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Problem: Given a right triangle $ABC$, with right angle at $C$ and side lengths $|AB|=c \ , |BC|=a$ and $|CA|=b,$ let $r$ denote the radius of the inscribed circle. Then it is true that:

a) $r=\dfrac{a+b-c}{2}$

b) $r=\dfrac{c-a-b}{2}$

c) $r=\dfrac{3a+2b-2c}{2}$

d) $r=\dfrac{2c-a-b}{2}$

This is a problem that one should be able to solve by eliminating the wrong answers without doing too much arithmetic and memorising different formulas. What is the most effective way of eliminating the wrong answers? I drew a picture of the problem but I can't really see how to relate $r$ to the sides of the triangle.

I'd also know how to rigorously prove the correct answer, but this is just a curios sidestep.

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4 Answers

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• The answer should be symmetric in $a,b\,$, which eliminates option c).

• The triangle inequality $c \le a+b$ gives a negative $r$ for option b) which eliminates it.

• When one the legs $a \to 0$ the other one $ b \to c\,$, and in that case $r \to 0\,$, which eliminates d).

Or, just use that $\,r = \cfrac{ab}{a+b+c}\,$ by the right triangle area argument, then easily verify a).

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HINT:

Using the relation of in & circum-radius,

$$r=R(\cos A+\cos B+\cos C-1)=R(\sin B+\sin A-\sin C)$$ as $C=90^\circ,\sin C,\cos C=?$

Now use Sine Law

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Here's a trigonograph from an answer to a related question:

$$( a - r ) + ( b - r ) = 2 R$$

(where $R$ is the circumradius). For your question, we can ignore the circumcircle and replace the right-hand side of the relation with $c$.

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There are very useful known identities for a general triangles that bind area $S$, semiperimeter $\rho=\tfrac12(a+b+c)$, inradius $r$ and corresponding angles $\alpha,\beta$ and $\gamma$:

\begin{align} S&=\rho\,r, \\ S&= \rho(\rho-a)\tan\tfrac\alpha2 = \rho(\rho-b)\tan\tfrac\beta2 = \rho(\rho-c)\tan\tfrac\gamma2 \end{align}

For the right triangle we have $\tan\tfrac\gamma2=\tan\tfrac\pi4=1$ and $r=\tfrac12(a+b-c)$ instantly follows.

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