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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.

ACB
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Parseval
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4 Answers4

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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).

dxiv
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  • What do you mean by "symmetric" and how does it eliminate c)? I understood point 2 and 3, excellent explanation! – Parseval May 10 '17 at 07:00
  • @Parseval Thanks. By "symmetric" I meant that right triangles with sides $a,b,c$ and $b,a,c$ are simply reflections of one another, and have the same inradius $r,$, therefore the expression for $r$ must be invariant when you swap the variables $a,b$. But c) is not symmetric in $a,b,$, since $3a + 2b \ne 3b + 2a$ in general. – dxiv May 10 '17 at 07:04
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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

lab bhattacharjee
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  • Should be able to solve this without having to memorize that relation. But thanks, this answers my second curiosity question! – Parseval May 10 '17 at 06:12
  • @Parseval, I think we need to memorize some relations like https://artofproblemsolving.com/community/c6t121f6h1342529s1_relationship_between_inradius_circumradius_and_sides – lab bhattacharjee May 10 '17 at 06:16
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Here's a trigonograph from an answer to a related question:

enter image description here

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

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

Blue
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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.

g.kov
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