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

#Geometry #Pre-RMO

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Comments

Q12 Q12

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

$r_ 2 + r_1 + 200 = 999$

$r_1+ r_2 = \boxed{799}$

Pranshu Gaba - 6 years, 8 months ago

799

Siddhartha Srivastava - 6 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.

$\Rightarrow 999-r=R+200$

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

Aneesh Kundu - 6 years, 8 months ago

Why 799???

Nitish Deshpande - 6 years, 8 months ago

499.5 sqrt 2

MAYYANK GARG - 6 years, 8 months ago

799

Shrutesh Patil - 6 years, 3 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$