Happy New Year! Thank you everyone for your support! It was a fun challenge to make so many videos in a short period of time. I really wanted to get it all 31 done by December 31! I intend to finish in the next day or 2. Love you all!
There is an assumption going on here that we're dealing with the center points of all the diameters. It's not explicitly stated, but solving this without that assumption would be impossible.
It is not an assumption, if they were not on the centre points you couldn't draw three semi-circles with vertices at the ends of the red lines. Move individual vertices out in your mind and you'll see the circles have to change diameter to accommodate.
In order to complete this problem, the vertices have to be in the center of the diameters and you must assume that since you couldn't complete it without that.
0:29: How can we just assume that the base of the orange traingle (or the blue triangle) is the radius of that semi circle? Since it's not given, the sides might as well be unequal.
@bowiez comments below "in the original Catriona Agg puzzle, the question is: 'The corners of the blue triangle are at the centres of the semicircles. What’s the area of the red triangle?'"
Only 3 left! Go, Andy! Congratulations for your dedication to math communication, and making the learning fun. Here's wishing 2025 brings your subscriber count up to a million.
Andy, I came here to note, as many others have, that we needed to know the given that the vertices of the blue triangle lay on the centers of the semicircles. I love your work and often will try to solve the problems before viewing your solution. Please give the givens in the intro so we're equipped with all we need at the start. Aside from that, thanks for your videos, and have a great new year!
This could have a much more elegant solution by quickly proving that because the height of each triangle is double the blue one then the area is found and so the total area is 7*6
For the first time, the solution to the problem was not as elegant as usual. It was enough to divide the red triangle into seven triangles of equal area, with the blue triangle in the middle, whose area is equal to 6, and the area of the red triangle is equal to 6*7=42.
@@raymondseeger4832You can divide the red triangle into the blue triangle, whose area is 6, and into three other triangles whose base is a diameter of the three circles. Each of these three triangles has a median in it, dividing the triangle into two median triangles in area because they have the same height and equal base. Also note that the area of each of them is 6, and thus in total there are seven triangles, each with an area of 6, which is the red triangle. So the area of the red triangle is 7*6=42.
@@raymondseeger4832You can divide the red triangle into the blue triangle, whose area is 6, and into three other triangles whose base is a diameter of the three circles. Each of these three triangles has a median in it, dividing the triangle into two median triangles in area because they have the same height and equal base. Also note that the area of each of them is 6, and thus in total there are seven triangles, each with an area of 6, which is the red triangle. So the area of the red triangle is 7*6=42.
@bowiez comments below "in the original Catriona Agg puzzle, the question is: 'The corners of the blue triangle are at the centres of the semicircles. What’s the area of the red triangle?'"
21 area=(base*height)/2 split the triangle into two right angled triangles with bases a and b and hypotenuses of c and d respectively left triangle a^2+4^2=c^2 and since you know the diameter of the inscribed circle, a+4-3=c a^2+16=(a+1)^2 which simplifies to a=7.5 right triangle b^2+4^2=d^2 b+4-2=d which combined gets b=3 so base is a+b=10.5 area=(10.5*4)/2=21
Tomorrow's solution -- are we assuming the line marked "4" is tangent to the two circles? If so, to solve I just used some very basic pythagorean algebra to arrive at 21 units squared.
Happy New Year! For the next question: The left-side of the big triangle is divided at the point of tangency into a & 2.5 The right-side of the big triangle is divided at the point of tangency into 3 & b. The base of the big triangle is divided by the two points of tangency into a, 1.5, 1, b So (a + 1.5)^2 + 4^2 = (a + 2.5)^2 2a = 1.5^2 + 4^2 - 2.5^2 = 12 => a = 6 (b + 1)^2 + 4^2 = (b + 3)^2 4b = 1 + 4^2 - 3^2 = 8 => b = 2 The big circle area = (a + b + 2.5)(4)/2 = 21
The large triangle is not a right triangle so make the two small triangles similar. Use the relationship a+b-c=2r in both triangles to get the two lengths that form the base of the triangle whose area is to be calculated.
Happy New Year! Enjoyed the series. I was rooting for you to beat the challenge. Say, got a video suggestion: some sort of graph that charts the various outfits and hairdos/hats used during the series. And - extra points here - hopefully we can work it somehow so something works out to 69 or 420. Haven't had those answers for a while. If so... how exciting!! All the best.
Mark all angles of triangles red and blue accordingly: a-a1, b-b1, c-c1. Connect points a-b1, b-c1, c-a1. You will get 6 triangles with the area 6(easy to find) 6x6=36 plus blue one=42
Let x and y be the parts formed by the height of the triangle with the base, and from this x+4-√(x²+4)=2 and y+4-√(y²+4)=3. By solving these two equations, we get x=3, and y=15/2. Therefore, the area of the triangle is equal to ((3+15/2)*4)/2=21.
i was wondedring if ther was a quicker way to get the fact all triangles wher 12? and so simuler? happy new year, thanks for makign maths fasinating again.
Clearly as drawn each "corner" of each semicircle is set at the midpoint of the adjacent semicircle. Your solution relies on that. However, the problem as stated does not set this as a requirement. If we slide the contact points to the extremes, the area of the two triangles is identical. As we slide the other way so that the three "corners" move toward a single point, the ratio of the large triangle to the small trends toward infinity. And since we don't have a constraint that all of the contact points are at any particular common ratio, any value from 6 to infinity is possible. If we assume that the contact points are the centers of the semicircles, then yes the area of the large triangle is 42.
True, but in the original Catriona Agg puzzle, the question is: "The corners of the blue triangle are at the centres of the semicircles. What’s the area of the red triangle?" I think it was omitted here as it's implied by the drawing.
@AndyMath Do you think there might be any relation to subtended triangles with this? I may be grasping at straws, but with each side being doubled I was sceptical.
There is an easier solution: Connect the centers of the semicircles with the vertices of the red triangle. This will make 6 triangles with an area of 6u². 7 * 6u² = 42 u².
We divide the red triangle into seven triangles of equal area, with the blue triangle in the middle, with an area of 6. Thus, the area of the red triangle is 7*6=42.
Without prior knowledge of the answer, why would you divide it into 7 triangles and not some other amount? If you have 9 triangles it would be a different answer.
Other than the blue triangle, there are three triangles, each of which has a base diameter of the circles, and the median in each triangle divides the base into two triangles of equal area, which is equal to 6, because they have the same height and are equal in base. Thus, in total, there are 7 triangles, each of which has an area of 6. Therefore, the area of the red triangle is equal to 7*6=42.@@square6293
Apparently it was stated in the original problem, but omitted in this video. Apart from that, it's the only way the the problem would "make sense", since it would basically be impossible to solve otherwise.
What if the edge of the semi circles are not at the mid point of the diameters? Then the bases of the triangles would not be equal. We won’t be able to use this method. I must be wrong in my thinking because your knowledge and ability far exceeds mine.
Happy New Year! Thank you everyone for your support! It was a fun challenge to make so many videos in a short period of time. I really wanted to get it all 31 done by December 31! I intend to finish in the next day or 2. Love you all!
Happy new year! I'll happily wait a few days for the last ones! Good luck :-D
You're so close! Id honestly be fine with just notes on paper
Thanks for all the hard work delivering quality videos!
Happy New Year! We'll wait Agg-citingly
Happy new year!
There is an assumption going on here that we're dealing with the center points of all the diameters. It's not explicitly stated, but solving this without that assumption would be impossible.
This, made me so angry it's not given information
I stopped watching at 47 seconds.
It is not an assumption, if they were not on the centre points you couldn't draw three semi-circles with vertices at the ends of the red lines. Move individual vertices out in your mind and you'll see the circles have to change diameter to accommodate.
Thank you. I wondered about this. This stuff is way beyond my skill level but I hope some of it is sinking in.
In order to complete this problem, the vertices have to be in the center of the diameters and you must assume that since you couldn't complete it without that.
0:29: How can we just assume that the base of the orange traingle (or the blue triangle) is the radius of that semi circle?
Since it's not given, the sides might as well be unequal.
Came here to ask that same question
@bowiez comments below "in the original Catriona Agg puzzle, the question is: 'The corners of the blue triangle are at the centres of the semicircles. What’s the area of the red triangle?'"
@@joeldcanfield_spinhead Oh, that solves that, then! Thanks!
@@joeldcanfield_spinhead Ah, fair enough. Thanks for pointing that out.
Also, if we don't assume that, then all the semicircles have no particular use and we have nothing to solve the puzzle.
Only 3 left! Go, Andy! Congratulations for your dedication to math communication, and making the learning fun. Here's wishing 2025 brings your subscriber count up to a million.
Andy, I came here to note, as many others have, that we needed to know the given that the vertices of the blue triangle lay on the centers of the semicircles. I love your work and often will try to solve the problems before viewing your solution. Please give the givens in the intro so we're equipped with all we need at the start.
Aside from that, thanks for your videos, and have a great new year!
I’ve really enjoyed this series. Thank you for creating such great content and making math challenges fun to watch! Happy New Year! 🎊
This could have a much more elegant solution by quickly proving that because the height of each triangle is double the blue one then the area is found and so the total area is 7*6
Happy New Year Andy!
He might do it, HE MIGHT DO IT
He's already failed
not in pacific time. It’s 11:54
@@Doggo2_cool oh wow, I completely forgot how different the times are
@@Doggo2_cool he actually has around 4 hours till the more farther place hits New Years!
@@Doggo2_coolif he travels to Kiribati and films the last two problems he would complete it this year and
Happy New Year to you too Andy! I've said it before, but I really do love your videos. Thanks!
For the first time, the solution to the problem was not as elegant as usual. It was enough to divide the red triangle into seven triangles of equal area, with the blue triangle in the middle, whose area is equal to 6, and the area of the red triangle is equal to 6*7=42.
How would you show that the orange, magenta and yellow triangles are all 2x the area of the blue one?
@@raymondseeger4832You can divide the red triangle into the blue triangle, whose area is 6, and into three other triangles whose base is a diameter of the three circles. Each of these three triangles has a median in it, dividing the triangle into two median triangles in area because they have the same height and equal base. Also note that the area of each of them is 6, and thus in total there are seven triangles, each with an area of 6, which is the red triangle. So the area of the red triangle is 7*6=42.
@@raymondseeger4832You can divide the red triangle into the blue triangle, whose area is 6, and into three other triangles whose base is a diameter of the three circles. Each of these three triangles has a median in it, dividing the triangle into two median triangles in area because they have the same height and equal base. Also note that the area of each of them is 6, and thus in total there are seven triangles, each with an area of 6, which is the red triangle. So the area of the red triangle is 7*6=42.
How do we know the sides of the blue triangle are radii of each of the semicircles?
thought the same thing. is that a fair assumption to make?
Agreed. How can we also assume that the base of orange and the blue triangle are equal to the radius of the semicircle 1? That isn't given either.
@bowiez comments below "in the original Catriona Agg puzzle, the question is: 'The corners of the blue triangle are at the centres of the semicircles. What’s the area of the red triangle?'"
@@phattjabba He left it out yesterday. If this were not the case, the answers (in plural) would be different.
21
area=(base*height)/2
split the triangle into two right angled triangles with bases a and b and hypotenuses of c and d respectively
left triangle a^2+4^2=c^2
and since you know the diameter of the inscribed circle, a+4-3=c
a^2+16=(a+1)^2
which simplifies to a=7.5
right triangle
b^2+4^2=d^2
b+4-2=d
which combined gets b=3
so base is a+b=10.5
area=(10.5*4)/2=21
Tomorrow's solution -- are we assuming the line marked "4" is tangent to the two circles?
If so, to solve I just used some very basic pythagorean algebra to arrive at 21 units squared.
Yes. Also, we can assume that that's the height of the triangle 👀
He still has time! Hawaii will be new year in 4 hrs
Happy New Year Andy, and everybody here, Please God a good one viewing more great Andy Math videos
Happy new year, Mr math! 🎉 ive loved these daily uploads, will miss them!
Happy New Year! For the next question:
The left-side of the big triangle is divided at the point of tangency into a & 2.5
The right-side of the big triangle is divided at the point of tangency into 3 & b.
The base of the big triangle is divided by the two points of tangency into a, 1.5, 1, b
So
(a + 1.5)^2 + 4^2 = (a + 2.5)^2
2a = 1.5^2 + 4^2 - 2.5^2 = 12 => a = 6
(b + 1)^2 + 4^2 = (b + 3)^2
4b = 1 + 4^2 - 3^2 = 8 => b = 2
The big circle area = (a + b + 2.5)(4)/2 = 21
The large triangle is not a right triangle so make the two small triangles similar. Use the relationship a+b-c=2r in both triangles to get the two lengths that form the base of the triangle whose area is to be calculated.
@@AzouzNacir Good catch! I have modified my answer. Thanks and happy 2025!
Happy New Year! Enjoyed the series. I was rooting for you to beat the challenge. Say, got a video suggestion: some sort of graph that charts the various outfits and hairdos/hats used during the series. And - extra points here - hopefully we can work it somehow so something works out to 69 or 420. Haven't had those answers for a while. If so... how exciting!! All the best.
What a gorgeous problem & beautiful solution.
Mark all angles of triangles red and blue accordingly: a-a1, b-b1, c-c1. Connect points a-b1, b-c1, c-a1. You will get 6 triangles with the area 6(easy to find) 6x6=36 plus blue one=42
Still 4 hours of time on Baker Island. Go for it!
So the triangle in this diagram is the universe, huh?
Thank you for such amazing videos.
Andy, you are rad af. thank you for turning math on a daily basis into something fun.
I transformed the yellow triangle into equilateral with no loss of generalization.
The red triangle is 7 copies of the yellow triangle. Neat visual.
Let x and y be the parts formed by the height of the triangle with the base, and from this x+4-√(x²+4)=2 and y+4-√(y²+4)=3. By solving these two equations, we get x=3, and y=15/2. Therefore, the area of the triangle is equal to ((3+15/2)*4)/2=21.
Set (x + y) = base,
x² + 4² = (x - 1.5 + 2.5)² = x² + 2x + 1
y² + 4² = (y - 1 + 3)² = y² + 4y + 4
=> x = 7.5 and y = 3
=> area = 42
i was wondedring if ther was a quicker way to get the fact all triangles wher 12? and so simuler? happy new year, thanks for makign maths fasinating again.
Clearly as drawn each "corner" of each semicircle is set at the midpoint of the adjacent semicircle. Your solution relies on that. However, the problem as stated does not set this as a requirement. If we slide the contact points to the extremes, the area of the two triangles is identical. As we slide the other way so that the three "corners" move toward a single point, the ratio of the large triangle to the small trends toward infinity. And since we don't have a constraint that all of the contact points are at any particular common ratio, any value from 6 to infinity is possible.
If we assume that the contact points are the centers of the semicircles, then yes the area of the large triangle is 42.
True, but in the original Catriona Agg puzzle, the question is: "The corners of the blue triangle are at the centres of the semicircles. What’s the area of the red triangle?" I think it was omitted here as it's implied by the drawing.
@ Ah! Makes sense.
Also, we can safely assume, that the puzzle should make sense, thus it has a single solution.
@AndyMath Do you think there might be any relation to subtended triangles with this? I may be grasping at straws, but with each side being doubled I was sceptical.
I know there's no reason to, but I was hoping to see 3 sig figs (nice) on the answer.
the answer to life, the universe, and everything!
Andy's on a roll, he's rollin'!
There is an easier solution:
Connect the centers of the semicircles with the vertices of the red triangle. This will make 6 triangles with an area of 6u². 7 * 6u² = 42 u².
LERN TO HANDWAVE, rather than repeating the same proof 3x. ;)
HAPPY NEW YEAR!
Day 30: 21 sq units
We divide the red triangle into seven triangles of equal area, with the blue triangle in the middle, with an area of 6. Thus, the area of the red triangle is 7*6=42.
Without prior knowledge of the answer, why would you divide it into 7 triangles and not some other amount? If you have 9 triangles it would be a different answer.
Other than the blue triangle, there are three triangles, each of which has a base diameter of the circles, and the median in each triangle divides the base into two triangles of equal area, which is equal to 6, because they have the same height and are equal in base. Thus, in total, there are 7 triangles, each of which has an area of 6. Therefore, the area of the red triangle is equal to 7*6=42.@@square6293
2:39 how does this work? Why can you move the 2? Like its 2h1 not 2r1h1?
Multiplication is commutative, so r1•2 = 2•r1
Ah, I see! Thanks@@brettappleton3882! Is this also true for when to divide, add or substract?
SO CLOSE!!!! he might do it!!!
Andy, Happy new year!!!
I love the "y" joke 😂
Happy new year Andy!
How do you prove that the ends of the blue triangle are the center of the 3 semi circles?
Apparently it was stated in the original problem, but omitted in this video. Apart from that, it's the only way the the problem would "make sense", since it would basically be impossible to solve otherwise.
what software do you use?
Happy new year andy!!
this one is hard as fuck
Nice speedrun man
he's speedrunning!
How exciting!
happy new year
lappy goo bear! :)
day30
1.(a+5/2)^2-(a+3/2)^2=4^2
→2a+4=16→a=6
2.(b+3)^2-(b+1)^2=4^2
→2(2b+4)=16→b=2
3.A=4(a+3/2+b+1)/2=21😊
A long one today!
your are a wizard
Why andy
42, had to do be 🎉
What if the edge of the semi circles are not at the mid point of the diameters? Then the bases of the triangles would not be equal. We won’t be able to use this method.
I must be wrong in my thinking because your knowledge and ability far exceeds mine.
Please adorn a baseball cap in your channel profile picture to mark this momentous occasion.
32 is tomorrows answer
asnwer=50cm ah ha hmm sam isit math matter isit
asnwer=42cm isit happy new sam
hi
👎 Didn't give all necessary information
Happy new year Andy!