Pre-RMO 2014/12

Let ABCDABCD be a convex quadrilateral with DAB=BDC=90\angle DAB = \angle BDC = 90^\circ. Let the incircles of triangles ABDABD and BCDBCD touch BDBD at PP and QQ, respectively, with PP lying in between BB and QQ. If AD=999AD = 999 and PQ=200PQ = 200 then what is the sum of the radii of the incircles of triangles ABDABD and BDCBDC?


This note is part of the set Pre-RMO 2014

Note by Pranshu Gaba
5 years ago

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799

Shrutesh Patil - 4 years, 8 months ago

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Let O1O_1 and O2O_2 be the incenters of DAB \bigtriangleup DAB and CDB \bigtriangleup CDB respectively. Draw perpendicular from O1O_1 to sides AD,DBAD, DB and ABAB at points S,PS, P and EE respectively. Draw perpendicular from O2O_2 to sides CDCD and DBDB at points FF and QQ respectively. Let RR and rr be the inradii of DAB \bigtriangleup DAB and CDB \bigtriangleup CDB respectively.

SO1EA SO_1EA and FO2QDFO_2QD are squares with side rr and RR respectively.

DS=ADR=999rDS=AD-R=999-r

DP=DQ+PQ=R+200DP=DQ+PQ=R+200

Since DPDP and DSDS are tangents from the same point they are equal in length.

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

R+r=999200=799 \Rightarrow R+r=999-200= \boxed{799}

Aneesh Kundu - 5 years ago

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Q12 Q12

From the figure we can see that SD=PD=PQ+QD=r1+200SD = PD = PQ + QD = r_1 + 200 and AD=AS+SD=r2+SD=999AD = AS + SD =r_ 2 + SD = 999.

r2+r1+200=999r_ 2 + r_1 + 200 = 999

r1+r2=799r_1+ r_2 = \boxed{799}

Pranshu Gaba - 5 years ago

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Why 799???

Nitish Deshpande - 5 years ago

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799

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499.5 sqrt 2

MAYYANK GARG - 5 years ago

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