AIME 2015 Problem 11

Geometry Level 5

The circumcircle of acute Δ A B C \Delta ABC has center O O . The line passing through point O O perpendicular to O B OB intersects lines A B AB and B C BC and P P and Q Q , respectively. Also A B = 5 AB=5 , B C = 4 BC=4 , B Q = 4.5 BQ=4.5 , and B P = m n BP=\dfrac{m}{n} , where m m and n n are relatively prime positive integers. Find m + n m+n .


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The answer is 23.

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1 solution

Let OD and OE be the perpendicular bisectors of BC and AB resp. B D = 2 , a n d E B = 5 2 . DOQ and OEP are right angles, draw O D Q o n d i a m e t e r Q O a n d P E O o n d i a m e t e r O P r e s p . s Q O B a n d B O P a r e r t . s . O B i s a T A N G E N T t o b o t h s . B O 2 = E B B D = 9 2 2 = 9 , i n O D Q . a n d B O 2 = P B B E = P B 5 2 , i n P E O . P B = 9 5 2 = 18 5 . = m n . m + n = 23 \text{Let OD and OE be the perpendicular bisectors of BC and AB resp. }~ \therefore ~ BD=2, ~and ~EB=\dfrac 52.\\ \because ~ \text{DOQ and OEP are right angles, draw } \bigodot ~ODQ ~ on ~ diameter ~ QO ~ and ~ \bigodot ~PEO ~ on ~ diameter ~ OP ~ resp.\\ \angle ~ s~QOB ~ and ~ BOP ~ are ~ rt. ~ \angle ~s.\\ \therefore ~ OB ~ is ~ a ~ TANGENT ~ to ~ both ~\bigodot ~s.\\ \therefore ~ BO^2=EB*BD=\dfrac 9 2 *2=9, ~ in ~ \bigodot ~ ODQ.\\ and ~ BO^2=PB*BE=PB*\dfrac 5 2, ~ in ~ \bigodot ~PEO.\\ \implies ~ PB=\dfrac 9 {\frac 5 2 }=\dfrac{18} 5.=\dfrac m n.\\ m+n=\Large \color{#D61F06}{23}\\

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