This may have been the hardest one yet! Not necessarily in terms of actual math work, but definitely in terms of creativity for how to get all of the sides you need. I could sit here for weeks and not come up with the ways you thought of to find the side lengths. This is why this channel is so cool to watch.
Here's a shortcut for the second part of the solution starting at 4:01. Looking at the green triangle, we know that the bottom angle is 60°, and has side lengths a, b and c with c opposite the 60° angle. So, using the law of cosines we have c² = a² + b² -2ab(cos 60°). cos 60° is 1/2, so c² = a² + b² - ab. This can be substitued for c² in the 2nd equation resulting in (√3/4)( a² + b² - ab + ab) = ? or simply (√3/4)( a² + b²) = ? which is the same as the top equation left-hand side. Hence ? = 100.
True, but I personally prefer a pure geometric method whenever possible trig-free, since for me, I'm here mainly to improve my pure geometry skills (which Andy is really helping me with here). Besides, there was no point in his going through drawing the height of the scalene triangle as b and a base of √3 * a / 2, since the right triangle with hypotenuse c was gonna be drawn anyway, we can just use it to find the area of the scalene triangle and add that to the purple triangle
@@haitianGK You don't. The law of cosines can work for any triangle. Technically the Pythagorean Theorem is just the Law of Cosines for the hypotenuse but because cos 90 is 0, the only part of the Law of Cosines we need to take into account is the a^2+b^2 part.
I couldn't solve this the hard way, but the easy way is to notice either a or b are zero then the green triangle collapses and c=a (or b) giving ?=100. If a=b then you have 4 congruent equilateral triangles and ?=100.
The ans was 💯 in 10 secs.... As yellow blue was 100 in total .... And as no more info was given, assume shrinking the yellow one till it disappears..... In such a condition, as yellow collapsed, one side of green is gone too... So, its just one blue remaining with area 100, and red has same side length as blue.
I immediately thought it was a 100, but I can't come up with a neat solution, nice work as always! How exciting I just imagined a scenario where all triangles would have equal area and since it will still satisfy the conditions then if one area is 50 then the sum of two triangles is 100. I reckoned it'd just be the same for all of them
I did something similar - I just copied from my main comment... I did mine more indirectly. It's not a proper proof. I started with the area formulas, but then tossed those for a more brute force approach: * Limit 1: If A = zero, then (1) B = 100, (2) c=b, so C=B, and (3) D isn't really there, so C+D=100. * Limit 2: If B = zero, then (1) A = 100, (2) c=a, so C=A, and (3) D isn't really there, so C+D=100. * Midpoint: If A = B, then A = B = 50, and both C and D are equilateral triangles congruent with A and B, so C=D=50, so C+D=100. * Since the relationship between all three points is (probably) linear (citation???), C+D=100 for every combination of A+B. If anyone wants to help out with that fourth bullet point (the part that seems intuitive but the math teacher in my subconscious is screaming SHOW YOUR WORK!), feel free to help out.
Since the only cue is that triangles A B & C are equilateral you can assume them being equal, and in that scenario D becomes a same size equilateral...so A+B = C+D It's great how you worked out the math !!
That's exactly how I knew the solution is 100 or yellow + blue = purple + green. But knowing and proving are two different things. Andy did a wonderful job at proving it!
That almost feels like an axiom to me: If two equilateral triangles that share only one vertex have parallel sides, then a third equilateral triangle with a base that is perpendicular to a set of parallel sides of the other two triangles will have half the combined area of those triangles. Im sure if I went digging, I'd find a rule or proof that says pretty much that, if not a simpler form of it.
Solution: So we know, that the area of any equilateral triangle is a²√3/4 or a²√(3/16). Let's call the base lengths of the equilateral triangls y(yellow), b(blue) and p(pink) We therefore have the equation: 100 = y²√3/4 + b²√3/4 = √3/4 * (y² + b²) |*4/√3 400/√3 = y² + b² We also have the law of cosines. With it, we can say on the green triangle, that: p² = y² + b² + 2yb*cos(gamma) Since gamma HAS to be 60° (because yellow and blue are on the same line and both have 60° angles, so 180° - 2 * 60° = 60°) and cos(60°) = 1/2, we get: p² = 400/√3 - yb |*√3/4 p²√3/4 = 100 - √3/4 * yb So the area of the pink triangle (p²√3/4) is 100 - √3/4 * yb Now let's focus on the area of the green triangle. Since it shares the same angle at the bottom, it shares it's height with both yellow and blue, depending which baseline we use. Lets use the larger b baseline and therefore have a height of √3y/2 (typical for equilateral triangles). It's area is therefore: 1/2 * √3y/2 * b = √3/4 * yb This is the unknown term from form earlier, so we have everything we need. The area of the pink and green triangle together is: A = 100 - √3/4 * yb + √3/4 * yp A = 100 Or in other words, the yellow and blue triangles together have the same area as the green and pink triangle together.
The diagonals of the two squares form a right triangle whose hypotenuse is the diameter of the circle. If we assume that a and b are the lengths of the sides of the two squares, then (a√2)²+(b√2)²=8², and from this a²+b²=32, so the sum of the areas of the two squares equals 32.
The diameter always subtends an angle of 90 degrees anywhere on the circle. From the diagram, the two angles in between the diagonals of the squares add upto 90 degrees (45 deg + 45 deg) so the chord indeed is the diameter of the circle.
@@Viktor_JohanssonThe diameter of a square divides the angle into two angles measuring 45°, and from there the two diagonals form an angle measuring 45°+45°=90°. The right triangle inscribed by a circle has a hypotenuse that is a diameter of the circle.
@@Viktor_JohanssonThe diameter of a square divides the angle into two angles measuring 45°, and from there the two diagonals form an angle measuring 45°+45°=90°. The right triangle inscribed by a circle has a hypotenuse that is a diameter of the circle.
@@Viktor_JohanssonThe diameter of a square divides the angle into two equal parts, each measuring 45°. From this, the diagonals of the two squares form an angle measuring 45° + 45° = 90°. The right triangle inscribed by a circle has a hypotenuse as a diameter of the circle.
fun one, immediately got 32 because if both squares are the same height, each square is 4 squared and the question implies it's the same all around Then I sat and stared at it for ages until it finally clicked side of yellow square is a and square red square is b the area of the two squares are a^2+b^2 the diagonals of the squares are aroot2 and broot2 if you connect all three corners that touch the circle together you get two diagonals, the diagonal of a square has 45 degrees, so the bottom corner of the triangle has two of them, so 90 degrees, it's a right angled triangle if you have a two chords meeting at a right angle, the line between the other points is the diameter, which in this case we know is 8 so we have a right angled triangle, using pythag, 8^2=(aroot2)^2+(broot2)^2 which simplifies to a^2+b^2=32 and this is what we were trying to find
oh my god brother , I also immediately got 32 after scaling both squares as equal and then took time to solve the problem traditionally by making a right angled triangle . How fugging exciting😂
when you declare the height of triangle b is identical to the height of triangle a, I think you need to prove that the two sides are parallel to each other by stating angle a and angle b are both 60 degree on that straight line.
Everytime we barely have any number, I just try the edge cases. If the green and yellow triangle had area zero, the pink would be equal do the blue, so it would be 100. If the blue and yellow would be equal, the four would be equal equilateral triangles, each with area 50, and the answer would be 100. I know this is no proof, but at least I would know ahead the answer I was looking for.
Answer to the next question: If you connect the upper-right and the lower-left corner of the large square, then the upper-left and the lower-right corner of the small square, you get two diagonals (d & D) that are perpendicular to each other. This means that the distance between the upper-right corner of the large square and the upper-left corner of the small square = the diameter = 2 x 4 = 8 So (d)^2 + (D)^2 = (8)^2 = 64 But the diagonal of a square = (√2)(side length) (diagonal of a square)^2 = (2)(side length)^2 = twice the area. So the combined area = 64/2 = 32
In this one, since nothing indicates a constraint, I made both the yellow and blue triangles the same size. Because they are equilateral, the pink and green triangles can also be equilateral. Therefore, the sum of the areas of the yellow and blue triangles equals the sum of the areas of the pink and green triangles.
Visual proof and how I solved it. It doesn't specify the size of yellow or blue which means that it doesn't matter for the solution. If both yellow and blue have the same area then all the triangles will be congruent equilateral triangles. Therefore they will have the same area.
Yeah, I thought that as immediate answer without needing to calculate anything. If the size difference between blue are yellow isn't needed in the question because red plus green is proportional to blue plus yellow, then if yellow and blue are same size, green must be an identical equilateral triangle too to fit the space inbetween. And as it shares a side with red, and red is equilateral, then red is identical too. It's not a proof, but it makes the calculations feel a bit meaningless.
Next day is 32 square units, if you can draw a diagonal from the corners of the squares that goes through the center of the circle so that; 8^2 = (a+b)^2 + (b-a)^2. Otherwise I dont think you can solve it?
I did it like that too. Also, another right triangle you could use is the diagonal of a and b with the hypotenuse still 2r. Then (sqrt2a)^2 + (sqrt2b)^2 = (2r)^2 2a^2 + 2b^2 = 4r^2 a^2 + b^2 = 2r^2 = 32
I did it another way: - Using the information in the problem and applying the formula for the area of an equilateral triangle, we have: a² + b² = 400/sqrt(3). - Applying the law of cosines to the middle triangle: c² = a² + b² -2ab*cos(60º) => c² +ab = 400/sqrt(3). - Applying the area formula to the middle triangle: A = (1/2).a.b.sen(60º) => A = ab*sqrt(3)/4 - The desired area is: ab*sqrt(3)/4 + c².sqrt(3)/4 = [sqrt(3)/4]*[c²+ab]= [sqrt(3)/4]*[400/sqrt(3)] = 100.
At 2:00 how do you know that the height of the green triangle is equal to the height of the yellow triangle? Just by the fact that they share the same side?
the base of the green triangle, as he is calculating it, is the side of the blue triangle and that intersects the horizontal base at 60 degrees the top of the green triangle touches the side of the yellow triangle which also intersects the horizontal base at 60 degrees this makes the two lines parallel and in an equilateral triangle the height is the same whichever side you call the base
The blue triangle and the yellow has the same orientation and are similar. Since two of the green triangle's sides share a side with them, we know that perpendicular to those sides are the heights of those triangles, and the heights are parallel because of the orientation.
I did mine more indirectly. It's not a proper proof. I started with the area formulas, but then tossed those for a more brute force approach: * Limit 1: If A = zero, then (1) B = 100, (2) c=b, so C=B, and (3) D isn't really there, so C+D=100. * Limit 2: If B = zero, then (1) A = 100, (2) c=a, so C=A, and (3) D isn't really there, so C+D=100. * Midpoint: If A = B, then A = B = 50, and both C and D are equilateral triangles congruent with A and B, so C=D=50, so C+D=100. * Since the relationship between all three points is (probably) linear (citation???), C+D=100 for every combination of A+B. If anyone wants to help out with that fourth bullet point (the part that seems intuitive but the math teacher in my subconscious is screaming SHOW YOUR WORK!), feel free to help out.
Intuition immediately told me the answer but I can’t prove it.. there are 3 equilateral triangles connected like this, forming a fourth upside down triangle in the middle. If you wiggle them around without disconnecting them or changing the equilateral, wouldn’t the sun area for the two bottom ones always be equal to the area of the top and center one? It is that way if you move them all into one large equilateral triangle.
Worked this one out without doing the math because I could tell from how little information we got that the answer had to be true for any size or shape of a triangle, including if all triangles were the same size.
To solve this problem, you can assume: a=0, or b=0, or a=b Which you will get c=b, or c=a, or c=a=b Then you will instantly get green + red =100 What Andy did was to prove in all cases of a and b, green + red will always be the same with blue + yellow. How exciting!
Here you can use the posing of the question to solve it quickly. Since the question was posed, presumably it has an answer. Since no information was given about the relative sizes of the yellow and blue squares, suppose they are the same (it shouldn’t matter, since the problem doesn’t specify). The pink and green triangles are now forced to be the same size as the yellow and blue, so of course they add to 100.
Similarly, suppose the area of the yellow triangle is zero (it is a single point). Then the pink triangle has the same side length as the blue, and the green triangle is also of area zero (it is a line segment). The pink and blue triangles now must each have an area of 100.
I should emphasize that this is not a proof - it relies on the fact that the information given is simultaneously enough to determine a numerical answer, and not enough to determine the relative sizes of the yellow and blue triangles - in other words, the relative sizes cannot impact the answer, so make them something simple and you can solve from there
I noticed that the two equations ended up being equal, and the 100 didn't matter. Can this be generalized? If you have some triangle, and an equilateral triangle on each of its edges, is the area of the original triangle and one of the equilateral equal to the area of the other 2 equalaterals?
I struggled with this one, think I made a simple error somewhere as was getting some horrible equations 😂 Glad it wasn't really straightforward though and it wasn't just that I was missing something really obvious. Loving this series!
Since the heights of blue and yellow are not provided, we cannot assume that the drawing is to scale, so I made an assumption that they were equal. It was immediately obvious after this assumption that the answer was 100. The solution will always be the same, regardless of what the heights are. Easy peasy!
How does we KNOW the height of the green triangle was the same as a side of yellow? Looked to me the side of yellow was slightly longer than the height of the green.
If I assume that the combined area of the two squares are always the same and I would say yellow is zero unitsquare than we have only a red square. Which would have a hypotenuse = 2r. So its 32unit squares?
isn't there some corollary on the Pythagorean theorem that you can do it with other shapes-incl. triangles? does that shorten this problem at all, I wonder?
Well, you havn’t been given any values so could you not just set the hight of the yellow and blue triangles to be equal? If you do that you have 4 identical triangles and the problem would be very easy.
Quick question: how is it a hundred as the radical fraction is still there? I don't know how to explain my question but like A plus b part isn't multiplied by the radical 3 over four so it can't be 100? Can it? I'm missing something 😭
I have no idea who created these questions, but he came up with the word aggvent calendar because Agg is the last name of the person who forwarded these questions to him 😊
This is the first time my solution was faster, and somewhat more elegant than Andy's. For reference I started with the right triangle of hypotenuse c that Andy drew, instead of drawing the base b and height √3 * a / 2
...and enhance. Lol, you're silly. Don't kill yourself getting all these done, man. It's a mistake to try to solve and upload every day; the correct way would be to have these all done in advance, then just upload them later. Content creators typically have weeks, sometimes months, of videos ready to go. That way they don't have to work so hard to get them out on time, as well as have stuff ready to go in case something happens (illness, equipment failure, etc).
We WILL catch up!
The 30/60/90 triangle rule is getting some serious mileage this month
I kinda surprised that in the last two ones he didnt directly used that the the height of the equilateral triangle is equal to 3 times r
I swear if I cant remember n, 2n, and root3n by the end of the month what even was the point!
Ikr
Three "how excitings", truly a Christmas miracle.
This may have been the hardest one yet! Not necessarily in terms of actual math work, but definitely in terms of creativity for how to get all of the sides you need. I could sit here for weeks and not come up with the ways you thought of to find the side lengths. This is why this channel is so cool to watch.
Among many things, I love your articulate way of speaking and math-solving.
Cool resolution with the right triangle to find c
Here's a shortcut for the second part of the solution starting at 4:01. Looking at the green triangle, we know that the bottom angle is 60°, and has side lengths a, b and c with c opposite the 60° angle. So, using the law of cosines we have
c² = a² + b² -2ab(cos 60°).
cos 60° is 1/2, so
c² = a² + b² - ab.
This can be substitued for c² in the 2nd equation resulting in
(√3/4)( a² + b² - ab + ab) = ?
or simply
(√3/4)( a² + b²) = ?
which is the same as the top equation left-hand side. Hence ? = 100.
Good catch there! :D
True, but I personally prefer a pure geometric method whenever possible trig-free, since for me, I'm here mainly to improve my pure geometry skills (which Andy is really helping me with here).
Besides, there was no point in his going through drawing the height of the scalene triangle as b and a base of √3 * a / 2, since the right triangle with hypotenuse c was gonna be drawn anyway, we can just use it to find the area of the scalene triangle and add that to the purple triangle
How did you prove that the green triangle was a right triangle in order to use this?
@@haitianGK You don't. The law of cosines can work for any triangle. Technically the Pythagorean Theorem is just the Law of Cosines for the hypotenuse but because cos 90 is 0, the only part of the Law of Cosines we need to take into account is the a^2+b^2 part.
@@haitianGK It's definitely not a right triangle. See reply from JennyBlaze for a nice answer.
I couldn't solve this the hard way, but the easy way is to notice either a or b are zero then the green triangle collapses and c=a (or b) giving ?=100. If a=b then you have 4 congruent equilateral triangles and ?=100.
The ans was 💯 in 10 secs....
As yellow blue was 100 in total .... And as no more info was given, assume shrinking the yellow one till it disappears.....
In such a condition, as yellow collapsed, one side of green is gone too... So, its just one blue remaining with area 100, and red has same side length as blue.
I immediately thought it was a 100, but I can't come up with a neat solution, nice work as always! How exciting
I just imagined a scenario where all triangles would have equal area and since it will still satisfy the conditions then if one area is 50 then the sum of two triangles is 100. I reckoned it'd just be the same for all of them
I did something similar - I just copied from my main comment...
I did mine more indirectly. It's not a proper proof. I started with the area formulas, but then tossed those for a more brute force approach:
* Limit 1: If A = zero, then (1) B = 100, (2) c=b, so C=B, and (3) D isn't really there, so C+D=100.
* Limit 2: If B = zero, then (1) A = 100, (2) c=a, so C=A, and (3) D isn't really there, so C+D=100.
* Midpoint: If A = B, then A = B = 50, and both C and D are equilateral triangles congruent with A and B, so C=D=50, so C+D=100.
* Since the relationship between all three points is (probably) linear (citation???), C+D=100 for every combination of A+B.
If anyone wants to help out with that fourth bullet point (the part that seems intuitive but the math teacher in my subconscious is screaming SHOW YOUR WORK!), feel free to help out.
This series have been the best holiday gifts.
This challenge was tougher than most of the other ones. How exciting.
Since the only cue is that triangles A B & C are equilateral you can assume them being equal, and in that scenario D becomes a same size equilateral...so A+B = C+D
It's great how you worked out the math !!
That's exactly how I knew the solution is 100 or yellow + blue = purple + green.
But knowing and proving are two different things. Andy did a wonderful job at proving it!
I love how Andy just loves maths and thinks they are beautiful, finally someone that understand me. When everything matches up ❤❤
How exciting indeed. I solved it in a similar way and it put a smile on my face. Thank you and have a wonderful Christmas!😀
How exciting!
That almost feels like an axiom to me:
If two equilateral triangles that share only one vertex have parallel sides, then a third equilateral triangle with a base that is perpendicular to a set of parallel sides of the other two triangles will have half the combined area of those triangles.
Im sure if I went digging, I'd find a rule or proof that says pretty much that, if not a simpler form of it.
Solution:
So we know, that the area of any equilateral triangle is a²√3/4 or a²√(3/16).
Let's call the base lengths of the equilateral triangls y(yellow), b(blue) and p(pink)
We therefore have the equation:
100 = y²√3/4 + b²√3/4 = √3/4 * (y² + b²) |*4/√3
400/√3 = y² + b²
We also have the law of cosines. With it, we can say on the green triangle, that:
p² = y² + b² + 2yb*cos(gamma)
Since gamma HAS to be 60° (because yellow and blue are on the same line and both have 60° angles, so 180° - 2 * 60° = 60°) and cos(60°) = 1/2, we get:
p² = 400/√3 - yb |*√3/4
p²√3/4 = 100 - √3/4 * yb
So the area of the pink triangle (p²√3/4) is 100 - √3/4 * yb
Now let's focus on the area of the green triangle.
Since it shares the same angle at the bottom, it shares it's height with both yellow and blue, depending which baseline we use.
Lets use the larger b baseline and therefore have a height of √3y/2 (typical for equilateral triangles).
It's area is therefore:
1/2 * √3y/2 * b = √3/4 * yb
This is the unknown term from form earlier, so we have everything we need.
The area of the pink and green triangle together is:
A = 100 - √3/4 * yb + √3/4 * yp
A = 100
Or in other words, the yellow and blue triangles together have the same area as the green and pink triangle together.
The diagonals of the two squares form a right triangle whose hypotenuse is the diameter of the circle. If we assume that a and b are the lengths of the sides of the two squares, then (a√2)²+(b√2)²=8², and from this a²+b²=32, so the sum of the areas of the two squares equals 32.
Can we be sure the diagonal is the diameter of the circle and not just a chord-line?
The diameter always subtends an angle of 90 degrees anywhere on the circle. From the diagram, the two angles in between the diagonals of the squares add upto 90 degrees (45 deg + 45 deg) so the chord indeed is the diameter of the circle.
@@Viktor_JohanssonThe diameter of a square divides the angle into two angles measuring 45°, and from there the two diagonals form an angle measuring 45°+45°=90°. The right triangle inscribed by a circle has a hypotenuse that is a diameter of the circle.
@@Viktor_JohanssonThe diameter of a square divides the angle into two angles measuring 45°, and from there the two diagonals form an angle measuring 45°+45°=90°. The right triangle inscribed by a circle has a hypotenuse that is a diameter of the circle.
@@Viktor_JohanssonThe diameter of a square divides the angle into two equal parts, each measuring 45°. From this, the diagonals of the two squares form an angle measuring 45° + 45° = 90°. The right triangle inscribed by a circle has a hypotenuse as a diameter of the circle.
fun one, immediately got 32 because if both squares are the same height, each square is 4 squared and the question implies it's the same all around
Then I sat and stared at it for ages until it finally clicked
side of yellow square is a and square red square is b
the area of the two squares are a^2+b^2
the diagonals of the squares are aroot2 and broot2
if you connect all three corners that touch the circle together you get two diagonals, the diagonal of a square has 45 degrees, so the bottom corner of the triangle has two of them, so 90 degrees, it's a right angled triangle
if you have a two chords meeting at a right angle, the line between the other points is the diameter, which in this case we know is 8
so we have a right angled triangle, using pythag, 8^2=(aroot2)^2+(broot2)^2
which simplifies to a^2+b^2=32 and this is what we were trying to find
oh my god brother , I also immediately got 32 after scaling both squares as equal and then took time to solve the problem traditionally by making a right angled triangle . How fugging exciting😂
when you declare the height of triangle b is identical to the height of triangle a, I think you need to prove that the two sides are parallel to each other by stating angle a and angle b are both 60 degree on that straight line.
Everytime we barely have any number, I just try the edge cases. If the green and yellow triangle had area zero, the pink would be equal do the blue, so it would be 100. If the blue and yellow would be equal, the four would be equal equilateral triangles, each with area 50, and the answer would be 100. I know this is no proof, but at least I would know ahead the answer I was looking for.
How colourful!
Answer to the next question:
If you connect the upper-right and the lower-left corner of the large square, then the upper-left and the lower-right corner of the small square, you get two diagonals (d & D) that are perpendicular to each other. This means that the distance between the upper-right corner of the large square and the upper-left corner of the small square = the diameter = 2 x 4 = 8
So (d)^2 + (D)^2 = (8)^2 = 64
But the diagonal of a square = (√2)(side length)
(diagonal of a square)^2 = (2)(side length)^2 = twice the area.
So the combined area = 64/2 = 32
In this one, since nothing indicates a constraint, I made both the yellow and blue triangles the same size. Because they are equilateral, the pink and green triangles can also be equilateral. Therefore, the sum of the areas of the yellow and blue triangles equals the sum of the areas of the pink and green triangles.
Visual proof and how I solved it.
It doesn't specify the size of yellow or blue which means that it doesn't matter for the solution.
If both yellow and blue have the same area then all the triangles will be congruent equilateral triangles. Therefore they will have the same area.
Yeah, I thought that as immediate answer without needing to calculate anything. If the size difference between blue are yellow isn't needed in the question because red plus green is proportional to blue plus yellow, then if yellow and blue are same size, green must be an identical equilateral triangle too to fit the space inbetween. And as it shares a side with red, and red is equilateral, then red is identical too. It's not a proof, but it makes the calculations feel a bit meaningless.
Next day is 32 square units, if you can draw a diagonal from the corners of the squares that goes through the center of the circle so that; 8^2 = (a+b)^2 + (b-a)^2. Otherwise I dont think you can solve it?
I did it like that too.
Also, another right triangle you could use is the diagonal of a and b with the hypotenuse still 2r.
Then (sqrt2a)^2 + (sqrt2b)^2 = (2r)^2
2a^2 + 2b^2 = 4r^2
a^2 + b^2 = 2r^2 = 32
Very clever and well done.
Y'all now know "how exciting" isn't just a catch-phrase.
I did it another way:
- Using the information in the problem and applying the formula for the area of an equilateral triangle, we have: a² + b² = 400/sqrt(3).
- Applying the law of cosines to the middle triangle: c² = a² + b² -2ab*cos(60º) => c² +ab = 400/sqrt(3).
- Applying the area formula to the middle triangle: A = (1/2).a.b.sen(60º) => A = ab*sqrt(3)/4
- The desired area is: ab*sqrt(3)/4 + c².sqrt(3)/4 = [sqrt(3)/4]*[c²+ab]= [sqrt(3)/4]*[400/sqrt(3)] = 100.
At 2:00 how do you know that the height of the green triangle is equal to the height of the yellow triangle? Just by the fact that they share the same side?
the base of the green triangle, as he is calculating it, is the side of the blue triangle and that intersects the horizontal base at 60 degrees
the top of the green triangle touches the side of the yellow triangle which also intersects the horizontal base at 60 degrees
this makes the two lines parallel
and in an equilateral triangle the height is the same whichever side you call the base
The blue triangle and the yellow has the same orientation and are similar. Since two of the green triangle's sides share a side with them, we know that perpendicular to those sides are the heights of those triangles, and the heights are parallel because of the orientation.
After he rotates the triangles you can see that the height of the green triangle equals the height of the yellow because the yellow is equilateral.
I did mine more indirectly. It's not a proper proof. I started with the area formulas, but then tossed those for a more brute force approach:
* Limit 1: If A = zero, then (1) B = 100, (2) c=b, so C=B, and (3) D isn't really there, so C+D=100.
* Limit 2: If B = zero, then (1) A = 100, (2) c=a, so C=A, and (3) D isn't really there, so C+D=100.
* Midpoint: If A = B, then A = B = 50, and both C and D are equilateral triangles congruent with A and B, so C=D=50, so C+D=100.
* Since the relationship between all three points is (probably) linear (citation???), C+D=100 for every combination of A+B.
If anyone wants to help out with that fourth bullet point (the part that seems intuitive but the math teacher in my subconscious is screaming SHOW YOUR WORK!), feel free to help out.
Intuition immediately told me the answer but I can’t prove it.. there are 3 equilateral triangles connected like this, forming a fourth upside down triangle in the middle. If you wiggle them around without disconnecting them or changing the equilateral, wouldn’t the sun area for the two bottom ones always be equal to the area of the top and center one? It is that way if you move them all into one large equilateral triangle.
Worked this one out without doing the math because I could tell from how little information we got that the answer had to be true for any size or shape of a triangle, including if all triangles were the same size.
To solve this problem, you can assume:
a=0, or b=0, or a=b
Which you will get c=b, or c=a, or c=a=b
Then you will instantly get green + red =100
What Andy did was to prove in all cases of a and b, green + red will always be the same with blue + yellow.
How exciting!
Here you can use the posing of the question to solve it quickly. Since the question was posed, presumably it has an answer. Since no information was given about the relative sizes of the yellow and blue squares, suppose they are the same (it shouldn’t matter, since the problem doesn’t specify). The pink and green triangles are now forced to be the same size as the yellow and blue, so of course they add to 100.
Similarly, suppose the area of the yellow triangle is zero (it is a single point). Then the pink triangle has the same side length as the blue, and the green triangle is also of area zero (it is a line segment). The pink and blue triangles now must each have an area of 100.
I should emphasize that this is not a proof - it relies on the fact that the information given is simultaneously enough to determine a numerical answer, and not enough to determine the relative sizes of the yellow and blue triangles - in other words, the relative sizes cannot impact the answer, so make them something simple and you can solve from there
Again with the obvious answer that takes some impressive math to prove. HOW EXCITING.
This was so interesting, I've learnt something new.
Surprised not to see the sin and cosine rules for area and side length respectively. Nice solution though!
this was so beautiful
I noticed that the two equations ended up being equal, and the 100 didn't matter. Can this be generalized?
If you have some triangle, and an equilateral triangle on each of its edges, is the area of the original triangle and one of the equilateral equal to the area of the other 2 equalaterals?
Bro... that was exhausting, but;
I'm excited
I struggled with this one, think I made a simple error somewhere as was getting some horrible equations 😂
Glad it wasn't really straightforward though and it wasn't just that I was missing something really obvious.
Loving this series!
Lion goes back to go ahead
Since the heights of blue and yellow are not provided, we cannot assume that the drawing is to scale, so I made an assumption that they were equal. It was immediately obvious after this assumption that the answer was 100. The solution will always be the same, regardless of what the heights are. Easy peasy!
I was worried that I overthought it, but my work was the same as yours! So maybe not!
I just supposed the yellow and the blue ones are equal, resulting on the green being 50 and the purple also 50.
How does we KNOW the height of the green triangle was the same as a side of yellow? Looked to me the side of yellow was slightly longer than the height of the green.
area of both square combined is 32
2:13 "Mm" 🙂
Making a right triangle instead of using the rule of cosines was really cool
mind blown.
If I assume that the combined area of the two squares are always the same and I would say yellow is zero unitsquare than we have only a red square.
Which would have a hypotenuse = 2r. So its 32unit squares?
or you can assume two squares of equal size, they would also add up to 32
These are great and we'll catch up. What takes longer? Solving the initial problem or animating the living daylight out of it?
isn't there some corollary on the Pythagorean theorem that you can do it with other shapes-incl. triangles? does that shorten this problem at all, I wonder?
Law of Cosines; c^2 = a^2 + b^2 - 2ab * (Cos C)
@@trelligan42
is that the same as:
√3/4c^2=√3/4a^2+√3/4b^2
?
"Let's get rid of this Pink Triangle" - Weezer fans in 1996
Well, you havn’t been given any values so could you not just set the hight of the yellow and blue triangles to be equal? If you do that you have 4 identical triangles and the problem would be very easy.
It was my bday on the 17th!
Wow, what a trip!
How do we know the height of green is the same as yellow?
Hint 1 for tomorrow's problem: Use the diagonals of the squares
Hint 2 : Whats the angle between the diagonals?
Quick question: how is it a hundred as the radical fraction is still there? I don't know how to explain my question but like
A plus b part isn't multiplied by the radical 3 over four so it can't be 100?
Can it?
I'm missing something 😭
In Andy math we trust
I was expecting some cool pythagorean proof given it's a triangle with similar shapes on all its sides. This was still cool though
Merry Christmas 2024!
All that work just to end up Yellow + Blue = Pink + Green
I tried this and thank god i didn't try any further because golly!
Wasn't this part of a proof you showed us in another video?
Amazing ❤🎉
You need to settle down a little there. You had a premature exclamation: "How exciting!"
🔥🔥🔥🔥
“Oh no. We need to do more work…” 😎
Who comes up with these clever problems? All these problems help explain why students learn Algebra first, then Geometry.
I have no idea who created these questions, but he came up with the word aggvent calendar because Agg is the last name of the person who forwarded these questions to him 😊
catriona agg
4:02 this is where I got stuck😢
day18
(a·sqrt2)^2+(b·sqrt2)^2
=(2R)^2=8^2=64
2(a^2+b^2)=64→a^2+b^2=32😊
other method
let a=0→b=4sqrt2
→a^2+b^2=32😊
Very clever
This is the first time my solution was faster, and somewhat more elegant than Andy's.
For reference I started with the right triangle of hypotenuse c that Andy drew, instead of drawing the base b and height √3 * a / 2
Nice video
Anyone else using Desmos and excel to solve these problems?😢
Day 18: 32 sq units
I always find it weird how Americans say "one fourth" when their sports are broken into 4 QUARTERS. What do you guys think a quarter is, exactly?
Football has four quarters. Baseball divides the game into innings and there are 9 innings except in certain special circumstances.
...and enhance. Lol, you're silly. Don't kill yourself getting all these done, man. It's a mistake to try to solve and upload every day; the correct way would be to have these all done in advance, then just upload them later. Content creators typically have weeks, sometimes months, of videos ready to go. That way they don't have to work so hard to get them out on time, as well as have stuff ready to go in case something happens (illness, equipment failure, etc).
hi
Next problem: 32
First comment
There was an easier way