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Triangle Circle (Posted on 2004-11-20) Difficulty: 3 of 5
Begin with a right triangle with hypotenuse h.

Inscribe a circle and label its radius, r.

What is the ratio of the area of the circle to that of the triangle?

See The Solution Submitted by SilverKnight    
Rating: 3.0000 (4 votes)

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Solution Ambiguity in Problem & a Simpler Solution | Comment 7 of 8 |
The problem asks one to "begin with a right triangle with hypotenuse h." and "inscribe a circle and label its radius, r." However, because there are an infinite number of right triangles that have hypotenuse h, you cannot inscribe a circle without first making an additional assumption of one side of the triangle (and hence, by Pythagoras' Theorum, the other side is determined). In the stated problem, this assumption of the lengths of legs of the triangle is implicit in the inscribing of the circle (although how that can be done prior to specifying the triangle seems iffy to me).

Therefore, even though np_rt comes up with a simple, correct solution (pi*r/[r+h]), I think Larry's approach (not using r in the answer) is more appropriate.

Moreover, combining np_rt's diagram with Larry's approach yields a result that is even simpler than Larry's.

Let's set up the diagram as np_rt did. Quoting:

>This is going be hard without a picture. Imagine a right triangle ABC where C is the right angle. Let the orientation be that C is at the bottom left and A is at the right and B is at the top. The sides are a, b, and h.

>Draw the inscribed circle along with the radii to the sides of the triangle. The picture can be seen as a circle with 3 pairs of external tangents to the circle. And external tangents originating from the same point are congruent (proved by SSS for congruent triangles).

>For the legs, the radii break them into (r, a-r) and (r, b-r). The hypotenuse is broken into two segments too. Because the external tangents are congruent, their lengths turn out to be a-r and b-r. So we have (a-r)+(b-r)=h or a+b=2r+h.

It is here I break with np_rt by recasting his equation as 2r=a+b-h and combining it with Larry's intermediate result, which is:

2*pi*r/(b+a+h)

Substituting for 2r in the numerator gives us a very simple solution for the ratio of the circle's area to the area of the triangle:

pi*(a+b-h)/(a+b+h)

There's something about the subtraction and addition of the hypotenuse that makes this solution more elegant, IMO.

Finally, I would remove any ambiguity in the problem by restating it as follows:

Begin with a right triangle with legs a and b and hypotenuse h. Inscribe a circle. What is the ratio of the area of the circle to that of the triangle?


  Posted by Steve Cotler on 2004-11-23 16:05:45
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