# 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
3 years, 11 months ago

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Q12

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

$$r_ 2 + r_1 + 200 = 999$$

$$r_1+ r_2 = \boxed{799}$$

- 3 years, 11 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.

$$\Rightarrow 999-r=R+200$$

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

- 3 years, 11 months ago

799

- 3 years, 11 months ago

Why 799???

- 3 years, 11 months ago

499.5 sqrt 2

- 3 years, 11 months ago

799

- 3 years, 6 months ago