Yes, I also worked like this, this is because the areas are proportional to their bases, because they have the same height. To work more comfortably, what I did was assign 2k to the base of the triangle of 8 and 8k to the one of 32, so that the radius is 5k.
Could reach the z=4x step quicker by comparing the areas of the left and right triangles. Heights are the same so the bases have the same ratio as the areas.
I know a faster way. The two traingles are similar (cuz big triangle is right triangle), so ratio of their sides is 1:2. the big side of the small triangle is equal to the small side of the big triangle, so the ratio of sides within the triangle is 1:2. So let the small side of small triangle be x, now we have the (2x*x/2) = 8, so x= 2*(root(2)). Now the diameter is 5x, so radius is 2.5x = 5root(2). Area => pi * 25
You could also construct another right angled triangle with side lengths 3/2x, y and radius of the circle as the hypotenuse to get your third equation.
The two triangles are similar since they each share two angles with the larger triangle they form. Sharing two angles means sharing all three. Since one's area is 1/4 the other, you can tessellate the larger triangle with 4 of the smaller one to find that the smaller triangle's height is 2 times its base. From there, you're done already as it's base has to be sqrt(8) to satisfy its area. Trivial from there.
This whole series of Aggvent calendar is helping me in solving any problem in geometry of any competitive exams for High Schoolers.Thanks,Keep it up 😊😊😊
My solution: Once you know that the 8 and 32 triangles are similar, you can conclude that the scale factor between them must be 2, since the areas are scaled by a factor of 4. This means the hypotenuse of the 32 triangle must be twice the hypotenuse of the 8. Since those hypotenuses are the legs of the big triangle whose hypotenuse is the diameter of the semicircle (whose area is 40), we can assign its short side a variable of x, its long side 2x and create the two equations: (1/2)(x)(2x) = 40 (x)^2 + (2x)^2 = (2r)^2 Then just solve for the radius.
I used a slightly different method. We know these are similar triangles, and one's area is 4x the other, so the lengths are 2x the other. So if the vertical line is h, the area of the big triangle is rh = 40. And the base of the whole thing is 2r = h/2 + 2h = 5h/2 (because the lengths are in 2:1 ratio). h = 40/r = 4r/5. Cross multiply: 4r^2 = 200, r^2 = 50. The area of the semicircle is (pi)(r^2)/2, so we get 25pi.
Well, I thought I'd see if I could derive the general expression for the radius of the semicircle in terms of the two areas given (call them A = 8 and B = 32). Didn't end up being that pretty: R = SQRT(2)/2 * [(1+B/A)] * [A^0.75 / B^0.25]. This tells you the radius will usually be some real multiple of SQRT(2); R = 5 *SQRT(2) in this example. It also tells you that you'd better choose the two areas carefully if you want a "clean" answer. You're probably more likely to avoid computational errors by going through the steps as shown by Andy.
Your solution was definitely cooler than mine, I created a system of equations with the third equation just 8+32=(Both of the area formulas added together), solving for y with respect to x, plugging that into the z equation to find out the ratio between x and z, and solving for x which just turns out to be 1. How exciting!
You are long-winded determining the dimensions. Once you recognize the 2 triangles are similar, you know relative scaling is square root of the area ratio, 1:4 or 1:2. You can assign base of 8 triangle as X, height of same 2X, & base of 32 triangle 4X. 8 = (1/2)(X)(2X). X² = 8 X = 2√2. Diameter = 2√2 + 8√2 = 10√2. Radius = 5√2. Radius² = 50. Area semicircle = (1/2)(50)pi = 25pi.
Once we've determined that triangles are similar, then we can use that (Base of smaller triangle / Base of larger triangle) = (Height of smaller triangle / Height of larger triangle) = √(Area of smaller triangle / Area of larger triangle) x/y = y/z = √(8/32) = √(1/4) = 1/2 y = 2x z = 2y = 4x Now we use area of smaller triangle to find x. Then we can find z. 1/2 * x * y = 8 1/2 * x * 2x = 8 x² = 8 x = 2√2 → z = 4x = 8√2 Now we can find radius and area of semicircle. r = (x+z)/2 = 5√2 A = πr²/2 = π(5√2)²/2 = π(50)/2 *A = 25π*
Strange, because how I started is noticing that these are 30-60-90 triangles, by completing the circle by mirroring everything on the diameter. This way the bigger triangle became an equilateral triangle. And since we know they're 30-60-90 triangle, that means y = x√3 and z = y√3 = 3x, but that's not what you get. Weird. Maybe I'm overlooked something.
@keith6706 Oh. I think I let myself fooled by the illustration. I initially thought it has to be a 30-60-90 triangle, but my argument doesn't really proves that. I was too biased. But since those triangles are similar and their areas ratio is 1:4, then their sides should be 1:2, so y=2x and z=2y=4x. Makes more sense I guess.
@@Epyxoid A quick way of seeing that is that areas are whole numbers. Looking at the smaller triangle, if the height was x√3, that means the base would also have to be something multiplied by √3 in order to get an area of 8. But if the base was a multiple of √3, then the height would end up being a whole number, which means the area would have to have a √3 in it. Therefore, it can't be a 30-60-90.
My solution used the fact that all three triangles are similar to rewrite the formula for the area of the largest triangle as 1/2 * d/sqrt(5) * 2d/sqrt(5) = 40 and solve for d
Answer to the puzzle for day 14: If the radius of each small circle is r, the radius of the large semicircle is 3r. Area of the semicircle = 9π(r^2)/2 Area of the three small circles = 3π(r^2) Fraction of the yellow area = (9π/2 - 3π)/(9π/2) = (3/2)/(9/2) = 1/3
@ You are looking at half of a large circle with six small circles inside. If you connect the six small circle centers you get a hexagon whose side length is 2r (r = radius of each small circle) and whose center is the center of the large circle. If you connect the center of a hexagon to two adjacent corners you get an equilateral triangle. The distance from each corner of the hexagon to the center of the hexagon = 2r = the distance from each small circle center to the large circle center. Now you extend the line between the centers of the large and a small circle further to the circumference of the large circle, you get 2r + r, which is the large circle radius.
I don't know if I liked your approach. I think it would be more didactic to call it h instead of y (height). Also, it would be interesting to start with the fact that z = 4x, since the triangles have the same height. Making the problem purely algebraic makes it difficult to understand what is happening and whether the values obtained make sense.
My answer for tomorrow's aggvent question was (just gonna hide it under the Read More thing so i dont spoil the answer for anyone) 1/3rd of the total area. I noticed that the circles made half of a hexagon, and i split it into rectangles and triangles to find the height (radius) of the semicircle which was 3 times the radius of the smaller circles. Then i found the area of the smaller circles, multiplied it by 3, then divided it by the area of the semicircle to get 2/3. The remainder was the yellow area, so 1/3
That's a terribly tortured way to do something that is something VERY simple! There are 3 tringles that are all the same in all but area. So Sq.Root the 8 and the 40 to find the angle that is about 26.565 degrees. From that angle and the larger triangles 40 units in area again find the Circumradius which is also simple trigonometry and you get 7.07... which is the Sq.Root of 50... half the Circle === 25pi. 1/3rd the time.
Didn't remember the right corner but, but following that: Z/y=y/x X=y2/z X=16/y Y3/16=z Y4/32=32 Y4=2^10 Y=4{2} Z=8{2} X=2{2} A=pi(x+z)^2/8=200pi/8=25pi
Solution: A = ((p + q)/2)² * π * 1/2 with p being the base of the small triangle and q the base of the big one Because the height is the same, q = 4p A = (5p/2)² * π * 1/2 A = 25p²/4 * π * 1/2 A = 25p²/8 * π A = 25π/8 * p² With Euclid's geometric mean theorem we know that h² = pq h² = p(4p) h² = 4p² h = 2p The area of the small triangle is therefore: 8 = p/2 * 2p 8 = p² The target area is therefore A = 25π/8 * p² A = 25π/8 * 8 A = 25π
I'm a computer engineering student that deals with math far more complex than this, but I still watch every single one of your videos because I find them entertaining!
It seems like you went through a lot of extra work to get z=4x. Using the area formula for a triangle, we can find that from the start. Since they share a side, the unknowns x and z must be a 1:4 ratio.
A premature "how exciting..."
He was probably nervous.
😂
He was excited. ;)
It doesn't usually happen this fast, I swear!
I think this is the first time I’ve seen your hair. Threw me off for a good two minutes
It's literally in his channel thumbnail...
Andy just woke up to answer math aggvert calendar for us. What a dedication
And without his hat
It's always so satisfying to see numbers cancel each other out
he is definitely not tired?
He even said his catch phrase and almost forgot he wasn’t done yet. Bro is exhausted. His hat is clearly the source of his power
@@DrFunkman nah I think he forgot to drink coffee in the evening or something
Put a box around it
@@DrFunkman Bros nerfed without his hat
base1=y
base2=4y
diameter=5y
Intersecting chords theorem
h•h=y•4y
h²=4y²
h=2y
8=(1/2)hy
hy=16
2y•y=16
y² =4/√2
y=4√2/2
y=2√2
diameter=5y
=5(2√2)
=10√2
radius= 5√2
Area(sc)= (1/2)(5√2)² π
Area=(1/2)(25)(2)π
Area=25π
Yes, I also worked like this, this is because the areas are proportional to their bases, because they have the same height.
To work more comfortably, what I did was assign 2k to the base of the triangle of 8 and 8k to the one of 32, so that the radius is 5k.
OMG, hair! So handsome.
The fakeout at 4:12!! 😅 Love it!
Hats off to this solution. Like, your hat, too.
Could reach the z=4x step quicker by comparing the areas of the left and right triangles. Heights are the same so the bases have the same ratio as the areas.
I know a faster way. The two traingles are similar (cuz big triangle is right triangle), so ratio of their sides is 1:2. the big side of the small triangle is equal to the small side of the big triangle, so the ratio of sides within the triangle is 1:2. So let the small side of small triangle be x, now we have the (2x*x/2) = 8, so x= 2*(root(2)). Now the diameter is 5x, so radius is 2.5x = 5root(2). Area => pi * 25
You could also construct another right angled triangle with side lengths 3/2x, y and radius of the circle as the hypotenuse to get your third equation.
I love your hair without the hat!
How exciting.
Now that's proper mathematician hair. 💪
The two triangles are similar since they each share two angles with the larger triangle they form. Sharing two angles means sharing all three.
Since one's area is 1/4 the other, you can tessellate the larger triangle with 4 of the smaller one to find that the smaller triangle's height is 2 times its base.
From there, you're done already as it's base has to be sqrt(8) to satisfy its area. Trivial from there.
"this seems important, let's put a box around it."
For tomorrow's problem remember that tangent lines to a circle are perpendicular to the radius and mirroring the other half circle will help
Premature excitation, bud!
☠️
Rough night for Andy (party on, dude!).
This whole series of Aggvent calendar is helping me in solving any problem in geometry of any competitive exams for High Schoolers.Thanks,Keep it up 😊😊😊
My solution:
Once you know that the 8 and 32 triangles are similar, you can conclude that the scale factor between them must be 2, since the areas are scaled by a factor of 4. This means the hypotenuse of the 32 triangle must be twice the hypotenuse of the 8. Since those hypotenuses are the legs of the big triangle whose hypotenuse is the diameter of the semicircle (whose area is 40), we can assign its short side a variable of x, its long side 2x and create the two equations:
(1/2)(x)(2x) = 40
(x)^2 + (2x)^2 = (2r)^2
Then just solve for the radius.
Wow in the end there was some hair below those caps!
I used a slightly different method. We know these are similar triangles, and one's area is 4x the other, so the lengths are 2x the other. So if the vertical line is h, the area of the big triangle is rh = 40. And the base of the whole thing is 2r = h/2 + 2h = 5h/2 (because the lengths are in 2:1 ratio). h = 40/r = 4r/5. Cross multiply: 4r^2 = 200, r^2 = 50. The area of the semicircle is (pi)(r^2)/2, so we get 25pi.
Well, I thought I'd see if I could derive the general expression for the radius of the semicircle in terms of the two areas given (call them A = 8 and B = 32). Didn't end up being that pretty:
R = SQRT(2)/2 * [(1+B/A)] * [A^0.75 / B^0.25]. This tells you the radius will usually be some real multiple of SQRT(2); R = 5 *SQRT(2) in this example. It also tells you that you'd better choose the two areas carefully if you want a "clean" answer. You're probably more likely to avoid computational errors by going through the steps as shown by Andy.
Your solution was definitely cooler than mine, I created a system of equations with the third equation just 8+32=(Both of the area formulas added together), solving for y with respect to x, plugging that into the z equation to find out the ratio between x and z, and solving for x which just turns out to be 1. How exciting!
You are long-winded determining the dimensions. Once you recognize the 2 triangles are similar, you know relative scaling is square root of the area ratio, 1:4 or 1:2. You can assign base of 8 triangle as X, height of same 2X, & base of 32 triangle 4X.
8 = (1/2)(X)(2X). X² = 8 X = 2√2. Diameter = 2√2 + 8√2 = 10√2. Radius = 5√2.
Radius² = 50. Area semicircle = (1/2)(50)pi = 25pi.
I used the intersecting chords method to get y² = xz rather than similar triangles.
*How exciting, no cap*
Once we've determined that triangles are similar, then we can use that (Base of smaller triangle / Base of larger triangle) = (Height of smaller triangle / Height of larger triangle) = √(Area of smaller triangle / Area of larger triangle)
x/y = y/z = √(8/32) = √(1/4) = 1/2
y = 2x
z = 2y = 4x
Now we use area of smaller triangle to find x. Then we can find z.
1/2 * x * y = 8
1/2 * x * 2x = 8
x² = 8
x = 2√2 → z = 4x = 8√2
Now we can find radius and area of semicircle.
r = (x+z)/2 = 5√2
A = πr²/2 = π(5√2)²/2 = π(50)/2
*A = 25π*
1/3 of the semicircle is not covered by the smaller circles
is your room a hexagon of a pentagon
I like your hair!
Strange, because how I started is noticing that these are 30-60-90 triangles, by completing the circle by mirroring everything on the diameter. This way the bigger triangle became an equilateral triangle. And since we know they're 30-60-90 triangle, that means y = x√3 and z = y√3 = 3x, but that's not what you get. Weird. Maybe I'm overlooked something.
That's because they're not 30-60-90s triangles. The angles are 26.565 - 63.435 - 90.
@keith6706 Oh. I think I let myself fooled by the illustration. I initially thought it has to be a 30-60-90 triangle, but my argument doesn't really proves that. I was too biased. But since those triangles are similar and their areas ratio is 1:4, then their sides should be 1:2, so y=2x and z=2y=4x. Makes more sense I guess.
@@Epyxoid A quick way of seeing that is that areas are whole numbers. Looking at the smaller triangle, if the height was x√3, that means the base would also have to be something multiplied by √3 in order to get an area of 8. But if the base was a multiple of √3, then the height would end up being a whole number, which means the area would have to have a √3 in it. Therefore, it can't be a 30-60-90.
@@keith6706 Nice! Good point.
My solution used the fact that all three triangles are similar to rewrite the formula for the area of the largest triangle as 1/2 * d/sqrt(5) * 2d/sqrt(5) = 40 and solve for d
_"Too much hair!"_ -- River Tam
"how exciting"🗣🗣🗣🔥🔥🔥🔥
Day 13? Why is this not day fifteen?
He is american it is 15 in europe but they just started 14th of december
@yaqp20
No. Just no.
They are at the most nine hours behind. 🙄
@@Peter_Riis_DK idk ther is some time for editing, uploading and stuff i guess
@@yaqp20
Thanks for your interest, but why answer when you don't know?
In a way, I think you proved the intersecting chords theorem at the start (which is what I used).
Ok, I want to know what ratio of the background is occupied by your hair this morning. 😁
day14
Symmentry→regular hexagon
→6 equilateral triangle
R-r=2r→R=3r
(丌R^2/2):[(丌R^2/2)-3丌r^2]
=1:3=1/3😊
Andy … can you identify any fractals in your haircut? I need it for personal stability as I can’t handle chaos ;) wow
Answer to the puzzle for day 14:
If the radius of each small circle is r, the radius of the large semicircle is 3r.
Area of the semicircle = 9π(r^2)/2
Area of the three small circles = 3π(r^2)
Fraction of the yellow area = (9π/2 - 3π)/(9π/2) = (3/2)/(9/2) = 1/3
But how did you come to the conclusion that R=3r?
@ You are looking at half of a large circle with six small circles inside. If you connect the six small circle centers you get a hexagon whose side length is 2r (r = radius of each small circle) and whose center is the center of the large circle. If you connect the center of a hexagon to two adjacent corners you get an equilateral triangle. The distance from each corner of the hexagon to the center of the hexagon = 2r = the distance from each small circle center to the large circle center. Now you extend the line between the centers of the large and a small circle further to the circumference of the large circle, you get 2r + r, which is the large circle radius.
1 - 6sin^2(18 degrees) ? or roughly 0.427?
What?
@ nvm im an idiot its 1/3
I don't know if I liked your approach. I think it would be more didactic to call it h instead of y (height). Also, it would be interesting to start with the fact that z = 4x, since the triangles have the same height. Making the problem purely algebraic makes it difficult to understand what is happening and whether the values obtained make sense.
My answer for tomorrow's aggvent question was (just gonna hide it under the Read More thing so i dont spoil the answer for anyone)
1/3rd of the total area. I noticed that the circles made half of a hexagon, and i split it into rectangles and triangles to find the height (radius) of the semicircle which was 3 times the radius of the smaller circles. Then i found the area of the smaller circles, multiplied it by 3, then divided it by the area of the semicircle to get 2/3. The remainder was the yellow area, so 1/3
No idea what happened there... But okay.
That's a terribly tortured way to do something that is something VERY simple! There are 3 tringles that are all the same in all but area.
So Sq.Root the 8 and the 40 to find the angle that is about 26.565 degrees. From that angle and the larger triangles 40 units in area again find the Circumradius which is also simple trigonometry and you get 7.07... which is the Sq.Root of 50... half the Circle === 25pi.
1/3rd the time.
This is art 🖼️
I got it but I used a complicated method haha😅
I did this in my head off the thumbnail and I got it right! Such a cool problem.
Are you asking "Y" if it makes sense to it? Or asking us "why it makes sense"
I'm guessing the answer is... "yes"
Didn't remember the right corner but, but following that:
Z/y=y/x
X=y2/z
X=16/y
Y3/16=z
Y4/32=32
Y4=2^10
Y=4{2}
Z=8{2}
X=2{2}
A=pi(x+z)^2/8=200pi/8=25pi
Why not just aply Pythagoras to get the diameter ?
By the way Andy, I think it’s Catriona not Catrina 😭
How exciting
grande!
Does that make sense y?
Solution:
A = ((p + q)/2)² * π * 1/2
with p being the base of the small triangle and q the base of the big one
Because the height is the same, q = 4p
A = (5p/2)² * π * 1/2
A = 25p²/4 * π * 1/2
A = 25p²/8 * π
A = 25π/8 * p²
With Euclid's geometric mean theorem we know that
h² = pq
h² = p(4p)
h² = 4p²
h = 2p
The area of the small triangle is therefore:
8 = p/2 * 2p
8 = p²
The target area is therefore
A = 25π/8 * p²
A = 25π/8 * 8
A = 25π
Bad hair day? How exciting.
🎉🎉🎉😮😮😮
This one was fun, tomorrow's one is a bit too easy. Spoiler below:
1/3
I've had a different method, same answer (I was alsi using geometry not trigonometry). Too bad the comment section is not suitable to share it😢
he said how exciting three times wtf
Im early
looks like so was andy
Yes i was correct yesterday
Couldn’t we just do ((x+z)*y)/2=40
too much boxes
You seem to be running a little late.
Next problem: 1/3
I'm a computer engineering student that deals with math far more complex than this, but I still watch every single one of your videos because I find them entertaining!
It seems like you went through a lot of extra work to get z=4x. Using the area formula for a triangle, we can find that from the start. Since they share a side, the unknowns x and z must be a 1:4 ratio.