## Poll

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39 members have voted

Gialmere Joined: Nov 26, 2018
• Posts: 2406
March 2nd, 2021 at 8:35:54 AM permalink
From circles in squares to circles in triangles, this is toughie Tuesday... A circle is inscribed inside an equilateral triangle and an infinite set of circles are nested inside such that each circle touches the previous circle and the edges of the triangle act as tangents.

What fraction of the large red circle do the infinite set of smaller circles represent?
Have you tried 22 tonight? I said 22.
chevy Joined: Apr 15, 2011
• Posts: 98
Thanks for this post from: March 2nd, 2021 at 10:55:30 AM permalink

1) Let red circle have radius R, thus area (pi*R^2)
2) Name some points
A = top of triangle, C = center of red circle, D = center of bottom side of triangle. ACD is a line and bisects the 60 degree angle of the triangle.
B = intersection of ACD with red circle (nearest A)
D= center of left side of triangle.

3) ACD is right triangle and
CD/(AB+BC) = R/(AB+R) = sin30=.5
So AB=R

4)Height of triangle = AB+BC+CD=R+R+R=3R

5) So for a triangle of height 3R, we can fit a circle of size R

6) Draw horizontal tangent to red circle at B (top tip of the triangle). That forms a triangle with height AB=R, so the first inscribed circle must have radius R/3. Then the next triangular tip has height R/3, and inscribed circle has height R/9........

So the set of smaller circles on the top have area
pi*(R/3)^2 + pi*(R/9)^2+....geometric series = pi*(R/3)^2 / (1-1/9) = (pi*R^2)/8

So the three sets of circles have area (3/8) (pi*R^2) = 3/8 Area of red circle

ThatDonGuy Joined: Jun 22, 2011
• Posts: 5452
March 2nd, 2021 at 2:55:54 PM permalink

Let ABC be the triangle, and O the center of the inscribed circle. Assume the radius of the big circle is 1.
Let D and E be the points on AB and BC tangent to the circle. ODB and OEB are congruent (hypotenuse-leg); since ABC = 60 degrees, ODB is a 30-60-90 triangle with OBD = 30 degrees.
Since OD = 1, this means OB = 2.
Let F be the point where the big circle is tangent to the largest of the smaller circles near B. OB = 2 and OF = 1, so FB = 1.
Draw the tangent to the two circles that touch at F, and let G and H be the points where this tangent intersects AB and BC.
BFG is a 30-60-90 triangle with BF = 1 and FBG = 30 degrees, so BG = 2 sqrt(3) / 3.
BGH and BHG are both 60 degrees, so triangles GBH and ABC are similar, with a length ratio of (2 sqrt(3) / 3) / 4 = sqrt(3) / 6.
The area ratio = the square of the length ratio = 1/12.
This process can be continued with each of the other small circles, so in terms of the area of the largest circle, the area of all of the small circles = 3 x (1/12 + (1/12)^2 + (1/12)^3 + ...)
= 3 x 1/12 x (1 + 1/12 + (1/12)^2 + ...)
= 1/4 x (1 / (1 - 1/12))
= 1/4 x 12/11 = 3/11

Wizard Joined: Oct 14, 2009
• Posts: 24506
Thanks for this post from: March 2nd, 2021 at 3:23:43 PM permalink
3/8.

I see I agree with Chevy. If we're both right, please give him credit for being first.

Good problem.
It's not whether you win or lose; it's whether or not you had a good bet.
chevy Joined: Apr 15, 2011
• Posts: 98
March 2nd, 2021 at 4:38:52 PM permalink

Quote: ThatDonGuy

BFG is a 30-60-90 triangle with BF = 1 and FBG = 30 degrees, so BG = 2 sqrt(3) / 3.
BGH and BHG are both 60 degrees, so triangles GBH and ABC are similar, with a length ratio of (2 sqrt(3) / 3) / 4 = sqrt(3) / 6.
The area ratio = the square of the length ratio = 1/12.

If I am following your argument, it seems you are saying ABC has a side of 4?

You have FB=1 (I agree) and the diameter of the circle is 2. So doesn't ABC has HEIGHT of 3, then side of 2sqrt(3) from 30-60-90 triangle???

If so, then your length ratio is (2sqrt(3)/3) / (2sqrt(3)) = 1/3.
Area scales as 1/9.

If so, then it agrees with me and Wiz, if not, I am missing something in your construction.

Wizard Joined: Oct 14, 2009
• Posts: 24506
March 2nd, 2021 at 5:28:07 PM permalink
Here is a sub-question.

A circle is inscribed in an equilateral triangle. What is the ratio of the area of the circle to the triangle?

Also, here is a diagram that may help in discussion. It's not whether you win or lose; it's whether or not you had a good bet.
chevy Joined: Apr 15, 2011
• Posts: 98
March 2nd, 2021 at 6:06:40 PM permalink

Yes, sub question as part of the work done in last question.

ADB is 30-60-90 with DB = R.

As in previous problem
So AD = 2R = ED
And EB = ED+DB=2R + R = 3R

Also
DB/AB=tan30 = 1/sqrt(3)
So AB = sqrt(3)*R

Area of triangle = .5 * EB * AC = .5*EB *(2AB) = 3sqrt(3) R^2

So If you mean a single (red) circle)
Circle/Triangle = pi/(3sqrt(3))

If you meant area of all the circles (red +blue)
All circles area = piR^2 + (3/8) piR^2 =11/8 pi*R^2 ....from previous problem

circle/triangle = 11 pi / (24 sqrt(3))

ThatDonGuy Joined: Jun 22, 2011
• Posts: 5452
March 2nd, 2021 at 6:16:01 PM permalink
Quote: chevy

If I am following your argument, it seems you are saying ABC has a side of 4?

You have FB=1 (I agree) and the diameter of the circle is 2. So doesn't ABC has HEIGHT of 3, then side of 2sqrt(3) from 30-60-90 triangle???

If so, then your length ratio is (2sqrt(3)/3) / (2sqrt(3)) = 1/3.
Area scales as 1/9.

If so, then it agrees with me and Wiz, if not, I am missing something in your construction.

No, you're right. I did have that distance as 4 by mistake.
Gialmere Joined: Nov 26, 2018
• Posts: 2406
March 2nd, 2021 at 8:14:54 PM permalink
Quote: chevy

1) Let red circle have radius R, thus area (pi*R^2)
2) Name some points
A = top of triangle, C = center of red circle, D = center of bottom side of triangle. ACD is a line and bisects the 60 degree angle of the triangle.
B = intersection of ACD with red circle (nearest A)
D= center of left side of triangle.

3) ACD is right triangle and
CD/(AB+BC) = R/(AB+R) = sin30=.5
So AB=R

4)Height of triangle = AB+BC+CD=R+R+R=3R

5) So for a triangle of height 3R, we can fit a circle of size R

6) Draw horizontal tangent to red circle at B (top tip of the triangle). That forms a triangle with height AB=R, so the first inscribed circle must have radius R/3. Then the next triangular tip has height R/3, and inscribed circle has height R/9........

So the set of smaller circles on the top have area
pi*(R/3)^2 + pi*(R/9)^2+....geometric series = pi*(R/3)^2 / (1-1/9) = (pi*R^2)/8

So the three sets of circles have area (3/8) (pi*R^2) = 3/8 Area of red circle

Quote: Wizard

3/8.

I see I agree with Chevy. If we're both right, please give him credit for being first.

Good problem.

Correct!!

Well done. -------------------------------

I was going to tell this joke about infinity...

...but I don't know how it ends.
Have you tried 22 tonight? I said 22.
teliot Joined: Oct 19, 2009
• Posts: 2376
March 3rd, 2021 at 10:21:47 AM permalink
Quote: teliot

Ok, here is a really easy follow-up question. The list in the spoiler gives all six solutions to "sum of the cubes of the digits equals the number."

My follow-up question is to find all integers that minimize the difference when it is greater than 0, that is:

minimize |(sum of cube of digits of number) - number| > 0

There are six solutions when this difference is equal to 0. I found 12 solutions to the next smallest difference.

I have a follow-up question to my follow-up question. And I don't know the answer to this.

We proved above that if N > 0 is an integer and S is the sum of the cubes of its digits, then the difference |N - S| is always divisible by 3. In my computer investigations, I've noticed that every multiple of 3 appears as the difference, at least as far up as I checked. I don't know if the following is true or false:

True or False? Let D >= 0 be divisible by 3. Then there is a positive integer N such that |N -S| = D.

Any takers?
End of the world website: www.climatecasino.net