2:10 For a right triangle with c^2 = a^2 + b^2 we always have that the radius r of the incircle is: r = (a+b-c)/2 So 5 = (12u + 5u - 13u)/2 --> 4u = 10 --> u = 2.5
I think this is an easier way to do the hard part of this problem, which is finding the length of the square. Define the point on the large circle where BE is tangent as point F. Define the point on BE that is tangent to the small circle as G. Define the point vertically up from the center of the large circle as H. Define the point to the right of the center of the small circle as I. Define the center of the large circle and small circle as C1 and C2 respectively. Define the length of the side of the square as S. Segment BH=S-12. Segment BI=S-5. By the equality of the two segments tangent to a circle from an exterior point, we also know that BF=S-12 and BG=S-5. Thus BG-BF=7. Now run a line segment between C1 and C2. Create a segment parallel to FB starting at C2 in the upper right direction. Extend segment C1F to the end of this segment. You now have a right triangle with sides from the right angle of 7 and 19. We then use Pythagorean to solve for C1C2=sqrt(2)*13. We note that C1C2 is actually a segment of the diagonal of the square and is at a 45 degree angle. The vertical and horizontal components of C1C2 are both just 1/sqrt(2) of the length of C1C2. Thus those horizontal components are 13 units in length. The side of the square is simply 12+5+ horizontal component of C1C2, thus it is 12+5+13. That gives us the side length = 30. That is the hard part of the problem and the rest is simple circle and square geometry calculations.
I think it's great that you took the time to think about it and type it with all the details Spa Fon. Thanks for this awesome comment. I understand how right I was to start a maths channel when awesome math fans leave these long, complicated comments on my videos. I could impress people just by showing them some of these comments I get. Thanks again. And make sure you check back from time to time.
Let s be the side of the square. Since angle B is a right angle, we have that: atan(12/(s-12))*2 + atan(5/(s-5))*2 = pi/2 Reformulated with complex numbers, we get: arg((s-12) + 12i) * ((s-5) + 5i)) = pi/4 So the real and imaginary part of the calculated product (s^2 - 17s) + i(17s - 120 ) must be the same, so: s^2 - 17s = 17s - 120 s^2 - 34s + 120 = 0 (s - 30)(s-4) = 0 --> s = 30 So the purple area = 900 - 12^2 - 5^2 - 3/4*π*(12^2 + 5^2) = 731 - (507 π)/4
Complex numbers to solve this?? I had to check wikipedia to understand some of the stuff in this comment... Sir, I have no idea who you are but now I feel really cool because I'm running a CZcams channel genius people are watching.
An alternative method to find the side of the square could be this: For the way in which the two circles are positioned inside the square we can say that their centers lie on the diagonal AC. Part of this diagonal is known because we can draw two square with the centers and the sides of the main square, so the first part of the diagonal in the upper side is 12√ 2, the last part of the diagonal in the lower side is 5√ 2, so we have to find the part in the middle between the 2 centers. Setting s = side of the square we can find the segment between the two points of tangency along the line BE in this way: s - 5 - (s - 12) = 7 (applying the two tangent theorem) Now we can draw a right triangle between the two center wich legs are 7 and R+r= 5+12 = 17 and the hypotenuse is: hyp.= √ 7² + 17² = 13√ 2 so the diagonal AC = 12√ 2 + 5√ 2 + 13√ 2 = 30√ 2 and being the diagonal of a square equal to side*√ 2 s = 30 what follows is well shown in the video
Can you explain further the part where you get the segment with s - 5 - (s - 12) = 7? I am having some trouble visualizing this. By the way, s is the side length of the square made by the two circular centers, right? Thanks in advance!
@@ibechane Sorry for the imprecision... s is the side of ABCD square. Calling F the point of tangency of the bigger circle with line BE and G the point of tangency of the smaller circle with BE and O and O' the two centers FG = s-5-(s-12)= 7 (applying the two tangent theorem from point B) now imagine to draw a rectangle extending each radius perpendicular to BE until the center of the other circle. This rectangle has base = sum of the two radii (5+12) and height = FG= 7 the diagonal of this rectangle is the line OO' that joins the two centers, that is what we need. Hope it's a little clearer
@@soli9mana-soli4953 No need to apologize! I'm sure many people could understand your original explanation. But yes, that is very clear to me now with your additional comment. Very cool! Thank you!!
Yup you’re right, those were some ugly numbers. You little guy had more in this one, niiiice. He’s doing so well. This was a very nice solution. Loving the graphics you’re using with the moving shapes. You put real time and effort into these videos. Sincerely hope this takes off for you. Always nice to get some sweet CZcams money in your account.
Oh Ron, it's nice to see your comment. I was starting to get worried that one of my best fans had quit. Thanks to all the supportive comments we received my son is very eager to speak in the maths videos. Oh I do hope my videos will do better. I am working with the most basic hardware and sofware and trying to do my best. I hope everyone enjoys my videos and don't mind the bad recording etc.
@@Maths-Paz Don’t blame him for being eager. A chance to work with dad and make him proud is ever little guys dream. Bad recording? You’re kidding, right? Being too hard on yourself. Nothing wrong with starting from the beginning. A good linear progression starts with the number 1. Ok it doesn’t have to but that would kill my analogy so I’m sticking to it. 😜 No fear, not going anywhere. If I don’t comment I either missed the video or have nothing snappy to say. 🫰
Hi Antony. There is a certain ratio between a triangle's sides and the radius of the incircle. And this ratio depends on how the sides are located vis-a-vis the other two sides i.e. the angles. So for every equilateral triangle the radius of the incircle is side A divided by 2 root 3. For a 3, 4, 5 right triangle the radius of the incircle is always hypotenuse divided by 5. And so on. Every "type" of triangle has its own ratio. And when two triangles are similar they are basically of the same "type" with different but proportional dimensions. That means the ratio between their sides and the radius of the incircle is also inevitably constant. I'm not a math teacher but I tried to explain it based on logic. I hope my explanation was helpful.
2:10
For a right triangle with c^2 = a^2 + b^2 we always have that the radius r of the incircle is:
r = (a+b-c)/2
So 5 = (12u + 5u - 13u)/2 --> 4u = 10 --> u = 2.5
Sir, I didn't even know there was a formula for that 😃 It just looked like something I could do with some algebra. Thanks for this information.
I think this is an easier way to do the hard part of this problem, which is finding the length of the square. Define the point on the large circle where BE is tangent as point F. Define the point on BE that is tangent to the small circle as G. Define the point vertically up from the center of the large circle as H. Define the point to the right of the center of the small circle as I. Define the center of the large circle and small circle as C1 and C2 respectively. Define the length of the side of the square as S.
Segment BH=S-12. Segment BI=S-5. By the equality of the two segments tangent to a circle from an exterior point, we also know that BF=S-12 and BG=S-5. Thus BG-BF=7. Now run a line segment between C1 and C2. Create a segment parallel to FB starting at C2 in the upper right direction. Extend segment C1F to the end of this segment. You now have a right triangle with sides from the right angle of 7 and 19. We then use Pythagorean to solve for C1C2=sqrt(2)*13. We note that C1C2 is actually a segment of the diagonal of the square and is at a 45 degree angle. The vertical and horizontal components of C1C2 are both just 1/sqrt(2) of the length of C1C2. Thus those horizontal components are 13 units in length. The side of the square is simply 12+5+ horizontal component of C1C2, thus it is 12+5+13. That gives us the side length = 30. That is the hard part of the problem and the rest is simple circle and square geometry calculations.
I see soly9mana seems to have come up with this approach also.
I think it's great that you took the time to think about it and type it with all the details Spa Fon. Thanks for this awesome comment. I understand how right I was to start a maths channel when awesome math fans leave these long, complicated comments on my videos. I could impress people just by showing them some of these comments I get. Thanks again. And make sure you check back from time to time.
Let s be the side of the square.
Since angle B is a right angle, we have that:
atan(12/(s-12))*2 + atan(5/(s-5))*2 = pi/2
Reformulated with complex numbers, we get:
arg((s-12) + 12i) * ((s-5) + 5i)) = pi/4
So the real and imaginary part of the calculated product
(s^2 - 17s) + i(17s - 120 )
must be the same, so:
s^2 - 17s = 17s - 120
s^2 - 34s + 120 = 0
(s - 30)(s-4) = 0 --> s = 30
So the purple area = 900 - 12^2 - 5^2 - 3/4*π*(12^2 + 5^2) = 731 - (507 π)/4
Complex numbers to solve this?? I had to check wikipedia to understand some of the stuff in this comment... Sir, I have no idea who you are but now I feel really cool because I'm running a CZcams channel genius people are watching.
@@Maths-Paz
Haha, I just like the problems you present, and the fact you are teaching your kid these skills. Keep up the good work!
An alternative method to find the side of the square could be this:
For the way in which the two circles are positioned inside the square we can say that their centers lie on the diagonal AC. Part of this diagonal is known because we can draw two square with the centers and the sides of the main square, so the first part of the diagonal in the upper side is 12√ 2, the last part of the diagonal in the lower side is 5√ 2, so we have to find the part in the middle between the 2 centers.
Setting s = side of the square
we can find the segment between the two points of tangency along the line BE in this way:
s - 5 - (s - 12) = 7 (applying the two tangent theorem)
Now we can draw a right triangle between the two center wich legs are 7 and R+r= 5+12 = 17
and the hypotenuse is:
hyp.= √ 7² + 17² = 13√ 2
so the diagonal AC = 12√ 2 + 5√ 2 + 13√ 2 = 30√ 2
and being the diagonal of a square equal to side*√ 2
s = 30
what follows is well shown in the video
That looks like a brilliant alternative way! Thank you for this. I never thought the possibility using the diagonals.
@@Maths-Paz I'm glad you like It! ❤️
Can you explain further the part where you get the segment with s - 5 - (s - 12) = 7? I am having some trouble visualizing this. By the way, s is the side length of the square made by the two circular centers, right? Thanks in advance!
@@ibechane Sorry for the imprecision... s is the side of ABCD square. Calling F the point of tangency of the bigger circle with line BE and G the point of tangency of the smaller circle with BE and O and O' the two centers
FG = s-5-(s-12)= 7 (applying the two tangent theorem from point B)
now imagine to draw a rectangle extending each radius perpendicular to BE until the center of the other circle. This rectangle has base = sum of the two radii (5+12) and height = FG= 7
the diagonal of this rectangle is the line OO' that joins the two centers, that is what we need. Hope it's a little clearer
@@soli9mana-soli4953 No need to apologize! I'm sure many people could understand your original explanation. But yes, that is very clear to me now with your additional comment. Very cool! Thank you!!
Great job as always!
Thanks for the comment. I appreciate that.
Yup you’re right, those were some ugly numbers. You little guy had more in this one, niiiice. He’s doing so well. This was a very nice solution. Loving the graphics you’re using with the moving shapes. You put real time and effort into these videos. Sincerely hope this takes off for you. Always nice to get some sweet CZcams money in your account.
Oh Ron, it's nice to see your comment. I was starting to get worried that one of my best fans had quit. Thanks to all the supportive comments we received my son is very eager to speak in the maths videos. Oh I do hope my videos will do better. I am working with the most basic hardware and sofware and trying to do my best. I hope everyone enjoys my videos and don't mind the bad recording etc.
@@Maths-Paz
Don’t blame him for being eager. A chance to work with dad and make him proud is ever little guys dream. Bad recording? You’re kidding, right? Being too hard on yourself. Nothing wrong with starting from the beginning. A good linear progression starts with the number 1. Ok it doesn’t have to but that would kill my analogy so I’m sticking to it. 😜
No fear, not going anywhere. If I don’t comment I either missed the video or have nothing snappy to say. 🫰
A math problem a day adds knowledge to the brain 🧠
Great video! Keep it up! 😃👍🏻🥜
Thanks a lot for the support Mellence!
Can someone explain to me why the ratio of similitude between the triangles AZB and BEC is equal to the ratio between the radii of their incircles?
Hi Antony. There is a certain ratio between a triangle's sides and the radius of the incircle. And this ratio depends on how the sides are located vis-a-vis the other two sides i.e. the angles. So for every equilateral triangle the radius of the incircle is side A divided by 2 root 3. For a 3, 4, 5 right triangle the radius of the incircle is always hypotenuse divided by 5. And so on. Every "type" of triangle has its own ratio. And when two triangles are similar they are basically of the same "type" with different but proportional dimensions. That means the ratio between their sides and the radius of the incircle is also inevitably constant.
I'm not a math teacher but I tried to explain it based on logic. I hope my explanation was helpful.
@@Maths-Paz ❤️