Thread: 3 Circles whose radii form a Triangle

1. 3 Circles whose radii form a Triangle

The Problem: In the diagram, the circles with centres P, Q, and R hav radii of 3, 2, and 1 respectively. Each circle touches the other two as shown. Find the area of triangle PQR.

I used the law of cosines to find angle P.

$c^2 = a^2 + b^2 - 2ac(cos \theta)$
$3^2 = 4^3 + 5^2 -2(4)(5)(cos P)$
$9 = 41 -40 (cos P)$
$-32 = -40 Cos P$
$-32/-40 = Cos P$
$0.8 = Cos P$
$0.8 (inverse cos) = P$
$P = 36.87^o$

I then thought of using the sine or cosine formula to find the height (I thought of drawing a height from the centre R to where the circles containing P and Q touch). But I don't know that it's a right-angle triangle. I am stuck at this point.

2. Hello, D. Martin!

This is a problem from the Dirty Tricks division.
. . Warning: It's a groaner!

In the diagram, the circles with centres $P, Q,R$ have radii of 3, 2, and 1, resp.
Each circle touches the other two as shown.
Find the area of $\Delta PQR.$

The answer is SO awful that I'll hide it . . .

Drag your cursor between the asterisks . . .

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The centers form a 3-4-5 right triangle!

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3. Thanks, but is it a right angle triangle because two lines two a circle centre result in a right angle iFF they both come from adjacent circles or something like that? I know there's probably a simple rule that explains it, but I can't quite put my finger on it.

But, anyways, thank you a lot for the help. I've always considered myself horrible at spatial questions.