Andy, I think I speak for all of your subscribers when I say that you absolutely crushed it for the last 31 videos. It was a delight watching each and every one of them, and I learned a few things about geometry in the process. Thank you so much for putting all of these videos together, this channel is one of the bright spots of TH-cam for me.
I think there is a faster solution for the "h". 1:44 right upper corner gives us 2 tangent lines to the right circle, which are "h-1". So we can apply the Pythagorean theorem. Rectangle's base is "6", it's hight is "h" and it's diagonal is "5+h-1". 6^2 + h^2 = (5+h-1)^2 h = 2.5
That's exactly what I did! I was surprised when Andy did the intersecting tangents thing for the bottom left corner... but then didn't do the exact same for the top right!
Exactly what I did. Just remember this theorem: If you have a circle with 2 tangent lines intersecting at a point, the distance from that point to the two points of tangency are equal. By using this fact twice on the problem, the answer can be reached very quickly. Congrats to Andy, though, for perservering through these 31 interesting geometry puzzles, and creating the wonderful animated illustrations. What a great math communicator!
Great Series. This problem is easier to solve if you use the three tangents of the circle in the right angle triangle and the “tangents from an external point” theorem to calculate h.
Very nicely done! I think your animations are underappreciated. I can relate to the difficulty in doing them! But it's worth the effort to make these solutions so understandable. Happy New Year, finally!
As Andy pointed out, we didn’t need to evaluate the value of x in the “little” triangle. But we don’t need to evaluate the value of y either (or even explicitly evaluate the value of h) in order to determine the area of the rectangle. We can just apply the appropriate scale factor (a couple of times). At 3:05, we established that the little triangle is 1/25 the area of the “larger” triangle. That larger triangle consists of the little triangle plus the two congruent triangles (each of area 5/2) previously drawn. Those two triangles hence comprise 24/25 the area of the larger triangle, which means the larger triangle has area 5 * 25/24 = 125/24. Now to get the area of the “whole” triangle, we simply multiply by the square of the linear scale factor going from 5 to 6, or 36/25. Then the whole triangle has area 125/24 * 36/25 = 7.5. Finally, the area of the rectangle is twice that, or 15. So you don’t need to figure out x, y or h to get that (instead of area rectangle = 6*h, we can use area rectangle = twice the area of the triangle formed by the diagonal of the rectangle).
Considering that the 12 days of Xmas extend through January 5th., I say you technically came in on time, theologically speaking. :) Congratulations, beautiful work.
That was a really slick finish on the last one. Great to have watched you through the whole thing. Hope you're having a great New Year, Andy. Looking forward to solving a lot more problems with you.
Thank you Andy for an exciting Andy Aggvent Month. Reading the comments became a thing too, you created a nice community. Your solutions with wonderful animations will be appreciated and live forever in cyber space for the next generations to enjoy. Happy new year… you can take a vacation now, you earned it!
For the record, this one can be solved with the same method from Day 30 where we we also had a right triangle with a known side and a known diameter of its inscribed circle. Definitely appreciate Andy mixing things up though.
At 1:40, we have two small right triangles with sides 1 and 5. Let the angle opposite the short side be Θ. We observe that tan(Θ) = 1/5. From our notes on the 5-12-13 Pythagorean triple right triangle, if tan(2Θ) = 5/12, then tan(Θ) = 1/5. The triangle above the small triangle has angle 2Θ opposite the short side h and the long side is 6. So, tan(2Θ) = h/6 but tan(2Θ) = 5/12, therefore h/6 = 5/12 and h = 2.5. Area = 6h = 6(2.5) = 15 u², as Andy Math also found.
Awesome solution! I did it by focusing on the small right triangle at the upper right. The lengths of its legs are 2 and h/3, and with the help of the equal tangents theorem (used twice), you can see that the hypotenuse is 5h/3 - 2. Now you solve for h using Pythagorean theorem: 2² + (h/3)² = (5h/3 - 2)². Some nice cancellations lead to h = 2.5, and so Area = 15.
2 วันที่ผ่านมา +2
Hello, Andy. I saw you solving all these "fun ones" problems, from 1 to 31. And, in fact, it was " How exciting". Thanks for doing it and for sharing. I can't wait for December 2025, so I can follow your solving for new 31 "fun ones" problems...Happy New Year to you.
Congrats on finishing your challenge, even with slight delay. How exciting! However I was hoping to see new problem. I love this format, where solution isn’t available. End up solving whole month by myself, and even got friends and family hooked as well :)
A simpler way to think about it is that the right angled triangle surrounding the blue circle forms 3 pairs of tangents. Both tangents in each pair have the same length. So the base is 5 + 1, the height is 1 + x (unknown) and the hypotenuse is 5 + x (unknown) (x+1)² + 6² = (x+5)² x² + 2x + 1 + 36 = x² + 10x + 25 8x = 12 x = 1.5 Area = base * height = 6(1+x) = 6 * 2.5 = 15
Happy new Year. Finally, the spherical object has changed its vertical position. ("The ball has dropped.") Wish you the best for the new year. I'm looking forward to fun and exciting videos.
This is waaaaay too complicated. Use the equal tangent distances from lower left/upper right vertices of the rectangle with the rightmost (third) circle. The left tangents to circle 3 have length 6 - 1 = 5, and the right tangents to circle 3 have length x. Then one gets 6² + (x + 1)² = (5 + x)² 36 + x² + 2x + 1 = 25 + 10x + x² 37 + 2x = 25 + 10x 12 = 8x x = 12/8 = 3/2 Therefore the height h = 1 + 3/2 = 5/2, and the area is 6·5/2 = 15.
woohoo! time for a well deserved break! I think it took you more time to edit than to actually solve it all ^^'. I really enjoyed all the problems in the calender, takes some outside the box thinking and lots of triangles and radiuses.
To solve, I used the Two Tangents Theorem, which states that if two tangents are drawn from the same external point to a circle, then the tangents are equal in length. Using the rightmost circle and the lower triangle in which it is inscribed, we have the following 3 pairs of tangents: Tangents from lower right corner have length = 1 Tangents from upper right corner have length = h − 1 (where h is height of triangle/rectangle) Tangents from lower left corner have length = 6 − 1 = 5 (where 6 is base of triangle/rectangle) Triangle has base = 6, height = h, and hypotenuse = (h−1) + 5 = h + 4. Using Pythagorean theorem, we get (h + 4)² = h² + 6² h² + 8h + 16 = h² + 36 8h = 20 h = 5/2 *Area of rectangle = 6 * 5/2 = 15*
You could also use the fact that the rightmost circle is the incircle of a right triangle, and you know a side that isn't the hypotenuse of that triangle, and that's already enough information to solve.
The way i did it was i applied the HL theorem to both "corners"(i guess) of the triangle. So the base of the triangle is 6, the height is x+1, and the hypotenuse is x+5. And with the pythagorean itll become 6² + (x+1)² = (x+5)². And we find x to be equal 3/2 and theredore the height in total will be 5/2.. dunno just thought this was a neater solution imo
Agreed. I was completely following until about 1:40 into the video. Then, I was thinking why is he making this so complicated with drawing other triangles and introducing two variables.
As a recent problem, we can use "the formula for the inradius of a right triangle." If we take the triangle formed by half of the rectangle, the full circle follows the formula... So, in this triangle A,B,C we are looking for A to solve the rectangle surface. We have B=6. The formula gives us the difference between the hypotenuse and a: ((a+6−c)/2)=1 Thus, c=a+4. Now we can use the Pythagorean equation: a²+6²=(a+4)² This simplifies to: a²+36=(a²+8a+16) Which gives us: a=2.5 That’s what we’re looking for. 2.5×6=15 It’s longer to describe than to actually do it! How exciting! 😊
The green circle is inscribed in a right triangle with legs 6 and h, so a formula for the radius of the circle is r = [6 + h - sqrt(36 + h^2)]/2. Because r = 1, you have an equation in the unknown height h. It is easy to solve for h. Simple.
You made this harder then necessary. Tangents to circle from common point are equal. From lower left corner are 2 tangents to right-hand circle. Lower one is 5, slanted one is also 5. Let remainder of slanted line be X long. Diagonal of rectangle is 5 + X. Tangents from lower right corner to right circle are both 1. Tangents from upper right corner are both X. Right triangle: bottom leg = 6, vertical leg = 1 + X, hypotenuse = 5 + X. X² + 10X + 25 = X² + 2X + 1 + 36. X² terms cancel. 8X = 12, X = 3/2. Area = (6)(1 + 3/2) = 30/2 = 15.
Thanks Andy, for making December enjoyable again. It’s obviously been a big undertaking for you so hope you can get some time away from YT for a bit now. Cheers.
To simplify the solution, the area of the rectangle is: A=(W)*(L)=(6)*(1+X). As the sides of the triangle with a right angle are tangent line of the circle, using the Pythagorean theorem: 6^2+(1+X)^2=(5+X)^2 36+1+2X+X^2=25+10X+X^2 8X=12, X=3/2 Plugin the results of X to the area: A= 6*(1+3/2)=15
Could've just used the inradius of a right triangle formula for that one, or even derived it, it'd be done in half the time or so. 1 = (6 + h - (h² + 36)^0.5)/2 It'll take some time but still get there fast. Anyways, awesome work with the aggvent calendar Andy, you were awesome (and still are of course) I get intrigued and excited by many of your videos, keep up the great work and keep being awesome
I did it differently. Since h equals 1 + x and the diagonal equals 5 + x (x are top right tangents of the blue circle) I just did 6^2 + (x + 1)^2 = (x + 5)^2 (Pythagorean theorem) and solved for x=1.5. Area width x length =2.5 x 6 = 15
In fact, you don't need either x or y. Just apply Pythagoras using the principle of secant lines: (5+(h-1))² = h² + 6² (4+h)² = h² + 6² h² + 8h + 16 = h² + 36 8h = 36 - 16 h = 20/8 = 5/2 A = 6h = 6*5/2 = 15
Andy, could you "redo" this Catriona Agg puzzle by using a faster method using the 2 tangent theorem from 1:00 onwards? This since you can divide the the base of the rectangle into 6=5+1, then (observing lower left angle) use the 2TT you can "reflect" the 5 on the part of the diagonal, then you divide the height of rectangle into 1+x, use the 2TT again (now observing the upper right angle) you get that the remainder part of the hypothenuse is x, thus hypothenuse is 5+x. Now you plug in the Pythagora those lenghts and you get the x: (5+x)^2 = 6^2 + (1+x)^2 x = 3/2 -> h = 1+x = 5/2 area=15 To me it seems much much neater than finding 3 triangles and having 2 variables?
You can see the two triangles in the video to see why not that be the case. The 5+x side of the smaller triangle is the hypotenuse. The length of the corresponding similar bigger triangle's hypotenuse would be 5+x+y, not 6, so the correct ratio using that side would be: (5+x) / (5+x+y) = (y+1) / h.
You could have done it a bit simpler, you get up to the hypotenuse being 5 plus a bit, the bit is the same bit as from the top corner of the triangle to the tangent of the circle since that makes an isosceles triangle then pythag says 6^2+(1+m)^2=(5+m)^2 which evens out at m=1.5 so the height is 2.5
Overcomplicated. Make another radius parallel to the base, then base will be 5 + 1, height will be 1 + y and hypotenuse will be 5 + y, then go Pythagorean (5 + y)^2 = (1 + y)^2 + 36 gives y == 1.5, making height 2.5 and the area 15.
Why all this work? It was enough to assume x is the width of the rectangle and we would have according to the Pythagorean theorem (x+4)²=x²+6² and from it x=5/2, so the area of the rectangle is equal to 6*5/2=15.
The right triangle that surrounds the circle with radius 1, its sides are a=6 and b=x and the hypotenuse c satisfies the relationship a+b-c=2r, i.e. 6+x-c=2, and from it c=x+4, and according to the Pythagorean theorem we have (x+4)²=6²+x². Isn't this method easier?@@Epyxoid
Let the fireworks go off, because the year hasn’t ended until he does the last problem. Happy ACTUAL new year
Happy new year 🥳 Love the series. Happy that he made it :)
Time is relative 😉
How eggciting!
How eggciting!
@@MAGNETO-i1i hmm yes very sceince
Andy, I think I speak for all of your subscribers when I say that you absolutely crushed it for the last 31 videos. It was a delight watching each and every one of them, and I learned a few things about geometry in the process. Thank you so much for putting all of these videos together, this channel is one of the bright spots of TH-cam for me.
Really enjoyed the past 31ish days of your puzzle solutions. Only 11 more months until we can do this again.
He should pre record them all now so he can schedule daily uploads for each one
I loved it too, it was so exciting
I like that you did not pre record all the problems. I felt part of the process and the reality of being human.
I think there is a faster solution for the "h". 1:44 right upper corner gives us 2 tangent lines to the right circle, which are "h-1".
So we can apply the Pythagorean theorem. Rectangle's base is "6", it's hight is "h" and it's diagonal is "5+h-1".
6^2 + h^2 = (5+h-1)^2
h = 2.5
I had the same idea, it's a fair bit faster and easier.
That's exactly what I did! I was surprised when Andy did the intersecting tangents thing for the bottom left corner... but then didn't do the exact same for the top right!
Yup, also did it this way. Andy probably a bit stressed after recording and editing all these puzzles 🙂
Exactly what I did. Just remember this theorem: If you have a circle with 2 tangent lines intersecting at a point, the distance from that point to the two points of tangency are equal. By using this fact twice on the problem, the answer can be reached very quickly.
Congrats to Andy, though, for perservering through these 31 interesting geometry puzzles, and creating the wonderful animated illustrations. What a great math communicator!
Great Series.
This problem is easier to solve if you use the three tangents of the circle in the right angle triangle and the “tangents from an external point” theorem to calculate h.
Yeah this seemed like a really simple approach that works
Very nicely done! I think your animations are underappreciated. I can relate to the difficulty in doing them! But it's worth the effort to make these solutions so understandable. Happy New Year, finally!
"...and that is the end of our Aggvent Calendar." [cut to black] Classic Andy Math!!!!!!!
did the cut to black seem important or something
Absolute cinema
@isuckatediting6898Fr
As Andy pointed out, we didn’t need to evaluate the value of x in the “little” triangle. But we don’t need to evaluate the value of y either (or even explicitly evaluate the value of h) in order to determine the area of the rectangle. We can just apply the appropriate scale factor (a couple of times). At 3:05, we established that the little triangle is 1/25 the area of the “larger” triangle. That larger triangle consists of the little triangle plus the two congruent triangles (each of area 5/2) previously drawn. Those two triangles hence comprise 24/25 the area of the larger triangle, which means the larger triangle has area 5 * 25/24 = 125/24. Now to get the area of the “whole” triangle, we simply multiply by the square of the linear scale factor going from 5 to 6, or 36/25. Then the whole triangle has area 125/24 * 36/25 = 7.5. Finally, the area of the rectangle is twice that, or 15. So you don’t need to figure out x, y or h to get that (instead of area rectangle = 6*h, we can use area rectangle = twice the area of the triangle formed by the diagonal of the rectangle).
Considering that the 12 days of Xmas extend through January 5th., I say you technically came in on time, theologically speaking. :) Congratulations, beautiful work.
That was a really slick finish on the last one. Great to have watched you through the whole thing. Hope you're having a great New Year, Andy. Looking forward to solving a lot more problems with you.
Thank you Andy for an exciting Andy Aggvent Month. Reading the comments became a thing too, you created a nice community. Your solutions with wonderful animations will be appreciated and live forever in cyber space for the next generations to enjoy. Happy new year… you can take a vacation now, you earned it!
Congratulations Andy, thank you for this series.
For the record, this one can be solved with the same method from Day 30 where we we also had a right triangle with a known side and a known diameter of its inscribed circle.
Definitely appreciate Andy mixing things up though.
At 1:40, we have two small right triangles with sides 1 and 5. Let the angle opposite the short side be Θ. We observe that tan(Θ) = 1/5. From our notes on the 5-12-13 Pythagorean triple right triangle, if tan(2Θ) = 5/12, then tan(Θ) = 1/5. The triangle above the small triangle has angle 2Θ opposite the short side h and the long side is 6. So, tan(2Θ) = h/6 but tan(2Θ) = 5/12, therefore h/6 = 5/12 and h = 2.5. Area = 6h = 6(2.5) = 15 u², as Andy Math also found.
Literally gasped with excitement, 2025 is officially here 👏🏼👏🏼👏🏼
enhance! I lol'd. Truly exciting series Andy.
Awesome solution! I did it by focusing on the small right triangle at the upper right. The lengths of its legs are 2 and h/3, and with the help of the equal tangents theorem (used twice), you can see that the hypotenuse is 5h/3 - 2. Now you solve for h using Pythagorean theorem: 2² + (h/3)² = (5h/3 - 2)². Some nice cancellations lead to h = 2.5, and so Area = 15.
Hello, Andy. I saw you solving all these "fun ones" problems, from 1 to 31. And, in fact, it was " How exciting". Thanks for doing it and for sharing. I can't wait for December 2025, so I can follow your solving for new 31 "fun ones" problems...Happy New Year to you.
Congrats on finishing your challenge, even with slight delay. How exciting!
However I was hoping to see new problem. I love this format, where solution isn’t available. End up solving whole month by myself, and even got friends and family hooked as well :)
Big congratulations, that was a fun ride. Keep being your awesome self, the intellectual honesty is massively appreciated. Huge fan.
A simpler way to think about it is that the right angled triangle surrounding the blue circle forms 3 pairs of tangents. Both tangents in each pair have the same length. So the base is 5 + 1, the height is 1 + x (unknown) and the hypotenuse is 5 + x (unknown)
(x+1)² + 6² = (x+5)²
x² + 2x + 1 + 36 = x² + 10x + 25
8x = 12
x = 1.5
Area = base * height = 6(1+x)
= 6 * 2.5 = 15
Thank you for this series, Andy!
5:42 did i get it wrong, or he actually needed to sum the x with the 5 in the minor triangle?
Great work
((5+(h-1))squared=h*squared+6*squared) would also work as the other side of the triangle also has tangentlines 90 degrees from the radius
King! Congrats on reaching the end. I thought it was exciting that you caught the efficiency gain with the x!
Way to go Andy!!! I enjoyed every bit of the past 31 days
Thanks Andy for this series, really enjoyed all the problems 👍
Happy new Year. Finally, the spherical object has changed its vertical position. ("The ball has dropped.")
Wish you the best for the new year. I'm looking forward to fun and exciting videos.
This is waaaaay too complicated. Use the equal tangent distances from lower left/upper right vertices of the rectangle with the rightmost (third) circle. The left tangents to circle 3 have length 6 - 1 = 5, and the right tangents to circle 3 have length x. Then one gets
6² + (x + 1)² = (5 + x)²
36 + x² + 2x + 1 = 25 + 10x + x²
37 + 2x = 25 + 10x
12 = 8x
x = 12/8 = 3/2
Therefore the height h = 1 + 3/2 = 5/2, and the area is 6·5/2 = 15.
woohoo! time for a well deserved break! I think it took you more time to edit than to actually solve it all ^^'. I really enjoyed all the problems in the calender, takes some outside the box thinking and lots of triangles and radiuses.
To solve, I used the Two Tangents Theorem, which states that if two tangents are drawn from the same external point to a circle, then the tangents are equal in length. Using the rightmost circle and the lower triangle in which it is inscribed, we have the following 3 pairs of tangents:
Tangents from lower right corner have length = 1
Tangents from upper right corner have length = h − 1 (where h is height of triangle/rectangle)
Tangents from lower left corner have length = 6 − 1 = 5 (where 6 is base of triangle/rectangle)
Triangle has base = 6, height = h, and hypotenuse = (h−1) + 5 = h + 4.
Using Pythagorean theorem, we get
(h + 4)² = h² + 6²
h² + 8h + 16 = h² + 36
8h = 20
h = 5/2
*Area of rectangle = 6 * 5/2 = 15*
You could also use the fact that the rightmost circle is the incircle of a right triangle, and you know a side that isn't the hypotenuse of that triangle, and that's already enough information to solve.
The way i did it was i applied the HL theorem to both "corners"(i guess) of the triangle. So the base of the triangle is 6, the height is x+1, and the hypotenuse is x+5. And with the pythagorean itll become 6² + (x+1)² = (x+5)². And we find x to be equal 3/2 and theredore the height in total will be 5/2.. dunno just thought this was a neater solution imo
Congrats! I used de same theorem!!!!
It is definitely neater, I did it that way too.
Agreed. I was completely following until about 1:40 into the video. Then, I was thinking why is he making this so complicated with drawing other triangles and introducing two variables.
Thanks for this series Andy! PS, don't forget to update the Aggvent playlist for those who want to celebrate throughout the year 😅
Crushed it! Thanks and Happy New Year, Andy!
As a recent problem, we can use "the formula for the inradius of a right triangle."
If we take the triangle formed by half of the rectangle, the full circle follows the formula...
So, in this triangle A,B,C we are looking for A to solve the rectangle surface.
We have B=6.
The formula gives us the difference between the hypotenuse and a:
((a+6−c)/2)=1
Thus, c=a+4.
Now we can use the Pythagorean equation:
a²+6²=(a+4)²
This simplifies to:
a²+36=(a²+8a+16)
Which gives us:
a=2.5
That’s what we’re looking for. 2.5×6=15
It’s longer to describe than to actually do it!
How exciting! 😊
Nice one! I loved these videos (along with your others of course). I feel it's truly the new year now 😀
It’s already New Year again 🎉 How exciting!
The green circle is inscribed in a right triangle with legs 6 and h, so a formula for the radius of the circle is r = [6 + h - sqrt(36 + h^2)]/2. Because r = 1, you have an equation in the unknown height h. It is easy to solve for h. Simple.
Better late than never...... congratulations you finished a long egg-citing journey 🎉
Congratulations Andy 🎉🎉🎉
You made this harder then necessary. Tangents to circle from common point are equal. From lower left corner are 2 tangents to right-hand circle. Lower one is 5, slanted one is also 5. Let remainder of slanted line be X long. Diagonal of rectangle is 5 + X. Tangents from lower right corner to right circle are both 1. Tangents from upper right corner are both X. Right triangle: bottom leg = 6, vertical leg = 1 + X, hypotenuse = 5 + X.
X² + 10X + 25 = X² + 2X + 1 + 36. X² terms cancel. 8X = 12, X = 3/2.
Area = (6)(1 + 3/2) = 30/2 = 15.
Thanks Andy, for making December enjoyable again. It’s obviously been a big undertaking for you so hope you can get some time away from YT for a bit now. Cheers.
Thank you, I needed that. Happy New Year 🎉
To simplify the solution, the area of the rectangle is: A=(W)*(L)=(6)*(1+X).
As the sides of the triangle with a right angle are tangent line of the circle, using the Pythagorean theorem:
6^2+(1+X)^2=(5+X)^2
36+1+2X+X^2=25+10X+X^2
8X=12, X=3/2
Plugin the results of X to the area:
A= 6*(1+3/2)=15
Could've just used the inradius of a right triangle formula for that one, or even derived it, it'd be done in half the time or so.
1 = (6 + h - (h² + 36)^0.5)/2
It'll take some time but still get there fast.
Anyways, awesome work with the aggvent calendar Andy, you were awesome (and still are of course) I get intrigued and excited by many of your videos, keep up the great work and keep being awesome
The similarity ratio is really cool! Good call!
ENHANCE!!! 😂😂😂
Congratulations! Well done!
Was a lot of work for you, but I enjoyed the event! Thank you.
I did it differently. Since h equals 1 + x and the diagonal equals 5 + x (x are top right tangents of the blue circle) I just did 6^2 + (x + 1)^2 = (x + 5)^2 (Pythagorean theorem) and solved for x=1.5. Area width x length =2.5 x 6 = 15
Congrats on completing the aggvent calendar🎉
It already has a box, like swiping the carpet beneath my feet
It was already in a box! Ah~ ❤
In fact, you don't need either x or y.
Just apply Pythagoras using the principle of secant lines:
(5+(h-1))² = h² + 6²
(4+h)² = h² + 6²
h² + 8h + 16 = h² + 36
8h = 36 - 16
h = 20/8 = 5/2
A = 6h = 6*5/2 = 15
ハッピーニューイヤー、アンディ兄貴
The new year has officially started now that Aggvent is complete.
Well... in Julian Calendar is still December 22nd... so he made it in time
You did it! 🎉
Andy, could you "redo" this Catriona Agg puzzle by using a faster method using the 2 tangent theorem from 1:00 onwards?
This since you can divide the the base of the rectangle into 6=5+1, then (observing lower left angle) use the 2TT you can "reflect" the 5 on the part of the diagonal, then you divide the height of rectangle into 1+x, use the 2TT again (now observing the upper right angle) you get that the remainder part of the hypothenuse is x, thus hypothenuse is 5+x.
Now you plug in the Pythagora those lenghts and you get the x:
(5+x)^2 = 6^2 + (1+x)^2
x = 3/2 -> h = 1+x = 5/2
area=15
To me it seems much much neater than finding 3 triangles and having 2 variables?
HAPPY NEW YEAR!!!
Thank you. I never like math but now I find it very interesting
How exciting!
That was cool, I was keeping up with you till day 18, I think I will have to catch up
Should the ratio not be (5+x) / 6 = (y +1) / h?
You can see the two triangles in the video to see why not that be the case.
The 5+x side of the smaller triangle is the hypotenuse. The length of the corresponding similar bigger triangle's hypotenuse would be 5+x+y, not 6, so the correct ratio using that side would be: (5+x) / (5+x+y) = (y+1) / h.
Oh yeah! Thanks - I had got myself confused.
Yeah I think he forgot because it was so small.
How EXCITING 🎉!! 🥳!!!
OMG, Andy is a replicant !
Happy New Year, guys!! Now we can celebrate!
Happy new year tomorrow
You could have done it a bit simpler, you get up to the hypotenuse being 5 plus a bit, the bit is the same bit as from the top corner of the triangle to the tangent of the circle since that makes an isosceles triangle
then pythag says 6^2+(1+m)^2=(5+m)^2
which evens out at m=1.5
so the height is 2.5
At long last the quest is done! Please enjoy at least a little bit of a break
woo hooo .. made it. Thanks.
happy true new year (he couldn't have finished it after new year because then that would be against the rules
You have an error here (sorry). Your final ratio should be 5+x over 6, not 5 over 6, so you did need to find x.
Happy New Year, Andy! That was a tricky one.
He did it 🗣🗣🗣
ENHANCE
Overcomplicated.
Make another radius parallel to the base, then base will be 5 + 1, height will be 1 + y and hypotenuse will be 5 + y, then go Pythagorean (5 + y)^2 = (1 + y)^2 + 36 gives y == 1.5, making height 2.5 and the area 15.
HE DID IT BOYS
Does that explain Y?
Okay! Now it's the 1st of Jan!
Happy new year everybody 🎉 Hope you all have an exciting 2025
this is MY new years
I got a much nicer (and more correct) answer when I read the question right and didn't confuse radius for diameter 😂
Hey we can use the relation (a+b-c)÷2=r
🎉🎉🎉
I was waiting for the "does that make sense why" the whooooooole time :(
Can you do one that doesn't look fun?
Why all this work? It was enough to assume x is the width of the rectangle and we would have according to the Pythagorean theorem (x+4)²=x²+6² and from it x=5/2, so the area of the rectangle is equal to 6*5/2=15.
How do you know that the length of the rectangle's diagonal is x+4?
The right triangle that surrounds the circle with radius 1, its sides are a=6 and b=x and the hypotenuse c satisfies the relationship a+b-c=2r, i.e. 6+x-c=2, and from it c=x+4, and according to the Pythagorean theorem we have (x+4)²=6²+x². Isn't this method easier?@@Epyxoid
@@AzouzNacir a+b-c=2r is what kind of relationship? Why is that given? I'm not familiar with it. Is that an inscribed circle related theorem?
Yes, this is a well-known and widely used relationship, and it can be easily proven using tangents emanating from a point to a circle.@@Epyxoid
Yes, this is a well-known and widely used relationship, and it can be easily proven using tangents emanating from a point to a circle.@@Epyxoid
I don't think the base is 6. The diameters aren't collinear.
I'm sad weve finally finished.
lol enhance
This. Has . Been. Great! Your arithmetic needs help though... 31 answers in 35 days... which calendar are YOU using???
5:45 you need to add the x to the 5
Hey to be fair he never said during December. He said FOR the month!
wohhhh
Of course, he didn't say 31 CONSECUTIVE days.
Enhance!
6 * tan( atan( 1/5 ) * 2 ) * 6 = 15
That's cheating! 😆