Wizard Joined: Oct 14, 2009
• Posts: 21208
February 16th, 2020 at 12:54:01 AM permalink A and B are semicircles in in larger semicircle enclosing A, B, and circle C. Circle C is tangent to A, B, and the larger semicircle.

The base of A has length 6
The base of B has length 4

What is the radius of C?

Usual rules:
1. Please don't just plop a URL to a solution elsewhere until a winner here has been declared.
2. All those who have won a beer previously are asked to not post answers or solutions for 24 after this posting.
3. Beer to the first satisfactory answer and solution, subject to rule 2.
It's not whether you win or lose; it's whether or not you had a good bet.
OnceDear Joined: Jun 1, 2014
• Posts: 4287
February 16th, 2020 at 2:30:45 AM permalink

Wizard, I've seen this before. Without pre-knowledge of Descartes' work, or google, I can't see anyone solving this on their own.
Take care out there. Spare a thought for the newly poor who were happy in their world just a few days ago, but whose whole way of life just collapsed..
DogHand Joined: Sep 24, 2011
• Posts: 344
February 16th, 2020 at 5:25:50 AM permalink
Wiz,

Here's my solution:

Let's call the centers of the lettered circles Ca, Cb, and Cc, and their intersection points Tac and Tbc. Let r be the unknown radius of C, while the radii of A and B are 3 and 2, respectively.

Now since Tac is tangent to both A and C, we have a straight line passing from Ca through Tac to Cc with a length of 3+r. Similarly, since Tbc is tangent to both B and C, we have another straight line passing from Cb through Tbc to Cc with a length of 2+r. We also have a straight line from Ca to Cb with a length of 2+3=5.

These three lines from a right triangle with legs 2+r and 3+r and hypotenuse 5. From the Pythagorean Formula we get:

(2+r)^2 + (3+r)^2 = 5^2

r^2 +4r + 4 + r^2 + 6r + 9 =25

2r^2 + 10r - 12 = 0

r^2 + 5r - 6 = 0

So r = 1 is the only feasible solution.

Fun problem!

Dog Hand
ssho88 Joined: Oct 16, 2011
• Posts: 358
February 16th, 2020 at 8:23:23 AM permalink
Doghand, For triangle Ca-Cc-Cb, Pythagorean Formula only apply if angle Cc is 90 degree, how to prove that angle Cc is 90 degree ?
Doc Joined: Feb 27, 2010
• Posts: 6955
February 16th, 2020 at 8:24:52 AM permalink
Even though it is almost noon here, my brain doesn't seem to be fully awake, so I have a question for Dog Hand (or others) about Dog Hand's solution. I'll hide it in a spoiler so I don't ruin things for others who want to work it out for themselves.

In Dog Hand's solution, he says, "These three lines from a right triangle with ...."

I interpreted his "from" as a typo for "form." If I interpreted that correctly, how is it established that these lines form a right triangle? How do we know that the Ca-Cc line is perpendicular to the Cb-Cc line?

Edit: Ooops! Someone else asked basically the same thing while I was typing.
ssho88 Joined: Oct 16, 2011
• Posts: 358
February 16th, 2020 at 9:07:33 AM permalink
Quote: Wizard A and B are semicircles in in larger semicircle enclosing A, B, and circle C. Circle C is tangent to A, B, and the larger semicircle.

The base of A has length 6
The base of B has length 4

What is the radius of C?

Usual rules:
1. Please don't just plop a URL to a solution elsewhere until a winner here has been declared.
2. All those who have won a beer previously are asked to not post answers or solutions for 24 after this posting.
3. Beer to the first satisfactory answer and solution, subject to rule 2.

I think can be solved as per attached image :-

https://ibb.co/D1T8vH1

Form 5 equations to solve for 5 unknown p, q, s, m and r ( radius of circle C ). L is center of larger semicircle.

Law of sines :-

1) LC/sin(p) = AL/sin(m)

2) LC/sin(q) = BL/sin(s)

3) AC/sin(q) = BC/sin(p)

4) AC/sin(q+s) = AL/sin(m)

5) p+q+m++s = 180
Last edited by: ssho88 on Feb 16, 2020
Wizard Joined: Oct 14, 2009
• Posts: 21208
February 16th, 2020 at 10:11:27 AM permalink
Quote: OnceDear

Wizard, I've seen this before. Without pre-knowledge of Descartes' work, or google, I can't see anyone solving this on their own.

Yes, this one is rather difficult and I admit I had to cheat myself. I'll give a hint if nobody can solve it without one.
It's not whether you win or lose; it's whether or not you had a good bet.
charliepatrick Joined: Jun 17, 2011
• Posts: 1883
February 16th, 2020 at 10:18:46 AM permalink
Quote: Wizard

Yes, this one is rather difficult and I admit I had to cheat myself. I'll give a hint if nobody can solve it without one.

I'm having fun (perhaps) trying to find an angle of attack without looking up the hint you gave in an earlier spoiler.
Had to go out today so will look at Heron's forumla for area of triangles and seeing if it helps solves for r. Area = SQRT ( S (S-a)(S-b)(S-c) ) where S = (a+b+c)/2 .
OnceDear Joined: Jun 1, 2014
• Posts: 4287
February 16th, 2020 at 10:36:38 AM permalink
I'll throw you a bone.... It boils down to solving a quadratic and does not need the use of trig functions.
Take care out there. Spare a thought for the newly poor who were happy in their world just a few days ago, but whose whole way of life just collapsed..
Wizard Joined: Oct 14, 2009
• Posts: 21208
February 16th, 2020 at 10:42:10 AM permalink
Quote: DogHand

Wiz,

Here's my solution:

Let's call the centers of the lettered circles Ca, Cb, and Cc, and their intersection points Tac and Tbc. Let r be the unknown radius of C, while the radii of A and B are 3 and 2, respectively.

Now since Tac is tangent to both A and C, we have a straight line passing from Ca through Tac to Cc with a length of 3+r. Similarly, since Tbc is tangent to both B and C, we have another straight line passing from Cb through Tbc to Cc with a length of 2+r. We also have a straight line from Ca to Cb with a length of 2+3=5.

These three lines from a right triangle with legs 2+r and 3+r and hypotenuse 5. From the Pythagorean Formula we get:

(2+r)^2 + (3+r)^2 = 5^2

r^2 +4r + 4 + r^2 + 6r + 9 =25

2r^2 + 10r - 12 = 0

r^2 + 5r - 6 = 0

So r = 1 is the only feasible solution.

Fun problem!

Dog Hand

I'm afraid I don't agree. Perhaps you're right and I'm wrong, but I don't think so.
It's not whether you win or lose; it's whether or not you had a good bet.