Suppose the sides of the yellow and blue triangles are a and b, then the sum of their areas is (a²√3)/4+(b²√3)/4=100, and the area of the green triangle is (ab*sin60)/2=(ab√3)/4, and the square of the length of the side of the pink triangle is a²+b²-2abcos60=a²+b²-ab, and the area is (a²+b²-ab)*√3/4, so the sum of the areas of the pink and green triangles is (ab√3)/4+(a²+b²-ab)*√3/4=(a²+b²)*√3)4=100
I love these examples where you isolate a variable, but then don't actually have to solve for the variable to answer the question. I would have solved for x, and then done 3πx^2/2, but your way is so much more elegant. How exciting!
Funny how they are so often integer answers. You would expect the answers to be irrational, some combinations of sqrt(2), sqrt(3) and pi and stuff like that.
the answer is obviously 100 But that is because the two bottom triangles are equal the top two triangles must be equal to them too and the question implies that the answer is always the same. But, doing it properly... the area of an equilateral triangle is length squared times root3 all over 4 yellow length is a, area is A blue length is b, area is B purple length is d, area is D green area is C (a^2*root3)/4+(b^2*root3)/4=100 length d is the hypotenuse of a right angled triangle with the bottom length (a+b)/2 and the other side being height of b-height of a, height of an equilateral triangle is (length*root3)/2 so using pythag... d^2=((a+b)/2)^2+(((b-a)root3)/2)^2 which simplifies to d=root(a^2+b^2-ab) which makes the area D=((a^2+b^2-ab)*root3)/4 D=(a^2*root3)/4+(b^2*root3)/4-(ab*root3)/4 as per the above the first section 100, so D=100-(ab*root3)/4 the area of the green triangle is half of length a times length b times sin lowest angle equilateral triangles have 60 degree angles, the angle is 180-two of these angles, which is 60, so C=(absin60)/2 which goes to C=(ab*root3)/4 which makes C+D=100-(ab*root3)/4+(ab*root3)/4=100
The one at the end, hmm. I can empirically see that it should also be 100. I mean, if yellow = blue, then green and purple are congruent. If yellow = 0 and blue = 100, then green = 0 and purple = 100. No matter how you look at it, it's 100. I tried using law of cosines but that turned into a nightmare. What we learn from that is if the side of the yellow triangle is y and the side of the blue one is b, the side of the purple one is sqrt((y^2 + b^2)/(yb)). Not yet seeing how that's useful in a proof though.
You are close. If p is the side of the purple triangle, p^2+yb = y^2 + b^2. If you express the purple area in function of p, the green area in function of y and b and the blue + yellow area in funtion of y and b you can make a substitution where all the sides cancel out and you get 100
For what we have learned in the past few equilateral triangles, can we assume that: if there's a circle whit radius x inscribed in a equilateral triangle, the height of the triangle will always be 3x?
First, love your content. Second, wondering if I could request a video on a puzzle after the agg-vent series. It is easy to prove that the angle bisectors of the equal angles of an isosceles triangle are equal - but can you prove given that the angle bisectors of a triangle are of equal length, that the triangle is isosceles?
Tomorrow's problem was a bit hard to get the correct method at the beginning but managed to get it Hint: Law of cosines with the lower angle of the green triangle and you need to express the area of the green triangle in function of the side lengths of the blue and yellow triangles. Answer: 100, more generally the red+green area is always the same as the yellow+blue area
Answer to the next question: Sin60° = (√3)/2 If the side lengths of the yellow and blue equilateral triangles are x and y (Sin60°)(x^2 + y^2)/2 = 100 (x^2 + y^2) = 400/(√3) Green area = (Sin60°)(xy)/2 = xy(√3)/4 Side length of the pink equilateral triangle = √[(x/2 + y/2)^2 + ((y/2)√3 - (x/2)√3)^2] = √[(1/4)(x^2 + y^2 + 2xy) + (3/4)(x^2 + y^2 - 2xy)] = √(x^2 + y^2 - xy) = √(400/(√3) - xy) Pink area = (Sin60°)(400/(√3) - xy)/2 = ((√3)/4)(400/(√3) - xy) = 100 - (√3)/4)xy Pink + green = 100
I'm amazed at how many times you've had to "figure" out that the height of an equilateral triangle with an inscribed angle is 3r. It's just funny that this ends up being a part of most of these! Id suggest making a video showing this to be the case, but all of your aggvents would be 30 seconds long :D Also, not calling it liTTle-Arrrr is the real crime here.
Tomorrow thw ans would be 100 I think We can find the sides of the two triangle which would be x and 2x and then we can get x^2 from there and then. When we can get the area of the green and pink in terms of x and then substitute the value of x^2 in there
Hi Andy, I wonder if you find this one fun: th-cam.com/video/5Yaq7O3D9dY/w-d-xo.html The problem says, KN = 9 cm; ABC is an isosceles triangle (AB=BC; AC is the base); BD and CK are medians. They want us to find MD. Thanks!
That jump scared me😂
He dropped out of the frame in a similar way in the previous video.
Hey Vsauce! Andy Math here.
i was thinking the same...lol
Andy is behind schedule: :(
Andy has to upload multiple times to catch up: :)
How exciting.
Ah, the VSauce open. Very clever.
It's all about that tangency.
Need to catch up.. It's the 21st and only at 16th puzzle.. l
Wdym? It's 22nd december
@@CuriousFragdifferent Timezones meaning this guy is in a timezone behind yours 👍
Suppose the sides of the yellow and blue triangles are a and b, then the sum of their areas is (a²√3)/4+(b²√3)/4=100, and the area of the green triangle is (ab*sin60)/2=(ab√3)/4, and the square of the length of the side of the pink triangle is a²+b²-2abcos60=a²+b²-ab, and the area is (a²+b²-ab)*√3/4, so the sum of the areas of the pink and green triangles is (ab√3)/4+(a²+b²-ab)*√3/4=(a²+b²)*√3)4=100
I love these examples where you isolate a variable, but then don't actually have to solve for the variable to answer the question. I would have solved for x, and then done 3πx^2/2, but your way is so much more elegant. How exciting!
When I get bored, I watch Andy's channel.
Funny how they are so often integer answers. You would expect the answers to be irrational, some combinations of sqrt(2), sqrt(3) and pi and stuff like that.
how exciting
0:45 X???? ITS NOT LITTLE R????
It's also a question mark instead of big R
the answer is obviously 100
But that is because the two bottom triangles are equal the top two triangles must be equal to them too and the question implies that the answer is always the same.
But, doing it properly... the area of an equilateral triangle is length squared times root3 all over 4
yellow length is a, area is A
blue length is b, area is B
purple length is d, area is D
green area is C
(a^2*root3)/4+(b^2*root3)/4=100
length d is the hypotenuse of a right angled triangle with the bottom length (a+b)/2 and the other side being height of b-height of a,
height of an equilateral triangle is (length*root3)/2 so using pythag...
d^2=((a+b)/2)^2+(((b-a)root3)/2)^2
which simplifies to
d=root(a^2+b^2-ab)
which makes the area
D=((a^2+b^2-ab)*root3)/4
D=(a^2*root3)/4+(b^2*root3)/4-(ab*root3)/4 as per the above the first section 100, so
D=100-(ab*root3)/4
the area of the green triangle is half of length a times length b times sin lowest angle
equilateral triangles have 60 degree angles, the angle is 180-two of these angles, which is 60, so
C=(absin60)/2
which goes to C=(ab*root3)/4
which makes C+D=100-(ab*root3)/4+(ab*root3)/4=100
The one at the end, hmm. I can empirically see that it should also be 100. I mean, if yellow = blue, then green and purple are congruent. If yellow = 0 and blue = 100, then green = 0 and purple = 100. No matter how you look at it, it's 100. I tried using law of cosines but that turned into a nightmare. What we learn from that is if the side of the yellow triangle is y and the side of the blue one is b, the side of the purple one is sqrt((y^2 + b^2)/(yb)). Not yet seeing how that's useful in a proof though.
You are close. If p is the side of the purple triangle, p^2+yb = y^2 + b^2. If you express the purple area in function of p, the green area in function of y and b and the blue + yellow area in funtion of y and b you can make a substitution where all the sides cancel out and you get 100
For what we have learned in the past few equilateral triangles, can we assume that: if there's a circle whit radius x inscribed in a equilateral triangle, the height of the triangle will always be 3x?
the sum of pink and green area is 100
your math is always prettier than mine lol
When will the ambush of the 30-60-90 triangles stop 😭
First, love your content. Second, wondering if I could request a video on a puzzle after the agg-vent series. It is easy to prove that the angle bisectors of the equal angles of an isosceles triangle are equal - but can you prove given that the angle bisectors of a triangle are of equal length, that the triangle is isosceles?
day17
Iet A=sqrt3/4
1.yellow+blue=A(a^2+b^2)
2.pink+green=A(c^2+ab)
3.c^2=(b·sqrt3/2)^2+(a-b/2)^2
=a^2+b^2-ab
yellow+blue=pink+green=100😊
Tomorrow's problem was a bit hard to get the correct method at the beginning but managed to get it
Hint: Law of cosines with the lower angle of the green triangle and you need to express the area of the green triangle in function of the side lengths of the blue and yellow triangles.
Answer: 100, more generally the red+green area is always the same as the yellow+blue area
Another beautiful solution! But you're falling behind on days?
Great!
really liking the beard action on my man
Answer to the next question:
Sin60° = (√3)/2
If the side lengths of the yellow and blue equilateral triangles are x and y
(Sin60°)(x^2 + y^2)/2 = 100
(x^2 + y^2) = 400/(√3)
Green area = (Sin60°)(xy)/2 = xy(√3)/4
Side length of the pink equilateral triangle = √[(x/2 + y/2)^2 + ((y/2)√3 - (x/2)√3)^2] =
√[(1/4)(x^2 + y^2 + 2xy) + (3/4)(x^2 + y^2 - 2xy)] = √(x^2 + y^2 - xy) = √(400/(√3) - xy)
Pink area = (Sin60°)(400/(√3) - xy)/2 = ((√3)/4)(400/(√3) - xy) = 100 - (√3)/4)xy
Pink + green = 100
How Aggciting!
I'm amazed at how many times you've had to "figure" out that the height of an equilateral triangle with an inscribed angle is 3r.
It's just funny that this ends up being a part of most of these! Id suggest making a video showing this to be the case, but all of your aggvents would be 30 seconds long :D
Also, not calling it liTTle-Arrrr is the real crime here.
He only says "little r" when "big R" also exists
Tomorrow thw ans would be 100 I think
We can find the sides of the two triangle which would be x and 2x and then we can get x^2 from there and then. When we can get the area of the green and pink in terms of x and then substitute the value of x^2 in there
nooo i almost got this but did xsqrt3 for the 30 60 90 cs i keep mixing them up
Bro looks like Quinn from Dexter
yay
Andy Math lore
where he has didn't post for 3 days btween day 14 and 15 because he was using Maths to save the world.
Is anyone with 60√3 or Am I the only one. All the other people are answering 100 so anyone else with 60√3
Can you do zebra puzzles it would be so fun to watch
What is a zebra puzzle?
First one i solved!
36? 9/4 pir2
Next problem: 100
I think tomorrow's answer is 100.
Too similar to one you just did.
36
Hi Andy, I wonder if you find this one fun: th-cam.com/video/5Yaq7O3D9dY/w-d-xo.html
The problem says, KN = 9 cm; ABC is an isosceles triangle (AB=BC; AC is the base); BD and CK are medians. They want us to find MD.
Thanks!