Pre-RMO 2014/12

Let $ABCD$ be a convex quadrilateral with $\angle DAB = \angle BDC = 90^\circ$. Let the incircles of triangles $ABD$ and $BCD$ touch $BD$ at $P$ and $Q$, respectively, with $P$ lying in between $B$ and $Q$. If $AD = 999$ and $PQ = 200$ then what is the sum of the radii of the incircles of triangles $ABD$ and $BDC$?

This note is part of the set Pre-RMO 2014 Note by Pranshu Gaba
5 years ago

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799

- 4 years, 8 months ago

Let $O_1$ and $O_2$ be the incenters of $\bigtriangleup DAB$ and $\bigtriangleup CDB$ respectively. Draw perpendicular from $O_1$ to sides $AD, DB$ and $AB$ at points $S, P$ and $E$ respectively. Draw perpendicular from $O_2$ to sides $CD$ and $DB$ at points $F$ and $Q$ respectively. Let $R$ and $r$ be the inradii of $\bigtriangleup DAB$ and $\bigtriangleup CDB$ respectively.

$SO_1EA$ and $FO_2QD$ are squares with side $r$ and $R$ respectively.

$DS=AD-R=999-r$

$DP=DQ+PQ=R+200$

Since $DP$ and $DS$ are tangents from the same point they are equal in length.

$\Rightarrow 999-r=R+200$

$\Rightarrow R+r=999-200= \boxed{799}$

- 5 years ago Q12

From the figure we can see that $SD = PD = PQ + QD = r_1 + 200$ and $AD = AS + SD =r_ 2 + SD = 999$.

$r_ 2 + r_1 + 200 = 999$

$r_1+ r_2 = \boxed{799}$

- 5 years ago

Why 799???

- 5 years ago

799

499.5 sqrt 2

- 5 years ago