally referred as de-nesting the radicals, sqrt is the most common case~
The point where the 2 circles touch is exactly halfway the two lines. Make a triangle there from the center of the circle, the point in the middle of the 2 and to a line and solve for the length of the line. (sqrt(1^2-0.5^2)) Then make a triangle from point A to the middle and to the same point straight above from the middle and solve the hypotenuse. (sqrt(1.866^2+0.5^2)). Then just times that by 2 and you get 3.86 and change.
I got it right in my first attempt. It does not matter. It matters what a fun it was to solve it. And I am really satisfied with the problem.
The process of solving the equation is really great.
I like it very much.And also learnt a new process of equation solving without calculator.
A more intriguing question would be: what is the area of the space between the two semicircles?
Thats a good question though.
If you are talking about the total white area between the semi circles then it'll be (√3-π/2) and if we consider any one part among those two white areas between the semicircles then it will be the half of the earlier answer i.e. (√3-π/2) /2 ... [ as both the empty areas are identical]
The rectangle formed by the horizontal lines and the verticals from each of the centre points has an area of 2. Within the rectangle you have 2 quarter circles = 2(pi/4) = pi/2. So the area between the semicircles = (2 - pi/2)
what a great problem, i liked how it all unwrapped to the solution
Draw to scale and measure. Simple. No complicated math needed. I can measure it right on my screen and get very close to this length, probably within about 2%. From there it is easy to find the soution of x=2 and y=6.
🤣🤣😂😂😂 thats what legends doo😂🤣
Is your ruler calibrated in √x + √y ticks? 🤔
@@batchrocketproject4720 It doesn't have to be. Take a big enough circle (such as a hoola hoop), and some string, then measure the circumference and the diameter, and you can probably get pi correct to 2 or even 3 decimal places which is good enough for many calculations.
@@davidjames1684 The question specified "give your answer in the form √x + √y" - zero points even for a measurement correct to 100 decimal places.
@@batchrocketproject4720 Easy, just measure first, then try all possible reasonable tuples of (x,y) such that they sum very close to that measurement, and choose the best tuple. Just try reasonable combinations for x and y such as (1,6), (1,7), (1,8), (1,9), (2,4), (2,5), (2,6). Then we are done since (2,6) will be very close, although (1,8) for (x,y) is also close at about 3.83, so measure accurately. The only other candidate solution would be (3,4) but that evaluates to 3.73, so if measured accurately, the only choices it could be are (1,8) = 3.83 or (2,6) = 3.86. We don't have to check any tuples where x=4 because Sqrt(x) is already 2 so Sqrt(y) has to be greater than 2 and 4 is already too big (more than our measurement). Absolutely nothing wrong with this solution since the geometric details are given as part of the problem, which means it can be drawn to scale and measured accurately. Zero points (my ass). A correct solution (exactly right, not an approximation). Bonus points for finding the most simple solution is more like it. I win, you lose (this is too easy).
I solved it differently: the semi-circle is given by the equation y = sqrt(1-x^2), so for symmetric reasons we are looking for a positive solution y = 1/2 which leads to x = sqrt(3/4) and the total length (not the diagonal) is 2 + 2*sqrt(3/4). = 2 + sqrt(3). The Pythagoras leads to the result AD = sqrt((2 + sqrt(3))^2 + 1^2) = sqrt(8+4sqrt(3)).
this is just a simple right angle triangle, the base is equal to Diameter of circle A plus Diameter of Circle B, the hypotenuse is AD and the height is the radius which is 1, so to find AD we can just do 1^2 + 4^2 = squareroot(AD) = squareroot(1) + squareroot(16) = squareroot(x) + squareroot(y) right?
That 30 degree angle is obvious, because the radius is 1 and the distance between the point of the tangency of both semicircles and CD is 1/2. This allows you to calculate the half of the horizontal leg of the right triangle which has hypotenuse AD. Multiply the leg (sqrt(3)/2 + 1) by 2, square it, add 1 (vertical leg squared), get the same answer.
I actually solved this problem using the trigonometric identity. At y = 1/2, x = +/- 1/2 * sqrt(3). You can use this to calculate the horizontal length, whereas the vertical length is of course 1 (the radius of the circle).
I tried doing a different method in the first part to find the horizontal length between A and D. The tangent point between circles is denoted E and the tangent point between the right circle and top line is F. The angle EBF is 90. By symmetry of the pseudo-triangle EBF that was created, angle BEF must be twice that of BFE. There values are then 60 and 30. The length from B to F is then cos(60) = sqrt(3)/2. The entire horizontal length, is then 2*(sqrt(3)/2)+2*radius = sqrt(3)+2.
from circle centre to circle centre it is exactly 2 units. From line to line it is 1 unit: therfor the centrese are SQRT(3) apart from each other in the x axis. drop a straight line down from A; A' is the crosspomit on the lowest parallel line. so A- A' = 1 and A'-D = 1+ SQRT(3) + 1.
this means that A-D^2 = (AA' )^2 + (A'D )^2 = 1^2 + (2+sqrt(3))^2 = 1 + SQRT(3) ^2 + 2*2* SQRT(3) + 2^2 = 1+ 3 + 4SQRT(3) + 4 =
8 + 4SQRT(3) = 4 (2+SQRT(3)) so
AD = SQRT( 4 (2+SQRT(3))) = 2 * SQRT(2+sqrt(3))
so thats the answer 2 * SQRT(2+SQRT(3))
Before watching the video AD = √2 + √6. Thanks for making my brain work!
Find in using trigonometry. Angle BAD = 15° then AD = 2×2×cos pi/12 = (sqrt2)(sqrt(3) +1)
√(8+4√3) can be simplified to √2+√6 by using a neat trick. First rewrite √(8+4√3) as √(8+2√12) and ask yourself what two numbers add to 8 and multiply to 12
So the diameters of the circles are equal to 2 units, therefore the radii are equal to 1 unit
If you construct a radius from both semi circles to the point where they touch, it will form a straight line of which the total length would be equal to 2 units
If you then construct a line from the midpoint of one of the circles, to the base, so that it is perpendicular, you form a right angle triangle,
You can then use pythagoras to calculate the horizontal distance between the two centres of the semi circles
Then you can use that distance to calculate the total horizontal distance from A to D if A is projected perpendicular down to the base level of D ,
You can then use pythagoras to calculate the distance AD
I just can't use the notation that is required in the question (AD = square root of X + square root of Y)
:)
Edit: I apologise if my terminology is not very accurate, I don't do maths using english terms very often
I got it! I am happy now.
Was able to solve it in 3 mins using the same method but didn't bother to convert the answer to a more readable form.
The middle length is unknown, but we can form a triangle of hypotense 2 and and a leg of length 1. That means the middle length is sqrt(3). So the entire bottom is 2 + sqrt 3. Now we can just pythagoras.
1² + (2 + sqrt(3)² = 8 + 4sqrt(3)
AD = sqrt(8+ 4sqrt(3))
= sqrt((sqrt(2) + sqrt(6))²)
= sqrt(2) + sqrt(6)
This was my way of doing it:
The point of intersection is at the half way point for height meaning that it is 30 degrees because sin(30)=1/2, then cos(30) is (sqrt3)/2, then subtract this from 1 to get half the distance and then multiply by 2. Then take the diameters and add them to get 4 and subtract the value we got before.
The height is one since it is a semi circle
Then use the Pythagorean theorem to solve for the distance.
Didn't know how to do it for x and y
I worked out the chord length for one circle at 150degrees and doubled it for two circles, no simultaneous equations required
There is another way to find AD. Let's label the centers of semicircles as O1 and O2, tangent point as T. From right triangle with sides 1, sqrt(3), 2 we obtain that angle O1O2C is 30°. Therefore angle O1O2D = 150°. Then we can calculate length of segment TD by law of cosines and multiply it by 2. Answer will be the same.
Ez. Two circles intersect in the middle, so we know that sinθ=1/2,
Thus, θ=30° and cosθ=√3/2, that's the distance from center of a circle to intersection.
Add +1 and multiply by x2, so we get AB+CD horizontal distance = √3+2.
Use √((√3+2)^2+1) to find AD, which is √(8+4√3), then rewrite as sum √(a+b)^2, which is just (a+b).
Then do a little guess work so that a^2+2ab+b^2 = 8+4√3, which is easy using whole numbers.
I solved it in a similar way, but a bit more trig heavy.
Both circles are unit circles. The distance between the two lines (top and bottom) is 1, so the height of the tangent point is 1/2 if we consider the height of the bottom line to be 0.
Therefore, the horizontal distance from either of the circle centers to the tangent point is cos(30) or sqrt(3)/2, via the unit cirlce. Therefore, the base of the tirangle is sqrt(3)+2. Pythagorean theorem for the rest, and we arrive in the exact same place.
Overall, very similar solution, just skipping some geometry and circle properties in favor of a little extra trig.
I would say ~4r
it's at most r+4r (the long and short side of rectangle AD)
so the hypotenuse of that triangle
Simpler solution : The figure is antisymmetrical - the two circles are thus tangent at Y = 0.5. The angle between the X axis and the line leading from the centre of a circle to that point is 30°. Half of AD is the base of the isosceles triangle with sides 1 and a vertex angle of 150°. Therefore AD = 2 x 2 x sin(75°) = 4 sin (30 +45°) = sqrt(2) + sqrt(6).
Yeahhhhhh I did it without your solution
My maths improved alot because of you
Love from india