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Indian National Mathematical Olympiad

Geometry Difficulty 6.3 National Olympiad Prove it India

Problem:

In the given figure, $ABCD$ is a square paper. It is folded along $EF$ such that $A$ goes to a point $A'$ ( $A' \neq C, B$ ) on the side $BC$ and $D$ goes to $D'$ . The line $A'D'$ cuts $CD$ in $G$ . Show that the inradius of the triangle $GCA'$ is the sum of the inradii of the triangles $GD'F$ and $A'BE$ .

Solution

Solution:

Observe that the triangles $GCA'$ and $A'BE$ are similar to the triangle $GD'F$ . If $GF = u$ , $GD' = v$ and $D'F = w$ , then we have
$A'G = p u, \quad CG = p v, \quad A'C = p w, \quad A'E = q u, \quad BE = q w, \quad A'B = q v$
If $r$ is the inradius of $\triangle GD'F$ , then $p r$ and $q r$ are respectively the inradii of triangles $GCA'$ and $A'BE$ . We have to show that $p r = r + q r$ . We also observe that
$AE = EA', \quad DF = FD'$
Therefore
$p w + q v = q w + q u = w + u + p v = v + p u$
The last two equalities give $(p-1)(u-v) = w$ . The first two equalities give $(p-q)w = q(u-v)$ . Hence
$\frac{p-q}{q} = \frac{u-v}{w} = \frac{1}{p-1}$
This simplifies to $p(p-q-1) = 0$ . Since $p \neq 0$ , we get $p = q + 1$ . This implies that $p r = q r + r$ .

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