I drew a quadrilateral using the tangent points on the baseline and the centres of the circles. The corners on the base line are 90 degrees each, or 180 degrees total, so the other two angles total 180 degrees. Those angles match the arcs the inscribed angles match, so they add up to half of the 180 degrees, or 90 degrees.
The teeth visualization was really nice. I tried to solve the puzzle but quickly realized I understood nothing because I couldn't understand what was going on with the angles and the inscribed angles in a circle.
What I did was construct a quadrilateral by connecting lines to the centre points of each circle. Since radii connecting to any tangent point id 90 degrees, we can construct a big circle radius and a small circle radius to connect to the bottom line. Then, we can construct a line with the big radius and small radius connecting the big centre to the small centre. Again, the line between the two centres will pass directly through the point where the two circles meet. We know that the sum of all interior angles of a quadrilateral is 360 degrees and since we have already constructed 2 right angles, this means the sum of the remaining interior angles is 180 degrees. We can call these two angles x and y and form the expression, x + y = 180. What we will then notice is that both x and y are angles formed when the radii connect to the centre of their respective circles and we realise that a and b are the same lines but connecting to the circumference. From here, we can use the rule that angles found at the circumference are half of the angle at the centre when enclosed by the same arc. This means a + b = (x + y)/2. We can substitute 180 for x + y and this gives us 180/2 = 90. Thus x + y is equal to 90 degrees.
Another way to solve - In the big circle, if i draw an angle using the same arc in the center,that will be 2a. Similarly, in the small circle, if i draw an angle in the center using the same arc,that will be 2b. Thus i can draw 3 Triangle which will be a quadrilateral. So sum of the angles will be: (2a+ 90-2a+90-2a)+(2b+90-b+90-b)+(a+b+180-a-b)=360° ⇨a+b=90°
Also you can find special case with circles the same size and line goes directly across them and then is 45 degrees to base tangent so for standardized test would be quick
The triangle formed by the point of contact of the two circles and the other two points of contact with the common tangent, will always be a right-angled triangle, and no matter how you move these chords, a+ b will always be equal to 90. Thank you.
I know i do this with every puzzle (mostly because it's a good exercise for me) but since no specific sizes or angles are given, without loss of generality, you can assume both circles are the same size, and that points A and B are both at the top of the circles. From there, it's extremely clear that each angle must be 45° therefore a+b=90° As always, this assumes a unique solution exists, and that we have enough information to find it.
For the next challenge, the area of the red triangle, visually it looks like the red triangle is composed of four triangles, the blue triangle whose area is 6 and three other triangles whose bases are equal to adjacent sides from the blue triangle as these are all radii of the same circle. The other sides of these three other red triangles are diameters, so they're twice the length of the corresponding side of the blue triangle. Therefore each of these three triangles has the same base and twice the height of the blue triangle. Conclusion, each of the three triangles has twice the area of the blue triangle. Total area of the red triangle = 6 + 3 * 12 = 42.
i believe he is traveling to the pacific in order to have more time 🤣🤣 from an andrew to another happy new year... and happy new year to all non andrews out there 😂🥳
For some reason, I'm hungry for pie right now... You can also use the Inscribed Angle Theorem (also know as the Double Angle Theorem). In the large circle, if you subtend the same arc from the center, the "central" angle will be 2 * a and one of the radii will be vertical, while the other will hit the tangent between the circles along the same line as the radius from the other circle (whose central angle will be 2 * b and will also have a vertical radius). So from the center of the large circle, to the tangent between circles, to the center of the small circle, down to the horizontal line, across to directly under the center of the large circle, to the center of the large circle - this draws out a trapezoid, with interior angles 2a + 2b + 90 + 90 = 360, so a + b = 90.
This one was fairly easy; although I didn't know the chord/tangent eating rule, I drew in some radii and knew the angle between the radii was 2a and 2b respectively, and eventually came to the same triangle of a+(a+b)+b and the same solution. The next one has me stumped though! Look forward to some clever AndyMath solution!
Solved the next one. I don't want to just blurt it out, it's not hard, but solving it will stimulate an insight into triangles at least I'd never considered.
Answer to the next question: I think you might have forgotten to mention that the three sides of the blue triangle correspond to the three radii of the semicircles (from left to right to the bottom: r1, r2, r3) and the three angles = θ1, θ2, θ3 (facing r1, r2, r3, respectively) Blue area = 6 = (sinθ1)(r2)(r3)/2 = (sinθ2)(r3)(r1)/2 = (sinθ3)(r1)(r2)/2 The red triangle is composed of the blue triangle and three triangles that warp around the blue triangle, their areas are (sin(180° - θ2)(2r1)(r3)/2 = sin(θ2)(r1)(r3) = 12 (sin(180° - θ3)(r1)(2r2)/2 = sin(θ3)(r1)(r2) = 12 (sin(180° - θ1)(r2)(2r3)/2 = sin(θ1)(r2)(r3) = 12 So the red triangle = 3 x 12 + 6 = 42
Andy, this math is not required. The subtended angles from the center are 2a and 2b respectively. Connecting the centers and extending their radii to the baseline creates a trapezoid for which the top angles are 2a and 2b, and the two bottom angles are each 90 degrees. From this you can quickly surmise that 2a+2b=180
Aggvent starts with’Agg’, which upside down is a divide slash and 66. 66 divided by 2 is 33, which is how many days Andy really had. So he can complete it by Jan 2. Andy is on track!
Andy, cutting it fine with the deadline. 😮
Reminds me of every math assignment I ever had. :D
Depends on what time zone Andy is in. It's already 2025 here!
new year for me
Living in Asia means I already watched this on first January
I drew a quadrilateral using the tangent points on the baseline and the centres of the circles. The corners on the base line are 90 degrees each, or 180 degrees total, so the other two angles total 180 degrees. Those angles match the arcs the inscribed angles match, so they add up to half of the 180 degrees, or 90 degrees.
I did it by the same method
Exactly what I did.
Yeah, same
Me too
Very clever method. 👏🏻
Andy's gonna pull through - I believe in him 🙏
It's been already New Year here, where I live 👀
The teeth visualization was really nice. I tried to solve the puzzle but quickly realized I understood nothing because I couldn't understand what was going on with the angles and the inscribed angles in a circle.
January is the new December. Let’s go, Andy, three more puzzles! 😊
This one being solvable with no given numbers is crazy. I gasped when i saw where it was going
What I did was construct a quadrilateral by connecting lines to the centre points of each circle. Since radii connecting to any tangent point id 90 degrees, we can construct a big circle radius and a small circle radius to connect to the bottom line. Then, we can construct a line with the big radius and small radius connecting the big centre to the small centre. Again, the line between the two centres will pass directly through the point where the two circles meet. We know that the sum of all interior angles of a quadrilateral is 360 degrees and since we have already constructed 2 right angles, this means the sum of the remaining interior angles is 180 degrees. We can call these two angles x and y and form the expression, x + y = 180. What we will then notice is that both x and y are angles formed when the radii connect to the centre of their respective circles and we realise that a and b are the same lines but connecting to the circumference. From here, we can use the rule that angles found at the circumference are half of the angle at the centre when enclosed by the same arc. This means a + b = (x + y)/2. We can substitute 180 for x + y and this gives us 180/2 = 90. Thus x + y is equal to 90 degrees.
He's so friggin handsome it hurts 🥲
Why would it hurt?
Another way to solve -
In the big circle, if i draw an angle using the same arc in the center,that will be 2a. Similarly, in the small circle, if i draw an angle in the center using the same arc,that will be 2b. Thus i can draw 3 Triangle which will be a quadrilateral. So sum of the angles will be: (2a+ 90-2a+90-2a)+(2b+90-b+90-b)+(a+b+180-a-b)=360°
⇨a+b=90°
Also you can find special case with circles the same size and line goes directly across them and then is 45 degrees to base tangent so for standardized test would be quick
The triangle formed by the point of contact of the two circles and the other two points of contact with the common tangent, will always be a right-angled triangle, and no matter how you move these chords, a+ b will always be equal to 90. Thank you.
I know i do this with every puzzle (mostly because it's a good exercise for me) but since no specific sizes or angles are given, without loss of generality, you can assume both circles are the same size, and that points A and B are both at the top of the circles. From there, it's extremely clear that each angle must be 45° therefore a+b=90°
As always, this assumes a unique solution exists, and that we have enough information to find it.
For the next challenge, the area of the red triangle, visually it looks like the red triangle is composed of four triangles, the blue triangle whose area is 6 and three other triangles whose bases are equal to adjacent sides from the blue triangle as these are all radii of the same circle. The other sides of these three other red triangles are diameters, so they're twice the length of the corresponding side of the blue triangle. Therefore each of these three triangles has the same base and twice the height of the blue triangle. Conclusion, each of the three triangles has twice the area of the blue triangle. Total area of the red triangle = 6 + 3 * 12 = 42.
The solution to the next problem is the same as the answer to the main question of life, the universe and everything. How exciting
i believe he is traveling to the pacific in order to have more time 🤣🤣
from an andrew to another happy new year... and happy new year to all non andrews out there 😂🥳
For some reason, I'm hungry for pie right now... You can also use the Inscribed Angle Theorem (also know as the Double Angle Theorem). In the large circle, if you subtend the same arc from the center, the "central" angle will be 2 * a and one of the radii will be vertical, while the other will hit the tangent between the circles along the same line as the radius from the other circle (whose central angle will be 2 * b and will also have a vertical radius). So from the center of the large circle, to the tangent between circles, to the center of the small circle, down to the horizontal line, across to directly under the center of the large circle, to the center of the large circle - this draws out a trapezoid, with interior angles 2a + 2b + 90 + 90 = 360, so a + b = 90.
It’s now 9:30pm EST. Three more to go. Oh, the anticipation…
Happy new year Andy 🎉
The pitch:
Square hole girl looking at those clips and coming up with simple solution and andy goes "thats right, solve it with triangles"
This one was fairly easy; although I didn't know the chord/tangent eating rule, I drew in some radii and knew the angle between the radii was 2a and 2b respectively, and eventually came to the same triangle of a+(a+b)+b and the same solution.
The next one has me stumped though! Look forward to some clever AndyMath solution!
Three days to go! I didn’t know about the rule you showed. I went to the simplest case of both circles the same size.
Happy New Year, everyone!!
The ball just dropped. Rats!
Hippo Gnu Deer!!!
Happy New Year Andy
Happy new yearrr from the uk
Solved the next one. I don't want to just blurt it out, it's not hard, but solving it will stimulate an insight into triangles at least I'd never considered.
2025 will not begin before this aggvent. I refuse to celebrate!
What time zone are you using?
Happy new years all whose from europe
Happy (20+25)(20+25) Andy!
Answer to the next question:
I think you might have forgotten to mention that the three sides of the blue triangle correspond to the three radii of the semicircles (from left to right to the bottom: r1, r2, r3) and the three angles = θ1, θ2, θ3 (facing r1, r2, r3, respectively)
Blue area = 6 = (sinθ1)(r2)(r3)/2 = (sinθ2)(r3)(r1)/2 = (sinθ3)(r1)(r2)/2
The red triangle is composed of the blue triangle and three triangles that warp around the blue triangle, their areas are
(sin(180° - θ2)(2r1)(r3)/2 = sin(θ2)(r1)(r3) = 12
(sin(180° - θ3)(r1)(2r2)/2 = sin(θ3)(r1)(r2) = 12
(sin(180° - θ1)(r2)(2r3)/2 = sin(θ1)(r2)(r3) = 12
So the red triangle = 3 x 12 + 6 = 42
A lot more references to teeth than I was expecting...but still exciting.
Five hours left (west coast US time) for 3 more problems.
Andy, this math is not required. The subtended angles from the center are 2a and 2b respectively. Connecting the centers and extending their radii to the baseline creates a trapezoid for which the top angles are 2a and 2b, and the two bottom angles are each 90 degrees. From this you can quickly surmise that 2a+2b=180
No arcs were eaten in the making of this video... Right?
Dunno what timezone Andy’s in, but I don’t think he’ll be able to finish the 31 puzzles before 2024 ends, sadly….
But what is the angle for a or b?
Aggvent starts with’Agg’, which upside down is a divide slash and 66.
66 divided by 2 is 33, which is how many days Andy really had.
So he can complete it by Jan 2.
Andy is on track!
I think he might be using Internet Explorer
"We got this" lol
You cutting it close bro.
Wait, what time zone are you in?
Happy new year
Go go go!
3 videos left and 4 hours to go!! Come onnn!!
Yes Andy, you do got this!
A lot to chew on in this one but I think I got it 😃
35 minutes left! 😮
your name ?
2 minutes left, he may not make it 😔
Beautiful! Go on - you got this 100%
Andy, I can see some red marks on your throat - all okay, right? Or is it just me, can anyone else see it?
We got thiss!
You’re the GOAT! You’ve got this!
I love this man more then my own family
Tf 😂😂
Pick up the pace lil bro we’ve only got a few hours 😅left
Thay is a lot of teeth just for fun as the clock ticks down!
First video of 2025 🎉🎉🎉
Goooooo
I BELIEVE
day29
6x5+12=42😊
2:47
i got physically hurt from you saying that the parentheses wasnt doing anything. at least show some remorse
asnwer=25+45=70 isit
asnwer=90 what her
hi
| am indian but you ?
6 + 12 × 3 = 42
You're so smart... First comment