He just read your comment and you are kinda right. He is more hyped than stressed, realizing he's got to time tomorrow perfectly to get it all done! I solved the next one. But I haven't even looked at the last 3 yet!
Next puzzle: 90 degrees Create angles from the centers of the circles to their tangent point, which will measure 2a and 2b. The centers connect through the common tangent point so it's a straight line, making a 4 sided figure. The bottom tangent points are at 90 degrees to the line, 180 degrees total, so 2a+2b=180, therefore a+b=90.
The intuition on this one is harder than just going through the algebra. It's not too hard to see that the long side of the area is a projection of the hypotenuse, but it was trickier for me to see how to define the side of the top triangle in terms of the area's short side.
The next one is fairly easy compared to the rest of the aggvent puzzles up until now The answer is 90° 9f course, but it's mostly easy to see once the radii and the connections between all the tangent points are drawn + drawing the 90° angles of the radii with the ground
The triangle whose vertices are the points of contact is right, and the sum of the two non-right angles is equal to 90°, and they are two inscribed angles that measure a and b, and therefore a+b=90°
Answer to the next question: If we draw the tangent line at the point of contact between the two circles, then connect each circle center to the two points of tangency (one with the bottom line, one with the other circle), we get two quadrilaterals. Each quadrilateral has two 90° corners (at the points of tangency) and the sum of the other two corners will be 360° - 90° - 90° = 180° = the sum of the central angles, one from each circle. Since the measure of an inscribed angle is always half the measure of the central angle that subtends the same arc, a + b = 180° ÷ 2 = 90°
Let a and b be the sides of the horizontal and diagonal squares on the right, and x is the width of the shaded rectangle. From the two right triangles, we conclude a/(b-x)=b/a, and from it bx=b²-a²=25, which is the shaded area.
90 degrees This really needs to be visual to make sense, so bare with me Basically, draw a line from the bottom tangent of the left circle to the centre of the circle, another line to the centre of the right circle and then a line down to the tangent at the bottom of the right circle and back to the start, that's a quadrilateral, all the angles add up to 360, the bottom two angles add up to 90 so the two inner angles x and y add up to 180 a is half it's inner angle(x) and b is half it's inner angle(y) just using a and x, the opposite angle of the inner angle is 360-x you then get two isosceles triangles with the centre angles adding up to 360-x the sum of the other four angles would be 360-(360-x) and a is half of the sum of these four angles so a=x/2 exact same for b and y so b=y/2 since x+y=180 a+b=180/2=90
Next one: As the dimensions of the circles are not given, WLOG set them to be congruent. The circle centres and the tangent point between the circles are colinear, and the base line is tangent to both circles, so the angle at each circle centre formed by the radii to the respective tangent points is 90°. So by the inscribed angle theorem, a and b are both half of this value, so a = b = 45° and so a + b = 90°
For the next one, the sum of the two angles is 90 degrees. Draw central angles to the tangent points. That will form a quadrilateral connecting the two center points with a straight line and the centers to the bottom tangent, which is the fourth leg of the quadrilateral. Since the bottom angles are 90 degrees each, the top two add up to 180 degrees. But the inscribed angles are each half of their corresponding central angle. Therefore, the sum of the two inscribed angles is 90 degrees.
You can always play the card "it's still December until the last date line is crossed". You may just need to upload via a VPN from a Pacific island 😂 This will have to go down in YT history if Andy succeeds!
Maybe we’ll see the last one with a new year’s party hat! I can just see Andy in a party room somewhere putting up the last post! Happy Year Andy, we’ll forgive you if you post the last in the new year!
Oh I see how to prove it. If you draw lines from the circle centers to the same tangent points, you end up with a trapezoid with two right angles, so the sum of the upper corner angles c+d are 180. A and b are subtended angles, so they must be half of whatever c and d are, and 180/2 is 90
I can't even imagine the stress Andy must be having knowing that he has to solve, record and edit 4 problems in less than 24 hours...
He just read your comment and you are kinda right. He is more hyped than stressed, realizing he's got to time tomorrow perfectly to get it all done! I solved the next one. But I haven't even looked at the last 3 yet!
Solving might be the easy part lmao
@@AndyMathhow exciting
@@AndyMathnot to put pressure but it's already the 31st in my country right now.
This looks important, let's put a box around it.
no matter what, triangles really are the GOAT...
Eh... Hexagons are the bestagons. 🙂(Mainly because they're made up of triangles?)
bro this man is speedrunning both these puzzles and his sanity
Next puzzle: 90 degrees
Create angles from the centers of the circles to their tangent point, which will measure 2a and 2b. The centers connect through the common tangent point so it's a straight line, making a 4 sided figure. The bottom tangent points are at 90 degrees to the line, 180 degrees total, so 2a+2b=180, therefore a+b=90.
Bro I slept for a nap after the 26th one and woke up to this
The grind is real
You can do this Andy! We believe in you.
The intuition on this one is harder than just going through the algebra. It's not too hard to see that the long side of the area is a projection of the hypotenuse, but it was trickier for me to see how to define the side of the top triangle in terms of the area's short side.
You got this Andy! Even if you dont get all 31 in under the wire I still really appreciate this series I've enjoyed it a lot
Original title before change: "YT Aggvent 27 HD 1080p"
Thank you!
@AndyMath You are welcome
I'm not going to get any sleep today waiting for the last 4
Andy gets it done! These videos are very satisfying to watch.
The next one is fairly easy compared to the rest of the aggvent puzzles up until now
The answer is 90° 9f course, but it's mostly easy to see once the radii and the connections between all the tangent points are drawn + drawing the 90° angles of the radii with the ground
The triangle whose vertices are the points of contact is right, and the sum of the two non-right angles is equal to 90°, and they are two inscribed angles that measure a and b, and therefore a+b=90°
Just 4 more fun ones to go! We're rooting for you, Andy! Oh, the anticipation!
Answer to the next question:
If we draw the tangent line at the point of contact between the two circles, then connect each circle center to the two points of tangency (one with the bottom line, one with the other circle), we get two quadrilaterals. Each quadrilateral has two 90° corners (at the points of tangency) and the sum of the other two corners will be 360° - 90° - 90° = 180° = the sum of the central angles, one from each circle. Since the measure of an inscribed angle is always half the measure of the central angle that subtends the same arc, a + b = 180° ÷ 2 = 90°
Let a and b be the sides of the horizontal and diagonal squares on the right, and x is the width of the shaded rectangle. From the two right triangles, we conclude a/(b-x)=b/a, and from it bx=b²-a²=25, which is the shaded area.
CMON ANDY YOU GOT THIS
90 degrees
This really needs to be visual to make sense, so bare with me
Basically, draw a line from the bottom tangent of the left circle to the centre of the circle, another line to the centre of the right circle and then a line down to the tangent at the bottom of the right circle and back to the start, that's a quadrilateral, all the angles add up to 360, the bottom two angles add up to 90
so the two inner angles x and y add up to 180
a is half it's inner angle(x)
and b is half it's inner angle(y)
just using a and x, the opposite angle of the inner angle is 360-x
you then get two isosceles triangles with the centre angles adding up to 360-x the sum of the other four angles would be 360-(360-x)
and a is half of the sum of these four angles
so a=x/2
exact same for b and y
so b=y/2
since x+y=180
a+b=180/2=90
Next one:
As the dimensions of the circles are not given, WLOG set them to be congruent.
The circle centres and the tangent point between the circles are colinear, and the base line is tangent to both circles, so the angle at each circle centre formed by the radii to the respective tangent points is 90°.
So by the inscribed angle theorem, a and b are both half of this value, so a = b = 45° and so a + b = 90°
For the next one, the sum of the two angles is 90 degrees. Draw central angles to the tangent points. That will form a quadrilateral connecting the two center points with a straight line and the centers to the bottom tangent, which is the fourth leg of the quadrilateral. Since the bottom angles are 90 degrees each, the top two add up to 180 degrees. But the inscribed angles are each half of their corresponding central angle. Therefore, the sum of the two inscribed angles is 90 degrees.
You can always play the card "it's still December until the last date line is crossed". You may just need to upload via a VPN from a Pacific island 😂
This will have to go down in YT history if Andy succeeds!
Assumption: shaded area is a rectangle
I might have missed it. Where do we know for sure the green shaded area is a rectangle?
absolutely cranking them out!
Ok, this is quest is amazing
He’s doing it!
Maybe we’ll see the last one with a new year’s party hat! I can just see Andy in a party room somewhere putting up the last post! Happy Year Andy, we’ll forgive you if you post the last in the new year!
Damn dude youre killing it!
Also, neat that the scaling intuition turned out to be true!
For tomorrow, i think it's going to be 90*
Oh I see how to prove it. If you draw lines from the circle centers to the same tangent points, you end up with a trapezoid with two right angles, so the sum of the upper corner angles c+d are 180. A and b are subtended angles, so they must be half of whatever c and d are, and 180/2 is 90
Look what you did! Groovy!
2 more hours to go add oil 🎉🎉
Bro. You need to finish 4 more challenges immediately. Because the year 2024 is going to finish.
2a + 2b = 180°
a+b = 90°
_____________
Let's assume that the two circles are equal, then a = b = 45° => a+b = 90°
Bros try harding for us :) ❤
Nice solution! I stared at it for a bit and didn't get there!
That is such creative solution.
Three in one day, awesome!
28: circle radiai form a shape with 2 right angles so 2 remaining angles add up to 180, divide it by 2 and next video please
day28
2a+2b+90+90=360
→a+b=90😊
you are really rushing these last bosses right 😂
Man, this area doesn't look to be 25 units sqr at all! It looks way too small
no need to rush 😭
He need to rush
He's got 12 hours, what do you him to do
The title was different
@@synexiasaturnds727yearsago7no he's from north america he still have 22hrs
Bro is trying to clutch before 2025
Just 7 hours and 30 minutes left.. in American EST
Andy days go by (water flowing underground)
Great video! Peace out
It's a CTR-R day. Good luck.
It's already 2025 in my country... I don't think he will make it =\
How Agg-citing!
You mean today's question ❓😂
This series is high key becoming a nail biter
19th!
Oh!
a plus b iqual to 90 degrees
Bro hurry up😂
He wont make it
this one is pretty hard to understand
just use a protractor to find the angles
angle doesn't matter, it could be any angle and you get the same answer.
no views in 39 seconds bro fell off