Yeah and it's an Egyptian 3-4-5 triangle, apparently famous to boot. I used to be sure he got on some weird social commentary kick, but now I'm thinking he may just be a pedantic schmuck trying to show off how much history he's read. Either way, annoying as it just stands out like a sore thumb when compared to the quality problems and explanations he presents (this problem is one of the best I've seen recently -- really cute problem).
You guys beat me to it. At 4:22, he refers to the Gougu Theorem and the "Egyptian" 3-4-5 Triangle in the same sentence. Now, if the Chinese proved the theorem first, I'm pretty sure they would have found the 3-4-5 and all those other "Gouguian" Triples, so let's at least be consistent on which nationality gets the credit.
@@zanti4132 Oof lol There's probably this thing before communication at such long distances tho ahahahaha they both find find it at questionable times but its more likely that the Greeks(?... I think idk) discovered it first
I drew it in Fusion360 and added a couple of dimensions for the AB and EF lines, then added tangency constraints for the circles and a vertical constraint on the EF line and it all snapped into place giving the BC length of 9 :-)
He said “rectangle” twice in the first few seconds. Perhaps you were still stirring up your glass of Ovalitine at that point instead of paying attention.
I'd already watched the solution some 5 odd months back but didn't remember it now. On relooking the problem, I solved it little differently but interesting way which I'd love to share here. Consider 'x' to be the tangent length of smaller circle at upper left. Then Radius of same smaller circle at upper left = x Chord length (in two circles) = 6, so half chord length = Vertical distance between centre of two circles = 3 Then, AB = Tangent length + 3 + Radius of larger circle at lower right Radius of larger circle at lower right = 5 - x Then, Distance of line joining centre of two circles = x + 5 - x = 5 Vertical distance between centre of two circles = 3 So by Pythagoras Theorem, Horizontal distance between centre of two circles = 4 So, BC = Tangent length of smaller circle + Horizontal distance between centre of two circles + Radius of larger circle BC = x + 4 + (5 - x) = 9 units
Written between 2000 and 1786 BC, the Middle Kingdom Egyptian Berlin Papyrus 6619 includes a problem whose solution is the Pythagorean triple 6:8:10, but the problem does not mention a triangle. The Mesopotamian tablet Plimpton 322, written between 1790 and 1750 BC during the reign of Hammurabi the Great, contains many entries closely related to Pythagorean triples. (Source : Wikipedia)
I did it in a simpler way. Since he didn't specify the radii of the two circles, I assumed that EF would always be 6 no matter the radii. So I just set the radius of the first one to 0. That means the second circle passes through point B and a point P between A and B where BP is 6 (it's coincident with EF) and AP is 2. From there you get the diameter of the circle to be 10 from symmetry. So the radius (5) plus the distance of the center to the chord BP (4) is equal to BC.
Nice idea! You are assuming that EF is independent of the radii of the two circles in order for this result to work, whereas he proved that it was independent while calculating the value.
I did also observed the sizes of the two circles is irrelevant. Therefore, I solved the puzzle assuming both circles have the same size (radius of 2.5). Symmetry will ease the solution when both circles have the same size. By the way, EF will have a maximum size of 6.627416998 . That is, there will be no solution of EF happened to be 6.7 instead of 6.
Wouldn't the maximum height of EF given any value of BC be 8? That's when both circles are underneath eachother, tangent to the opposite sides, with a radius of 2 and a length BC of 4, with EF cleanly dividing the rectangle lengthwise.
excellent.... however please note that ancient India has very rich mathematical treasure more ancient than Pyramid era... in this treasure 3,4,5 units of right triangle is mentioned very often.
Let 𝑟 be the radius of the upper circle and 𝑅 the radius of the bottom circle. Using the bottom circle as a scaled up unit circle, the point where the two circles touch will have the coordinates (𝑅 cos 𝜃, 𝑅 sin 𝜃). This means that the combined length of the vertical chords is 2𝑅 |sin 𝜃| + 2𝑟 |sin 𝜃| = 6, which gives us 𝑅 + 𝑟 = 3 ∕ |sin 𝜃|. Similarly, the height of the rectangle is 𝑟 + 𝑟 |sin 𝜃| + 𝑅 + 𝑅 |sin 𝜃| = 8 ⇒ 𝑅 + 𝑟 = 8 ∕ (1 + |sin 𝜃|) Thus, 3 ∕ |sin 𝜃| = 8 ∕ (1 + |sin 𝜃|) ⇒ |sin 𝜃| = 3 ∕ 5 Also, |cos 𝜃| = √(1 − |sin 𝜃|²) = 4 ∕ 5 Finally, the width of the rectangle is 𝐴𝐵 = 𝑟 + 𝑟 |cos 𝜃| + 𝑅 + 𝑅 |cos 𝜃| = (𝑟 + 𝑅)(1 + |cos 𝜃|) = (𝑟 + 𝑅)(1 + 4 ∕ 5) = 9 ∕ 5(𝑅 + 𝑟) = 9 ∕ 5(3 ∕ |sin 𝜃|) = 9 ∕ 5(3 ∕ (3 ∕ 5)) = 9
Solved it thinking about the problem in terms of trigonometry. (r + R)(sin B) = 6 (using the chord) and (r+R)(sin B) = 8 (using the side of the rectangle). From those two, we know that sin B = 3/5 and r + R = 5. The other side of the rectangle is (r + R) (1 + cos B) = 5(1 + 4/5) = 9. Pretty much the same thing as the video solution, but thinking about the problem in trigonometry.
Align the whole diagram w/ the Cartesian plane, so that (1) the only point at where the 2 circles meet is the origin O(0,0); (2) the centre of C₁ & that of C₂ are at P₁(-r,0) & P₂(R,0), respectively, where C₁ is the circle touching AB & BC w/ radius r, & C₂ is the circle touching AD & DC w/ radius R. The eqn. of C₁ is (x+r)² + y² = r² x² + y² = -2rx ...(i). The eqn. of C₂ is (x-R)² + y² = R² x² + y² = 2Rx ...(ii). The eqn. of L, the line containing EF, is y = kx ...(iii) for some k E is at ( -2r/(k²+1), -2kr/(k²+1) ); (ii) & (iii) => F is at ( 2R/(k²+1), 2kR/(k²+1) ). |EF| = 6 => [2R/(k²+1) + 2r/(k²+1)]² + [2kR/(k²+1) + 2kr/(k²+1)]² = 6² => 4(R+r)² + 4k²(R+r)² = 36(k²+1)² => (R+r)² = 9(k²+1) ...(iv') => R+r = 3√(k²+1) ...(iv) (as R+r > 0) The eqn. of L₁₁, the line containing BC (⊥EF), is y = (-1/k)x + c₁₁' x + ky + c₁₁ = 0 for some c₁₁' & c₁₁ (c₁₁ = -kc₁₁'); so dist(P₁,L₁₁) = r => | [(-r)+k(0)+c₁₁] / √(k²+1) | = r => c₁₁ = r√(k²+1) + r ...(v) ("+ve √ branch" is taken: between the 2 tangents to C₁ w/ slope -1/k, L₁₁ has the larger y-int. (i.e. c₁₁'), so c₁₁ takes the larger value as -k>0). The eqn. of L₁₂, the line containing AD (⊥EF), is: y = (-1/k)x + c₁₂' x + ky + c₁₂ = 0 for some c₁₂' & c₁₂ (c₁₂ = -kc₁₂'); so dist(P₂,L₁₂) = R => | [(R)+k(0)+c₁₂] / √(k²+1) | = R => c₁₂ = -R√(k²+1) - R ...(vi) ("-ve √ branch" is taken: between the 2 tangents to C₂ w/ slope -1/k, L₁₂ has the smaller y-int. (i.e. c₁₂'), so c₁₂ takes the smaller value as -k>0). dist(L₁₁,L₁₂) = |BA| = 8 => dist(L₁₁,O) + dist(O,L₁₂) = 8 => |c₁₁/√(k²+1)| + |c₁₂/√(k²+1)| = 8 => |c₁₁| + |c₁₂| = 8√(k²+1) => [r√(k²+1) + r] + [R√(k²+1) + R] = 8√(k²+1) (by (v) & (vi)) => (R+r) = [8-(R+r)] √(k²+1) => (R+r) = [8-(R+r)] [(R+r)/3] (by (iv)) => R+r = 5 ...(*) (N.B. R+r > 0) The eqn. of L₂₁, the line containing BA (//EF), is: y = kx + c₂₁ kx - y + c₂₁ = 0 for some c₂₁; so dist(P₁,L₂₁) = r => | [k(-r)-(0)+c₂₁] / √(k²+1) | = r => c₂₁ = -r√(k²+1) + kr ...(vii) ("-ve √ branch" is taken: between the 2 tangents to C₁ w/ slope k, L₂₁ has the smaller y-int., i.e. c₂₁). The eqn. of L₂₂, the line containing CD (//EF), is: y = kx + c₂₂ kx - y + c₂₂ = 0 for some c₂₂; so dist(P₂,L₂₂) = R => | [k(R)-(0)+c₂₂] / √(k²+1) | = R => c₂₂ = R√(k²+1) - kR ...(viii) ("+ve √ branch" is taken: between the 2 tangents to C₂ w/ slope k, L₂₂ has the larger y-int., i.e. c₂₂). |BC| = dist(L₂₁,L₂₂) = dist(L₂₁,O) + dist(O,L₂₂) = |c₂₁/√(k²+1)| + |c₂₂/√(k²+1)| = (|c₂₁| + |c₂₂|) / √(k²+1) = {[r√(k²+1) - kr] + [R√(k²+1) - kR]} / √(k²+1) (by (vii) & (viii); N.B. k
It is so awesome i figured this out just using simple trig i learned 30 years ago, and didnt even remember all of these other theorems... just the trig.... AWESOME! (Totally tooting my horn, I know, but was still fun)
@@vashon100 In North America, yes; not elsewhere. We saved up and bought the letter s to add to the end. Besides, is there something called "Mathematic"?
The 3-4-5 triangle is only because the given lengths are 6 and 8. When you start with different lengths or generalize, you don't always get a 3-4-5 triangle; but you always do get a right triangle.
There is a much easier way to solve this by using a specific example rather than generic use case. Since the relation should hold up for any two circles with those properties, assume those circles to have the same radius (r). Now it is clear that the vertical distance between the centers of the circles is = 8 - 2r = half of the length EF (by symmetry) = 6 / 2 = 3 This gives r = 2.5 Now consider the (horizontal-vertical) right triangle with the hypotenuse between the two centers. The hypotenuse is clearly = 2r = 5 Vertical distance computed before was 8 - 2r = 3 By Pythagoras, the horizontal distance between the two centers (lets call it x) = 4 The side BC = r + x + r = 2.5 + 4 + 2.5 = 9
Nice, I was able to solve this with a slightly different equation: 2xsin(Θ)+2ysin(Θ)=6 x+xsin(Θ)+y+ysin(Θ)=8 x+xcos(Θ)+y+ycos(Θ)=? Divide first equation by 2 to get: xsin(Θ)+ysin(Θ)=3 If you subtract the above from the 2nd equation you will see terms cancel out: x+y+=5 Then factor out sin(Θ): (x+y) sin(Θ)=3 Plugging the known sum gives us: 5 sin(Θ)=3, thus: sin(Θ)=3/5 Knowing that sin is opposite/hypotenuse ratio, we can plug 5 for hypotenuse and 3 for a side of a triangle. This triangle starts to look familiar, in fact it is a Pythagorean triangle, so the other side is 4, which gives us: cos(Θ)=4/5 Now we can solve the third equation by plugging the known terms: x+xcos(Θ)+y+ycos(Θ) = (x+y) + (x+y)cos(Θ) 5 + 5 * 4/5 = 9 Anyone else used trigonometry and a system of equations like I did?
I solved the problem similarly, but introduced the angle a at ADB and z:=BC. Then: 2*(x+y)*sin(a)=6 and x+y+(x+y)*sin(a)=8 -> x+y=5. z=x+x*cos(a)+y+y*cos(a)=5+5*cos(a) -> z-5=5*cos(a) square the last and first equation, and add them together: (z-5)^2+3^2=(5*sin(a))^2+(5*cos(a))^2=5^2. Then solve for z - which comes out to 9.
With contents known much earlier, but in surviving texts dating from roughly the 100 years before common era, the Chinese text Zhoubi Suanjing (The Arithmetical Classic of the Gnomom) gives a reasoning for the Pythagorean theorem for the (3, 4, 5) triangle-in China it is called the "Gougu theorem" In India, the Baudhayana Shulba Sutra, the dates of which are given variously as between the 800 years before common era contains a list of Pythagorean triples and a statement of the Pythagorean theorem, both in the special case of the isosceles right triangle and in the general case, as does the Apastamba Shulba Sutra (c. 600 BC). Van der Waerden believed that this material "was certainly based on earlier traditions". Carl Boyer states that the Pythagorean theorem in the Śulba-sũtram may have been influenced by ancient Mesopotamian math, but there is no conclusive evidence in favor or opposition of this possibility. THEN WHY YOU CALLED PYTHAGORAS THEORAM AS GONGU; MENTION IT AS BAUDHAYANA THEORAM 😏😏
If H is the height of the rectangle (given as 8), and L is the sum of the two chords (given as 6); then the general formula for W (Width of rectangle) is W=H-L/2+sqrt (H^2-H*L)
Also I remind you of my puzzle which i wrote it in my comment in video of the "Largest Cone puzzle " " pentagon inside triangle" thanks for this lecture ... and I wait answer of puzzle . with my best regards .
By the way, the graphics does not show a rectangle, but a square. the diagonal line of a rectangle is not a symmetry axis. have fun 10 lba=8:lef=6:sw=.1:goto 60 20 r2=(lba-r1*(1+sqr(2)/2))/(1+sqr(2)/2) 30 xs=r1*(sqr(2)/2+1):ys=lba-r1*(sqr(2)/2+1) 40 ym1=lba-r1:ym2=ys-r2/sqr(2):ye=ym1+r1/sqr(2):yf=ym2+r2/sqr(2) 45 dlu1=(ye-yf)/lef:dl=1-dlu1:return 60 r1=sw:gosub 20 61 dl1=dl:ru1=r1:r1=r1+sw:if r1>10*lef then stop 62 ru2=r1:gosub 20:if dl1*dl>0 then 61 70 r1=(ru1+ru2)/2:gosub 20:if dl1*dl>0 then ru1=r1 else ru2=r1 80 if abs(dl)>1E-10 then 70 90 print "r1=";r1;"r2=";r2 www.imagebanana.com/s/1877/7bNnPfHU.html
One of the best question till now
Totally agree
Yes
Math
Geometry problem
One time see
th-cam.com/video/j9wY3hso1WQ/w-d-xo.html
Yup
Bhai ese questions SSC main normal hai. 🤷🏼♂️
I think the key piece of insight is using EF/2 instead of just EF. I constructed all the same guide lines as the solution but I just didn't click.
Woo
Btw i liked your channel, that's why i consider myself that catgirl, cats are my life😍👌🐱
Exactly, because only then you can use the fact that AB=8 to eventually get x+y
@@thomvandenhil4717 yupp
Same
Even when a right triangle is clearly a 3-4-5 Pythagorean Triple... Gotta name drop Gougu.
Oh No, A gOuGu TrIpLe. AaAaAaAaAaAa.........
Yeah and it's an Egyptian 3-4-5 triangle, apparently famous to boot. I used to be sure he got on some weird social commentary kick, but now I'm thinking he may just be a pedantic schmuck trying to show off how much history he's read. Either way, annoying as it just stands out like a sore thumb when compared to the quality problems and explanations he presents (this problem is one of the best I've seen recently -- really cute problem).
@@cauchysintegral3713 EEEEEEEE
You guys beat me to it. At 4:22, he refers to the Gougu Theorem and the "Egyptian" 3-4-5 Triangle in the same sentence. Now, if the Chinese proved the theorem first, I'm pretty sure they would have found the 3-4-5 and all those other "Gouguian" Triples, so let's at least be consistent on which nationality gets the credit.
@@zanti4132 Oof lol
There's probably this thing before communication at such long distances tho ahahahaha
they both find find it at questionable times but its more likely that the Greeks(?... I think idk) discovered it first
Both this problem and the way you phrased and graphed the solution were absolutely beautiful. Thank you!
(solved it by myself, not that it matters)
As soon as he said it was allways equal to nine, I paused the video, went for a long walk in the woods and smoked a pack of cigars whilst alone.
I'm NOT going to even pretend to pretend that I understood this.
It *is* tho
There's an injury lawyer in SC named George Sink. All 9's? Sounds like a George Sink commercial.
clearly someone does not understand basic maths(it's me)
I drew it in Fusion360 and added a couple of dimensions for the AB and EF lines, then added tangency constraints for the circles and a vertical constraint on the EF line and it all snapped into place giving the BC length of 9 :-)
Am I the only one who at first thought it is a square 😅
Same here 😂
same
Me too. And not only that, I thought the 8 is only the distance between A and the intersection between AB and a circle
Me too. I didn't notice the 8. My answer was BC = 3 + 6/√2
He said “rectangle” twice in the first few seconds. Perhaps you were still stirring up your glass of Ovalitine at that point instead of paying attention.
That 3 4 5 is so beautiful although i see it almost in every puzzle the person who made this must have intended that
Il problema, la grafica, la spiegazione, sono stupendi.
Grazie! ❤️
haven't started the video yet.. and I'm already hearing the gougu theorem..
Math
Geometry problem
One time see
th-cam.com/video/j9wY3hso1WQ/w-d-xo.html
Lol why am i not?😅😆
Really too awesome.... And mind-blowing also...and the presentation is really praiseworthy
Math
Geometry problem
One time see
th-cam.com/video/j9wY3hso1WQ/w-d-xo.html
Yess i like it rlly👌😍
My Solution: Assume ABCD is a square and say that BC is 8.
Actual Solution: Well you were pretty close.
My dad is an highschool math teacher and I love to send him your videos. He likes to try and solve them before watching the entire video
I'd already watched the solution some 5 odd months back but didn't remember it now. On relooking the problem, I solved it little differently but interesting way which I'd love to share here.
Consider 'x' to be the tangent length of smaller circle at upper left.
Then Radius of same smaller circle at upper left
= x
Chord length (in two circles) = 6, so half chord length = Vertical distance between centre of two circles = 3
Then,
AB = Tangent length + 3 + Radius of larger circle at lower right
Radius of larger circle at lower right = 5 - x
Then,
Distance of line joining centre of two circles
= x + 5 - x = 5
Vertical distance between centre of two circles = 3
So by Pythagoras Theorem,
Horizontal distance between centre of two circles = 4
So,
BC = Tangent length of smaller circle + Horizontal distance between centre of two circles + Radius of larger circle
BC = x + 4 + (5 - x)
= 9 units
oh beautiful puzzle!
4:22 famous theorem!!!
The problem is attractive, the solution is clever, but the theorem is.... no no no no
Presh: Let's use an ancient trick!
Pythagoras: Finally...
Presh: GOUGU!!!
😂😂😂😂.
Presh : Famous Egyptian 3-4-5 right triangle...
Pythagoras : Famous as Pythagorean triples !
The problem, as well as the solution, are both too good. Its a pleasure seeing both. Congratulations to Mr. Talwalkar.
Now our mind think that we could have solved it
But before watching the solution, it was like 😵🤯
Math
Geometry problem....
One time see
th-cam.com/video/j9wY3hso1WQ/w-d-xo.html
Presh is kinda obsessed with Gougu right now.
With any luck, maybe the CCP will find a cure!
Written between 2000 and 1786 BC, the Middle Kingdom Egyptian Berlin Papyrus 6619 includes a problem whose solution is the Pythagorean triple 6:8:10, but the problem does not mention a triangle. The Mesopotamian tablet Plimpton 322, written between 1790 and 1750 BC during the reign of Hammurabi the Great, contains many entries closely related to Pythagorean triples. (Source : Wikipedia)
Math
Geometry problem..
One time see
th-cam.com/video/j9wY3hso1WQ/w-d-xo.html
Gougu theorem: you are weak
Pythagoras theorem: i'm you
Great job using the Pythagorean Theorem to solve the problem!
What the heck is Pythagorean Therom
@@Zembie1lol😂
THIS IS THE BEST AND CLEVEREST QUESTION OF THIS CHANNEL TILL DATE ACCORDING TO ME
0:19 : I paused. I expect GoUgU theorem.
Gougu was there, in the 3-4-5 side triangle, length 4 was determined by famous Gougu theorem. 😄😂
Math
Geometry problem
One time see
th-cam.com/video/j9wY3hso1WQ/w-d-xo.html
@@ayazuit gougu is which thoerem that idk😕i know only its one of name is Piphagor😄
Perfect explanation and visualization! :) thank you very much for your great vids
All hail Presh's PowerPoint skills .
Also, it is an interesting question on geometry 👍👍👍👌👌👌
Math
Geometry problem
One time see
th-cam.com/video/j9wY3hso1WQ/w-d-xo.html
I did it in a simpler way. Since he didn't specify the radii of the two circles, I assumed that EF would always be 6 no matter the radii. So I just set the radius of the first one to 0. That means the second circle passes through point B and a point P between A and B where BP is 6 (it's coincident with EF) and AP is 2. From there you get the diameter of the circle to be 10 from symmetry. So the radius (5) plus the distance of the center to the chord BP (4) is equal to BC.
Nice idea! You are assuming that EF is independent of the radii of the two circles in order for this result to work, whereas he proved that it was independent while calculating the value.
I think you mean BP is 6 and the diameter of the circle is 10. Otherwise, very nice answer.
@@tomdekler9280 I did indeed, thanks for catching that.
I really love this channel
I wish my math class had had fun problems like this one occasionally.
you are my favorite youtuber, never miss your videos
The only question that I’m able to solve, made my day.
You are jussttt
A M A Z I N G
This problem was very unique
Love your videos
Brilliant question. Brilliant designer.
You are my inspiration
It’s not a Phresh video unless the Gogou theorem is mentioned
Yeah, a pain in the butt!
This has such a beautiful solution. You are actually the best Maths teacher on youtube🤩🤩🤩
This dude is on a whole other level
Math
Geometry problem
One time see
th-cam.com/video/j9wY3hso1WQ/w-d-xo.html
@@innocentzero9814 lol😶
I did also observed the sizes of the two circles is irrelevant. Therefore, I solved the puzzle assuming both circles have the same size (radius of 2.5). Symmetry will ease the solution when both circles have the same size. By the way, EF will have a maximum size of 6.627416998
. That is, there will be no solution of EF happened to be 6.7 instead of 6.
Wouldn't the maximum height of EF given any value of BC be 8?
That's when both circles are underneath eachother, tangent to the opposite sides, with a radius of 2 and a length BC of 4, with EF cleanly dividing the rectangle lengthwise.
excellent.... however please note that ancient India has very rich mathematical treasure more ancient than Pyramid era... in this treasure 3,4,5 units of right triangle is mentioned very often.
Awesome work presh talwalkar 🔥
Let 𝑟 be the radius of the upper circle and 𝑅 the radius of the bottom circle.
Using the bottom circle as a scaled up unit circle, the point where the two circles touch will have the coordinates (𝑅 cos 𝜃, 𝑅 sin 𝜃).
This means that the combined length of the vertical chords is 2𝑅 |sin 𝜃| + 2𝑟 |sin 𝜃| = 6, which gives us 𝑅 + 𝑟 = 3 ∕ |sin 𝜃|.
Similarly, the height of the rectangle is 𝑟 + 𝑟 |sin 𝜃| + 𝑅 + 𝑅 |sin 𝜃| = 8 ⇒ 𝑅 + 𝑟 = 8 ∕ (1 + |sin 𝜃|)
Thus, 3 ∕ |sin 𝜃| = 8 ∕ (1 + |sin 𝜃|) ⇒ |sin 𝜃| = 3 ∕ 5
Also, |cos 𝜃| = √(1 − |sin 𝜃|²) = 4 ∕ 5
Finally, the width of the rectangle is
𝐴𝐵 = 𝑟 + 𝑟 |cos 𝜃| + 𝑅 + 𝑅 |cos 𝜃|
= (𝑟 + 𝑅)(1 + |cos 𝜃|)
= (𝑟 + 𝑅)(1 + 4 ∕ 5)
= 9 ∕ 5(𝑅 + 𝑟)
= 9 ∕ 5(3 ∕ |sin 𝜃|)
= 9 ∕ 5(3 ∕ (3 ∕ 5))
= 9
That is a truly elegant solve.
Incredible Maths Puzzle. Really Impressed!
How about this: Express the side length BC in terms of side length AB for any valid side length for AB.
Math
Geometry problem
One time see
th-cam.com/video/j9wY3hso1WQ/w-d-xo.html
Solved it thinking about the problem in terms of trigonometry. (r + R)(sin B) = 6 (using the chord) and (r+R)(sin B) = 8 (using the side of the rectangle). From those two, we know that sin B = 3/5 and r + R = 5. The other side of the rectangle is (r + R) (1 + cos B) = 5(1 + 4/5) = 9.
Pretty much the same thing as the video solution, but thinking about the problem in trigonometry.
Did exactly the same thing
Could you explain more your reasoning? It's plenty interesting!!!
The solution is aesthetically pleasing.
Align the whole diagram w/ the Cartesian plane, so that
(1) the only point at where the 2 circles meet is the origin O(0,0);
(2) the centre of C₁ & that of C₂ are at P₁(-r,0) & P₂(R,0), respectively, where C₁ is the circle touching AB & BC w/ radius r, & C₂ is the circle touching AD & DC w/ radius R.
The eqn. of C₁ is
(x+r)² + y² = r² x² + y² = -2rx ...(i).
The eqn. of C₂ is
(x-R)² + y² = R² x² + y² = 2Rx ...(ii).
The eqn. of L, the line containing EF, is
y = kx ...(iii) for some k E is at ( -2r/(k²+1), -2kr/(k²+1) );
(ii) & (iii) => F is at ( 2R/(k²+1), 2kR/(k²+1) ).
|EF| = 6
=> [2R/(k²+1) + 2r/(k²+1)]² + [2kR/(k²+1) + 2kr/(k²+1)]² = 6²
=> 4(R+r)² + 4k²(R+r)² = 36(k²+1)²
=> (R+r)² = 9(k²+1) ...(iv')
=> R+r = 3√(k²+1) ...(iv) (as R+r > 0)
The eqn. of L₁₁, the line containing BC (⊥EF), is
y = (-1/k)x + c₁₁' x + ky + c₁₁ = 0
for some c₁₁' & c₁₁ (c₁₁ = -kc₁₁');
so dist(P₁,L₁₁) = r
=> | [(-r)+k(0)+c₁₁] / √(k²+1) | = r
=> c₁₁ = r√(k²+1) + r ...(v)
("+ve √ branch" is taken: between the 2 tangents to C₁ w/ slope -1/k, L₁₁ has the larger y-int. (i.e. c₁₁'), so c₁₁ takes the larger value as -k>0).
The eqn. of L₁₂, the line containing AD (⊥EF), is:
y = (-1/k)x + c₁₂' x + ky + c₁₂ = 0
for some c₁₂' & c₁₂ (c₁₂ = -kc₁₂');
so dist(P₂,L₁₂) = R
=> | [(R)+k(0)+c₁₂] / √(k²+1) | = R
=> c₁₂ = -R√(k²+1) - R ...(vi)
("-ve √ branch" is taken: between the 2 tangents to C₂ w/ slope -1/k, L₁₂ has the smaller y-int. (i.e. c₁₂'), so c₁₂ takes the smaller value as -k>0).
dist(L₁₁,L₁₂) = |BA| = 8
=> dist(L₁₁,O) + dist(O,L₁₂) = 8
=> |c₁₁/√(k²+1)| + |c₁₂/√(k²+1)| = 8
=> |c₁₁| + |c₁₂| = 8√(k²+1)
=> [r√(k²+1) + r] + [R√(k²+1) + R] = 8√(k²+1) (by (v) & (vi))
=> (R+r) = [8-(R+r)] √(k²+1)
=> (R+r) = [8-(R+r)] [(R+r)/3] (by (iv))
=> R+r = 5 ...(*) (N.B. R+r > 0)
The eqn. of L₂₁, the line containing BA (//EF), is:
y = kx + c₂₁ kx - y + c₂₁ = 0 for some c₂₁;
so dist(P₁,L₂₁) = r
=> | [k(-r)-(0)+c₂₁] / √(k²+1) | = r
=> c₂₁ = -r√(k²+1) + kr ...(vii)
("-ve √ branch" is taken: between the 2 tangents to C₁ w/ slope k, L₂₁ has the smaller y-int., i.e. c₂₁).
The eqn. of L₂₂, the line containing CD (//EF), is:
y = kx + c₂₂ kx - y + c₂₂ = 0 for some c₂₂;
so dist(P₂,L₂₂) = R
=> | [k(R)-(0)+c₂₂] / √(k²+1) | = R
=> c₂₂ = R√(k²+1) - kR ...(viii)
("+ve √ branch" is taken: between the 2 tangents to C₂ w/ slope k, L₂₂ has the larger y-int., i.e. c₂₂).
|BC| = dist(L₂₁,L₂₂)
= dist(L₂₁,O) + dist(O,L₂₂)
= |c₂₁/√(k²+1)| + |c₂₂/√(k²+1)|
= (|c₂₁| + |c₂₂|) / √(k²+1)
= {[r√(k²+1) - kr] + [R√(k²+1) - kR]} / √(k²+1) (by (vii) & (viii); N.B. k
Also the formula will be: the square root of b^2+ab plus b-a/2 (b is the length of AB while a is the length of EF)
It is so awesome i figured this out just using simple trig i learned 30 years ago, and didnt even remember all of these other theorems... just the trig.... AWESOME! (Totally tooting my horn, I know, but was still fun)
I haven't done much maths since Engineering school 40 years ago; should I be worried that I'm enjoying this channel so much?
Back then it was called math.
@@vashon100 In North America, yes; not elsewhere. We saved up and bought the letter s to add to the end. Besides, is there something called "Mathematic"?
I bet this uses the Gougu Theorem.
cash in that bet 4:21
Why not"physagor theorem"
@@talhakagantosun6341 pythagoras.
@@hishamsameh3496 i m from Turkish i not know very well English but learn this (:
@@talhakagantosun6341 it's okay. I thought that u didn't know, so may be I could contribute to your knowledge, that's all.
The quandary swam
before my eyes.
The 3-4-5 triangle is only because the given lengths are 6 and 8. When you start with different lengths or generalize, you don't always get a 3-4-5 triangle; but you always do get a right triangle.
And BC will be constant, assuming your AB and EF are possible pairs.
There is a much easier way to solve this by using a specific example rather than generic use case. Since the relation should hold up for any two circles with those properties, assume those circles to have the same radius (r).
Now it is clear that the vertical distance between the centers of the circles is = 8 - 2r = half of the length EF (by symmetry) = 6 / 2 = 3
This gives r = 2.5
Now consider the (horizontal-vertical) right triangle with the hypotenuse between the two centers.
The hypotenuse is clearly = 2r = 5
Vertical distance computed before was 8 - 2r = 3
By Pythagoras, the horizontal distance between the two centers (lets call it x) = 4
The side BC = r + x + r = 2.5 + 4 + 2.5 = 9
It's so entertaining and your voice is somehow lovely to hear ;(
Nice, I was able to solve this with a slightly different equation:
2xsin(Θ)+2ysin(Θ)=6
x+xsin(Θ)+y+ysin(Θ)=8
x+xcos(Θ)+y+ycos(Θ)=?
Divide first equation by 2 to get:
xsin(Θ)+ysin(Θ)=3
If you subtract the above from the 2nd equation you will see terms cancel out:
x+y+=5
Then factor out sin(Θ):
(x+y) sin(Θ)=3
Plugging the known sum gives us:
5 sin(Θ)=3, thus:
sin(Θ)=3/5
Knowing that sin is opposite/hypotenuse ratio, we can plug 5 for hypotenuse and 3 for a side of a triangle.
This triangle starts to look familiar, in fact it is a Pythagorean triangle, so the other side is 4, which gives us:
cos(Θ)=4/5
Now we can solve the third equation by plugging the known terms:
x+xcos(Θ)+y+ycos(Θ) = (x+y) + (x+y)cos(Θ)
5 + 5 * 4/5 = 9
Anyone else used trigonometry and a system of equations like I did?
The most beautiful thing I have seen in a while
Math
Geometry problem
One time see
th-cam.com/video/j9wY3hso1WQ/w-d-xo.html
Love your channel
One of your better puzzles - no trig involved. Nice.
Shows you don't need fancy things to have great problems
I solved the problem similarly, but introduced the angle a at ADB and z:=BC. Then:
2*(x+y)*sin(a)=6 and x+y+(x+y)*sin(a)=8 -> x+y=5.
z=x+x*cos(a)+y+y*cos(a)=5+5*cos(a) -> z-5=5*cos(a)
square the last and first equation, and add them together:
(z-5)^2+3^2=(5*sin(a))^2+(5*cos(a))^2=5^2. Then solve for z - which comes out to 9.
Good ..thanks
With contents known much earlier, but in surviving texts dating from roughly the 100 years before common era, the Chinese text Zhoubi Suanjing (The Arithmetical Classic of the Gnomom) gives a reasoning for the Pythagorean theorem for the (3, 4, 5) triangle-in China it is called the "Gougu theorem"
In India, the Baudhayana Shulba Sutra, the dates of which are given variously as between the 800 years before common era contains a list of Pythagorean triples and a statement of the Pythagorean theorem, both in the special case of the isosceles right triangle and in the general case, as does the Apastamba Shulba Sutra (c. 600 BC). Van der Waerden believed that this material "was certainly based on earlier traditions". Carl Boyer states that the Pythagorean theorem in the Śulba-sũtram may have been influenced by ancient Mesopotamian math, but there is no conclusive evidence in favor or opposition of this possibility.
THEN WHY YOU CALLED PYTHAGORAS THEORAM AS GONGU; MENTION IT AS BAUDHAYANA THEORAM
😏😏
Reference from Wikipedia
Wow I couldn't even think.
It's amazing, thank you sir .
Such an interactive and mind boggling question !
If H is the height of the rectangle (given as 8), and L is the sum of the two chords (given as 6); then the general formula for W (Width of rectangle) is W=H-L/2+sqrt (H^2-H*L)
Why u talking about squirtle on math
1:26 Because a 6 always needs a 9
Got the joke🤣🤣 u need both 6 and 9 to make 69 😂LMAO
The Idea is to think lines if you want length.
So Circles ... .: Radiuses
The rest is a box of mathmatical tricks!
Math
Geometry problem...
One time see
th-cam.com/video/j9wY3hso1WQ/w-d-xo.html
Beautiful solution
Math
Geometry problem.
One time see
th-cam.com/video/j9wY3hso1WQ/w-d-xo.html
The second you said rectangle, I knew we were in for trouble
I want someone to love me like Presh Tall Walker likes calling the “Pythagorean Theorem” any name except that
Nice video editing...
Which software is that.
Explanation is very good.
This solution is so amazing, i love it
Beautiful solution.
This is crazy i mean!!! Wow. I love the channel mind your decisions!
Very nice sir
Very interesting problem, with a neat solution.
But I have a question: what do you use for the animations?
i follow
Math
Geometry problem
One time see
th-cam.com/video/j9wY3hso1WQ/w-d-xo.html
U mean which programme? I wonder it too😉
I’d guess it’s after effects, although I have no proof
I guess GeoGebra
so this is where skeppy finds the big words...
Nice to see you
Great video man
And great question too
Yes . 3 4 5 😉😀
4:20
Math
Geometry problem
One time see
th-cam.com/video/j9wY3hso1WQ/w-d-xo.html
As piphagore😆i found it with closed eyes👌😍
Key: Resolve Center to Center distance to Horizontal and Vertical arms.
Wonderful solution sir.
Hint for those scrolling down:
- the diameter of the two circles adds up to 10
sum of radii is 5
Math
Geometry problem
One time see
th-cam.com/video/j9wY3hso1WQ/w-d-xo.html
Also I remind you of my puzzle which i wrote it in my comment in video of the "Largest Cone puzzle " " pentagon inside triangle" thanks for this lecture ... and I wait answer of puzzle . with my best regards .
Math
Geometry problem
One time see
th-cam.com/video/j9wY3hso1WQ/w-d-xo.html
Really great. I appreciate your knowledge about math puzzles. Could you please tell me which animation software are you using for these videos?
obviously powerpoint
*My Brain Before Answer* : You Can Do It Kid!
*My Brain After Answer* : Wait Thats a Square!🤔
😂😂😂😂😂😂😂😂😂
Great video. All the videos in this channel are great . thanks for great videos
A collection of golden questions, like ur idea
By the way, the graphics does not show a rectangle, but a square. the diagonal line of a rectangle is not a symmetry axis. have fun
10 lba=8:lef=6:sw=.1:goto 60
20 r2=(lba-r1*(1+sqr(2)/2))/(1+sqr(2)/2)
30 xs=r1*(sqr(2)/2+1):ys=lba-r1*(sqr(2)/2+1)
40 ym1=lba-r1:ym2=ys-r2/sqr(2):ye=ym1+r1/sqr(2):yf=ym2+r2/sqr(2)
45 dlu1=(ye-yf)/lef:dl=1-dlu1:return
60 r1=sw:gosub 20
61 dl1=dl:ru1=r1:r1=r1+sw:if r1>10*lef then stop
62 ru2=r1:gosub 20:if dl1*dl>0 then 61
70 r1=(ru1+ru2)/2:gosub 20:if dl1*dl>0 then ru1=r1 else ru2=r1
80 if abs(dl)>1E-10 then 70
90 print "r1=";r1;"r2=";r2
www.imagebanana.com/s/1877/7bNnPfHU.html
Thank you for sharing
My take away is that x+y is always the same constant no matter what size the circles are. In other words the radii of two circle add up to a constant.
My english is not enough to understand the terms but my math is strong enough to understand the things he do.
Great problem and impressive solution.
Thanks for this nice problem
If this guy was less attention seeking it would be one of the best channel on TH-cam. Call things with their actual name
Thanks!
4:22 the what?
The Pythagorean theorem does he mean? What was that word.