Somehow, I always think that there's going to be several more steps coming and then I hear Andy say, " and that's the answer to our problem," and it comes as a surprise every single time.
I solved this one by making both circles equal in size (because there is no indication that they can't be equal), and the answer is 135 (90+45, because the triangle formed in the right-hand circle is an isosceles triangle with a 90 degree central angle). You can also arrive at 135 with simple geometry (90+22.5+22.5) by making the left circle as small as possible (touching the left and bottom sides of the rectangle, or should I say square, in this case). It's always 135 no matter the arrangement. But Andy's solution is of course the more elegant one that is true no matter the apparent dimensions.
The day 12 puzzle is simpler than it looks at first. First thing I do is I label the radius of the small circle r and the radius of the semicircle R. Next, draw radii for the small circle and semicricle to one of the edges of the triangle (these will be perpendicular to the edge), and then a perpendicular through the height of the triangle intersects with the centers of both circles creating 2 right triangles that are both 30:60:90 triangles. One has r as it's smaller leg, the other has R as the smaller leg. The hypotenuse of a 30:60:90 triangle is always twice the length of the smaller leg, so 2r and 2R in these cases. However, the hypotenuse of the larger triangle can also be found by adding up the segments making it up which are the hypotenuse of the small triangle (2r) the radius of the small circle (r) and the radius of the semicircle (R). This gives us that 2R = 2r + r + R, subtract R from both sides and combine the other terms gives us R=3r. Now we can use the fact that multiplying the radius of a circle by any scalar will multiply the area of the circle by the square of that scalar, so since R is 3r that means that the area of a circle with radius R will be 9*6=54. Lastly, the area of a semicircle is half the area of the circle, so the semicircle's area is 27.
The puzzle for day 12 at the end of this video: R is the radius of the semicircle (R - r)/2R = r/R R - r = 2r R = 3r Area of the semicircle = 9(r^2)π/2 = 27
This solution is very fast and elegant, but does not use any coordinate geometry concepts, and is a bit hard to come up with. I've come up with an alternative using the coordinate geometry mindset and incorporating concepts from both pure geometry and coordinate geometry. To locate the centres of both circles, I draw the normal lines starting from the tangent points of both circles to the sides. These normal lines then become the radii of both circles, and the intersection points of the radii within a circle is the centre of the circle. I put the origin at the small circle's centre, and manage to find the coordinates of the large circle's centre. Letting the small circle radius = 1 and the large circle radius = n, I extend the horizontal line passing through the origin (x-axis) until it touches the large circle's radius. When both centres are connected together, a right triangle is formed, and the connection line will pass through the tangent point of both circles. The line has length = 1 + n, and the vertical rise of the larger circle's centre = n - 1, i.e., y-coordinate of the centre is n - 1. At this stage, I am still managing to find the x-coordinate of the centre, but I will focus on the inclination of the connection line so that its equation can be built. I denote the inclination = a, then the missing angle = 180° - a - 90° = 90° - a. Here is the point the solution switches to complete pure geometry. The solution will get faster than if I don't switch, and the question focuses on an angle size and does not give any lengths or areas initially. I noticed that the adjacent angles of the "?" angle can be found using the property of isosceles triangle, as they are inside isosceles triangles with angles inside the right triangle (a and 90° - a) as exterior angles. Therefore, I found that the two adjacent angles are a/2 and 45° - a/2 respectively. Finally, using the fact that all the angles on a straight line add u to 180°, I find the "?" angle = 135°. It took me so long to come up with this hybrid solution. I am averse to solving geometry puzzles with only pure geometry and no coordinate geometry, and managed to solve the puzzle only using coordinate geometry concepts and tools. I managed to get the coordinates of all the three points of the angle to find the lengths of its 2 arms, and then find the "?" angle size using the cosine rule or the definition of dot product. However, if I do so, I would be wasting time doing a lot of unnecessary algebraic simplifications. That's why I have to incorporate pure geometry concepts to come up with this accelerated solution instead.
P.S. I know the "squeezing the smaller circle" and "letting the circles be identical" are much faster solutions, as the angle is independent of the sizes of both circles. 😄
If we move the yellow circle's tangent point with the rectangle to the right side instead of the upper side, then the angle would be 180 degrees. As the point was turned by 45 degrees (plus the 135). And two of the tangent lines are parallel, so there's congruence. Still, the chords are not diameters. A new way for me to cut two cakes with one chop. It's funny how triangles and circles can make up intriguing puzzles, when each of them simply consists of three points. Doesn't make my brain feel too smart. 3 dots here, 3 dots there, and then a question about how they go together that one must put in an effort to figure out.
I just realised that the first thing you do, is construct segments and declare some naming to define what you actually looking for. Then the solution unfolds through it
the line in the blue circle can extend to the other side of the black triangle and its the diameter of the black circle draw a line from the top part of the line inside the black circle to the right side of the black circle's diameter, it will make a right triangle a right triangle is made up of 90° + 45° + 45°, the line is 180° so ? = 180° - 45° = 135°
The tangent assumption is very consistent in Catriona Agg puzzles, but usually drawing is a bit more accurate. The leftmost tangent should be perpendicular to the side and above the center of the circle. I did this as a construction in a geometry app but I used the chord perpendicular bisector goes through the center of the circle.
@@KristoforSelden No, the chord in the lefthand circle isn't tangent to the side of the rectangle, it simply touches the point where the side of the rectangle is tangent to the circle (just as the other chord is not tangent to the top of the rectangle, it just touches the point where the top of the rectangle is tangent to the circle). In fact, what you describe is impossible. A chord cannot be perpendicular to the tangent of a circle and not pass through the center. The only way for that chord to be perpendicular to the box is if both circles were the same size, which would make the tangent line between them parallel to the side of the box, making the red triangle he constructs impossible. The answer in that case is still 135 degrees, but a different approach would be necessary for that case.
Does this mean that the ? angle is always 135 degrees for any circle/rectangle configuration (given that they all "touch" each other)? 🤔 That in itself is pretty exciting!
(edited because I made a stupidly obvious mistake pointed out by grizzly01, yes I am back to being a dumbass) 27 easy enough once you realise two things 1, you can put a horizontal line anywhere in the triangle and the new smaller triangle will still be equilateral &2, a circle inscribed in an equilateral triangle has a radius one third of the triangle's height so the semicircle has a radius (R) of half the height of the triangle and the upper half of the triangle is three times the radius (r) of the little circle pi*r^2=6 so r=root(6/pi) so R=3r=3root(6/pi) the area of the big circle(y) is pi*R^2 y=pi*(3root(6/pi))^2 y=pi*9*(6/pi) y=9*6=54 the area of the semicircle is half that, so answer=27
Mostly right, but you've gone wrong with saying that r = √(π/6) which has led you to the wrong answer. The case is that r = √(6/π) and when you get to the final calculation of the area of the semicircle, you'll find that π's will cancel, giving you an integer answer.
For tomorrow's problem I got 27 (Edited because I wrote 54, the area of the whole circle) Maybe there is an easier way to do it but a hint on how I did it: Split the triangle in half and take one half and turn it sideways. You should be able to make two similar triangles to relate the radii of both circles. Also the incenter of a equilateral triangle is at 1/3 of its height.
No. You could arbitrarily increase or decrease the size of the left circle, and nothing about the solution would change, but the sizes of the angles would.
Man how long does it take to make these videos?!, I mean the solving steps, the time it takes, and the whole editing stuff and how you keep up with it Im truly curious!.
Im struggling with something here. If you draw two radii of the large circle to each of the points of tangency, the one on the left has a 45 degree angle with the diagonal line segment, so that means the other radius must also have a 45 degree angle because they form an isosceles triangle. That means it's always a 45-45-90 triangle, regardless of the dimensions of the circle on the left, but the top point of tangency should always be directly above the large circle's center. What am I missing? It seems to be a contradiction.
No. The tangent point looks to be at 270 degrees but it is not as evidenced by the non parallel diameter of the small circle. The angle you refer to is more than 90 degrees.
@@Glenn-o7f Yes, I realize the angle I'm referring to must be more than 90 degrees, but we also just showed that the other two angles in the triangle it forms (from drawing the two radii of the large circle) are each 45 degrees. There must be some error in that.
@@goseigentwitch3105”The one on the left forms a 45 degree angle with the diagonal line”. It does not. Both angles labeled “a” are congruent but not 45 degrees as well.
OK, I solved it by assuming that the puzzle has a unique solution that is independent of the ratio r2/r1, since this is not specified. So, I just set r2 = r1 and solved a very simple geometrical problem. Is this cheating?! I am assuming that the puzzle setter hasn't messed up.
Two boxes? Making up for the ellipse yesterday? How exciting!
How Exciting !🔥🔥🔥
@@charanrenesh9922How exciting! 🔥🔥🔥🗣️🗣️🗣️📣📢🔊🔈🔉✍️
Somehow, I always think that there's going to be several more steps coming and then I hear Andy say, " and that's the answer to our problem," and it comes as a surprise every single time.
I solved this one by making both circles equal in size (because there is no indication that they can't be equal), and the answer is 135 (90+45, because the triangle formed in the right-hand circle is an isosceles triangle with a 90 degree central angle). You can also arrive at 135 with simple geometry (90+22.5+22.5) by making the left circle as small as possible (touching the left and bottom sides of the rectangle, or should I say square, in this case). It's always 135 no matter the arrangement.
But Andy's solution is of course the more elegant one that is true no matter the apparent dimensions.
I took a solid 10 minutes trying this puzzle and had no clue where to even start. This was exciting.
Bro , I am watching ur videos almost every day so I can go from a mathematical looser to an trigonometrical professor
How exciting
Loosely speaking that is. 😀
But still a literary loser!
The day 12 puzzle is simpler than it looks at first. First thing I do is I label the radius of the small circle r and the radius of the semicircle R. Next, draw radii for the small circle and semicricle to one of the edges of the triangle (these will be perpendicular to the edge), and then a perpendicular through the height of the triangle intersects with the centers of both circles creating 2 right triangles that are both 30:60:90 triangles. One has r as it's smaller leg, the other has R as the smaller leg. The hypotenuse of a 30:60:90 triangle is always twice the length of the smaller leg, so 2r and 2R in these cases. However, the hypotenuse of the larger triangle can also be found by adding up the segments making it up which are the hypotenuse of the small triangle (2r) the radius of the small circle (r) and the radius of the semicircle (R).
This gives us that 2R = 2r + r + R, subtract R from both sides and combine the other terms gives us R=3r. Now we can use the fact that multiplying the radius of a circle by any scalar will multiply the area of the circle by the square of that scalar, so since R is 3r that means that the area of a circle with radius R will be 9*6=54. Lastly, the area of a semicircle is half the area of the circle, so the semicircle's area is 27.
Box is back! and with a vengeance ! Doubling down to stablish it's dominance..
Bigger circle fits right into rectangle, but the intersting fact is no matter the size of small circle the angle always remains 135°
The puzzle for day 12 at the end of this video:
R is the radius of the semicircle
(R - r)/2R = r/R
R - r = 2r
R = 3r
Area of the semicircle = 9(r^2)π/2 = 27
Also thank you for using a box and not an ellipse this time ❤
Independent of the circle diameters. Humm. Didn’t expect that How exciting!!
Fascinating. I was sure we didn't have enough information at the beginning!
This solution is very fast and elegant, but does not use any coordinate geometry concepts, and is a bit hard to come up with. I've come up with an alternative using the coordinate geometry mindset and incorporating concepts from both pure geometry and coordinate geometry.
To locate the centres of both circles, I draw the normal lines starting from the tangent points of both circles to the sides. These normal lines then become the radii of both circles, and the intersection points of the radii within a circle is the centre of the circle.
I put the origin at the small circle's centre, and manage to find the coordinates of the large circle's centre. Letting the small circle radius = 1 and the large circle radius = n, I extend the horizontal line passing through the origin (x-axis) until it touches the large circle's radius. When both centres are connected together, a right triangle is formed, and the connection line will pass through the tangent point of both circles. The line has length = 1 + n, and the vertical rise of the larger circle's centre = n - 1, i.e., y-coordinate of the centre is n - 1.
At this stage, I am still managing to find the x-coordinate of the centre, but I will focus on the inclination of the connection line so that its equation can be built. I denote the inclination = a, then the missing angle = 180° - a - 90° = 90° - a.
Here is the point the solution switches to complete pure geometry. The solution will get faster than if I don't switch, and the question focuses on an angle size and does not give any lengths or areas initially.
I noticed that the adjacent angles of the "?" angle can be found using the property of isosceles triangle, as they are inside isosceles triangles with angles inside the right triangle (a and 90° - a) as exterior angles. Therefore, I found that the two adjacent angles are a/2 and 45° - a/2 respectively.
Finally, using the fact that all the angles on a straight line add u to 180°, I find the "?" angle = 135°.
It took me so long to come up with this hybrid solution. I am averse to solving geometry puzzles with only pure geometry and no coordinate geometry, and managed to solve the puzzle only using coordinate geometry concepts and tools. I managed to get the coordinates of all the three points of the angle to find the lengths of its 2 arms, and then find the "?" angle size using the cosine rule or the definition of dot product. However, if I do so, I would be wasting time doing a lot of unnecessary algebraic simplifications. That's why I have to incorporate pure geometry concepts to come up with this accelerated solution instead.
P.S. I know the "squeezing the smaller circle" and "letting the circles be identical" are much faster solutions, as the angle is independent of the sizes of both circles. 😄
If we move the yellow circle's tangent point with the rectangle to the right side instead of the upper side, then the angle would be 180 degrees. As the point was turned by 45 degrees (plus the 135). And two of the tangent lines are parallel, so there's congruence. Still, the chords are not diameters. A new way for me to cut two cakes with one chop.
It's funny how triangles and circles can make up intriguing puzzles, when each of them simply consists of three points. Doesn't make my brain feel too smart. 3 dots here, 3 dots there, and then a question about how they go together that one must put in an effort to figure out.
Whew, I was worried about the stock price of Consolidated Boxes, but they rallied today.
I just realised that the first thing you do, is construct segments and declare some naming to define what you actually looking for. Then the solution unfolds through it
A question involving my favourite polygon on my birthday?? How exciting!
I love this guy.
Never knew i actually like math. How exciting
I love you, Andy
1:20 The first thing came to my mind is that he is *thinking outside the box*
the line in the blue circle can extend to the other side of the black triangle and its the diameter of the black circle
draw a line from the top part of the line inside the black circle to the right side of the black circle's diameter, it will make a right triangle
a right triangle is made up of 90° + 45° + 45°, the line is 180° so ? = 180° - 45° = 135°
It never stated it was a rectangle. How could we assume that? Well, to be fair, we also assumed everything was tangent.
The tangent assumption is very consistent in Catriona Agg puzzles, but usually drawing is a bit more accurate. The leftmost tangent should be perpendicular to the side and above the center of the circle. I did this as a construction in a geometry app but I used the chord perpendicular bisector goes through the center of the circle.
@@KristoforSelden No, the chord in the lefthand circle isn't tangent to the side of the rectangle, it simply touches the point where the side of the rectangle is tangent to the circle (just as the other chord is not tangent to the top of the rectangle, it just touches the point where the top of the rectangle is tangent to the circle). In fact, what you describe is impossible. A chord cannot be perpendicular to the tangent of a circle and not pass through the center.
The only way for that chord to be perpendicular to the box is if both circles were the same size, which would make the tangent line between them parallel to the side of the box, making the red triangle he constructs impossible. The answer in that case is still 135 degrees, but a different approach would be necessary for that case.
So good
Thanks 🎉
What an exciting solution!
Omg I thought this was impossible and you solved it so fast and easy.
We got two boxes. How exciting
you could count up 1, 2, 3 in the beginning to really shake things up
How exciting
Ok Andy, lets make this one quick. I gotta poop.
Does this mean that the ? angle is always 135 degrees for any circle/rectangle configuration (given that they all "touch" each other)? 🤔 That in itself is pretty exciting!
Answer for day 12: 27 sq units
area of semi circle = 27 (if area of yellow circle is 6 )
how?
@@Zlyeak how? exciting
@@Zlyeak Tune in tomorrow to watch Andy solve it.
I also got 27
(edited because I made a stupidly obvious mistake pointed out by grizzly01, yes I am back to being a dumbass)
27
easy enough once you realise two things
1, you can put a horizontal line anywhere in the triangle and the new smaller triangle will still be equilateral
&2, a circle inscribed in an equilateral triangle has a radius one third of the triangle's height
so the semicircle has a radius (R) of half the height of the triangle and the upper half of the triangle is three times the radius (r) of the little circle
pi*r^2=6 so r=root(6/pi)
so R=3r=3root(6/pi)
the area of the big circle(y) is pi*R^2
y=pi*(3root(6/pi))^2
y=pi*9*(6/pi)
y=9*6=54
the area of the semicircle is half that, so
answer=27
Mostly right, but you've gone wrong with saying that r = √(π/6) which has led you to the wrong answer.
The case is that r = √(6/π) and when you get to the final calculation of the area of the semicircle, you'll find that π's will cancel, giving you an integer answer.
@@Grizzly01-vr4pn
...I am dumb
thanks for that
The double box!
Please consider doing a two-fer? Feels like you are a day behind on your calendar.
For tomorrow's problem I got 27 (Edited because I wrote 54, the area of the whole circle)
Maybe there is an easier way to do it but a hint on how I did it:
Split the triangle in half and take one half and turn it sideways. You should be able to make two similar triangles to relate the radii of both circles.
Also the incenter of a equilateral triangle is at 1/3 of its height.
Is there enough info to find out aº and bº on their own?
No. You could arbitrarily increase or decrease the size of the left circle, and nothing about the solution would change, but the sizes of the angles would.
for tomorrow's puzzle, 18 is what I got edit: 27 🤦 I had 9pi/2 *6/pi and cancelled it to 3*6 instead of 3*9
Man how long does it take to make these videos?!, I mean the solving steps, the time it takes, and the whole editing stuff and how you keep up with it
Im truly curious!.
Very clever.
wowwww
how are you assuming that the 2 tangents for the blue circle will intersect?
Any non parallel tangents of a circle will Intersect at a point
and theres the answer, lets put a frame around it
Im struggling with something here.
If you draw two radii of the large circle to each of the points of tangency, the one on the left has a 45 degree angle with the diagonal line segment, so that means the other radius must also have a 45 degree angle because they form an isosceles triangle.
That means it's always a 45-45-90 triangle, regardless of the dimensions of the circle on the left, but the top point of tangency should always be directly above the large circle's center.
What am I missing? It seems to be a contradiction.
No. The tangent point looks to be at 270 degrees but it is not as evidenced by the non parallel diameter of the small circle. The angle you refer to is more than 90 degrees.
@@Glenn-o7f Yes, I realize the angle I'm referring to must be more than 90 degrees, but we also just showed that the other two angles in the triangle it forms (from drawing the two radii of the large circle) are each 45 degrees.
There must be some error in that.
@@goseigentwitch3105”The one on the left forms a 45 degree angle with the diagonal line”. It does not. Both angles labeled “a” are congruent but not 45 degrees as well.
Why is there no excitement in his "how exciting"?
5 th comment how exciting
Did we just prove this true for the general case?
Andy actually did prove it about a year ago I believe in another puzzle. The angle is always 135°.
Bro also makes a video on this topic. Why negative into negative is positive?
Next problem: 27
OK, I solved it by assuming that the puzzle has a unique solution that is independent of the ratio r2/r1, since this is not specified.
So, I just set r2 = r1 and solved a very simple geometrical problem. Is this cheating?! I am assuming that the puzzle setter hasn't messed up.
do you know mathematical legend ramanujan
Wdym 13 minutes ago???
Tomorrow? I'ma say 48,
day12
r:R:1:3→6:A=1:9/2→A=27😊
oh it isnt r:R:1:4
Tomorrow’s ans might be 54
Then again, it might not be, seeing as it's a semicircle, not a circle.... (hint)
@ wait yeah the area would be halved so I guess 27
Hi
FIRST ONE WOOOOOOOOOOOOOOOO
How exciting
There are way too many assumptions