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Thread: 3 Circles whose radii form a Triangle

  1. #1
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    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.

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

    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.
    Attached Thumbnails Attached Thumbnails 3 Circles whose radii form a Triangle-triangle-formed-circles.png  
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  2. #2
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    Hello, D. Martin!

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


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

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


    Drag your cursor between the asterisks . . .

    *

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

    *


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  3. #3
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    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.
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