Aggvent Calendar Day 26

[@pie8021]

I appreciated the "notes right here" 😁 0:45

[@AndyMath]

I laughed out loud when I thought of the pun! "Right here", as in "Right angle here!". It was subtle but I thought it was funny.

[@dylanowens6742]

I had no idea this series would provide so much tension 😂

[@miamoberg827]

Guess he has a few more hours to spare in December. In Europe we're headed for 2025 pretty soon. Go Andy Go!

[@FlintStryker]

Whew! He’s catching up, but will he make it?!?! I’m counting on you, Andy. Just imagine “How Exciting” new year’s eve is going to be. Boy oh boy!

[@scp-1-2-3]

Andy, you’re on a roll!! You’ve got this!

[@m.h.6470]

Solution: If you draw lines from the semicircle's centerpoint to the end points of the blue line, you get an isosceles triangle with the two legs both being r = 8/2 = 4 long. Because the green triangle is tangential to the semicircle on both sides, it too is a an isosceles triangle, and its angles HAVE to be 45°-45°-90°, as it shares its top angle with the big right angle triangle. Since the triangle created by the two extra lines has two legs who - because they are radii - HAVE to be perpendicular to the legs of the green triangle, it too HAS to have 45°-45°-90° angles and it shares its hypotenuse with the green triangle. Therefore the two triangles are identical, and as such, the two legs of the green triangle also have to be of length 4. The requested green area is therefore 1/2 * 4 * 4 = 8 units².

[@rileypeterson4708]

We gonna get to 31 by the end of the year??

[@rileypeterson4708]

How exciting!

[@Bosiolio]

My prediction is that he's going to drop day 31 at 11:59 pm on the 31st, wherever his local time is.

[@90Rush]

This looks like an easy one since the circle is tangent to both legs, meaning if you draw a line from the center to the two tangent points the angle is alway 90%. Resulting triangle from this drawing is equivalent to the green shaded area.

[@blintzy6969]

Keep on the grind Bro 💪 Also no biggie if you don’t finish in time I’ve learnt so much from you

[@etlam3]

Never thought that a math serie could get this intense

[@KrytenKoro]

Im going to take advantage of the next one having to work for all scales, and shove that other corner all the way to the right, so one square is 5x5. That makes the other square 5{2}x5{2}, and makes the shaded area 5{2}x5{2}/2=25. Not sure how to solve it analytically but im guessing thats the solution.

[@5gearz]

He’s locked in

[@alexjimenez8088]

Thank you Andy.

[@JennyBlaze253]

2 released within the span of 2 hours?! Is he gonna catch up?! 😮

[@JenishTheCrafter]

We're gonna enter new year's with Math and I'm not complaining

[@MarieAnne.]

For Day 27 Let x = side length of larger dark blue square, y = side length of smaller light blue square. Right triangle on left side of diagram has base = 5, height = y, and hypotenuse = x. By Pythagorean Theorem: x² = y² + 5² → *x² − y² = 25* Now right triangle at top of diagram has one leg formed by black dashed line, another leg formed by part of dark blue line (call it z), and hypotenuse formed by light blue line (y). This triangle is similar to the previous right triangle by AA, because they both have a right angle, and the angles of both triangles located at top left vertex are both complementary angles to the angle between both squares, so they are congruent. By similar triangles we get z/y = y/x → z = y²/x Shaded rectangle has one side = x, and other side = x − z = x − y²/x *Area of shaded rectangle = x (x − y²/x) = x² − y² = 25*

[@JamCliche]

Let's goooooo! Andy on a rolllllll!

[@michaelengelby732]

Pythagorean Theorem is the star (or at least a willing participant) of nearly all these problems!

[@Qermaq]

Not the next one, though. Well, it could be.

[@rossbrashear535]

I am such a nerd. I Love These!!!!

[@MYCROFTonX]

Best advent calendar ever... so much fun... how ecks-ziting!

[@therealrickycat]

You got this, Andy!

[@cyruschang1904]

Answer to the next question: If the side length of the smaller square = x, the side length of the bigger, tilted square = √(5^2 + x^2). Their respective area = x^2 and (5^2 + x^2) The four triangles are all similar. The hypotenuses of the two largest are √(5^2 + x^2) & x, their longer legs are x & (x^2)/√(5^2 + x^2) So the width and length of the green area are √(5^2 + x^2) - (x^2)/√(5^2 + x^2) and √(5^2 + x^2) The area = (√(5^2 + x^2) - (x^2)/√(5^2 + x^2))(√(5^2 + x^2)) = (5^2 + x^2) - x^2 = 25

[@DaHaiZhu]

Go, Andy, Go!!!

[@realnazarene5379]

For the next puzzle, the overlapping squares, a quick back of the napkin solution gave me 25 square units.

[@Syfes]

same

[@nathanc6516]

I don't even know where to start with the next one. I'm assuming they are 30-60-90 triangles, but I don't know how to prove it. But if 30-60-90, I get 25 too.

[@eikebehrmann3493]

same, but i assumed the total cyan side length was 5 (it should check out but isn’t mathematically rigorous)

[@joeydifranco0422]

⁠@@nathanc6516 If the light square had side length L and the dark square has side length D, and the side length of the dark square minus the shorter length of the shaded rectangle is X, then the area is D*(D-X)= D²-DX. In the top left, the angle made by the dark lines is 90°. If the angle made by the dark line and horizontal light lines is a, then the angle made by the horizontal light line and other dark line (middle angle) is 90-a. The angle between the light lines is also 90°, and the middle angle, as we have said, is 90-a, so the angle between the vertical light line and dark line is also a. That means that the triangle with a side length 5 and the upper triangle with a dotted line side are similar. This means the ratio of corresponding sides is also equal, so L/D=X/L -> L²=DX. We can also see that L²+25 = D² -> D²-L²= 25, but L²= DX, so D²-DX = 25, and area=D²-DX, so area=25.

[@nathanc6516]

@@joeydifranco0422 Thanks, I see you don't need to prove it's a 30-60-90 after all.

[@whippersnap8497]

I was really worried you wouldn't put a box around it at the end

[@JasonMoir]

i'm guessing the next one will not be as clean or quick to solve.

[@brettappleton3882]

It’s going to be super easy, barely an inconvenience.

[@m.h.6470]

For tomorrow's question: Flip the light blue square to the left and you can see, that the blue square is the square over the hypotenuse and the light blue square is the square over the long leg of the bottom left triangle. We therefore have the relationship b² + 5² = c², where b is the sidelength of the light blue square and c is the sidelength of the blue square. The top triangle - through symmetry - shares all three angles with the bottom left triangle, as such they are similar and they share the same ratios. Calling the two legs (c - x) and y, we are looking for the area cx and have the following equations: b² + 5² = c² (c - x)² + y² = b² → c² - 2cx + x² + y² = b² (c - x)/b = b/c → c² - cx = b² Putting the last two together, we get b² - cx + x² + y² = b² |-b² +cx x² + y² = cx Now elongate the right side of the light blue square to the top of the blue square. This creates two other right angle triangles: One large one, which is identical to the bottom left triangle, just turned by 90° around the top left corner of the squares, as well as a tiny triangle... with the sides x, y and an hypotenuse of 5 (due to the symmetry). As such, x² + y² = 5² and that is the final piece to get our answer: cx = 5² = 25 units²

[@crunnifle9114]

We got like 5 left we can do it

[@seahawk124]

How exciting!

[@raheemmiah6992]

Let's put a box around it

[@theobserver9066]

This could be a fun one

[@AniedoAniedo]

May i ask what app/website do you use for your demonstrations?

[@VolS-oj5md]

To take half of square area probably would be simpler? (4*4) / 2 ? Just because most people remember formula for area of the square but not triangle :)

[@ianbrooks6816]

Wow, a simple one!

[@mahmoudrashidy6735]

why the two tangent lines made 90 degree (what is the law)

[@MarieAnne.]

Because we are *told* (in the statement of the problem) that the triangle is a right triangle. It's clear that the right angle of the triangle is not located at one of the two vertices at the base of the triangle (otherwise the inscribed semicircle would have one side of its diameter located at that vertex), so the right angle of the triangle must be located at the top vertex, where the two tangent lines intersect.

[@tunneloflight]

Duh. I got caught thinking in triangles and solved it the harder way. Radius 4, so smaller triangles side lengths = 4/root(2). So since there are two, area = (4/root(2)) squared = 16/2 = 8. Far simpler to see the 4x4 square and divide by 2. Though functionally they are the same.

[@kurisu.senpai]

Day 27, ans: A = 25 units squared (???)

[@KrytenKoro]

Its 8 isnt it

[@davidedelpapa7706]

How Agg-citing!

[@tellerhwang364]

day27 a:b=(b-x):a→bx=b^2-a^2=5^2 →A=25😊

[@chrishelbling3879]

Yay, a Geometry problem with no Algebra.

[@YaztromoX]

I hate to be a pedant who picks at nits, buttering your final step of resolving the area of the triangle 1/2 * 4 * 4, you violated the order of operations by multiplying the 4's together first. I know in this instance as all of the operations are multiplications ultimately it doesn't change this particular solution (hence why this is a nit-pick), but considering the staggering number of people online who don't know the order of operations IMO it would be good to set a proper example by following them strictly, even in trivial situations like this.

[@matthijs22]

This stuff is too easy for a KZbin video. Come on.