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Problem based on Triangle | PRMO-2016 | Problem 10

Try this beautiful problem from PRMO, 2016 based on Triangle

Triangle | PRMO | Problem 10

In a triangle ABC right angled at vertex B, a point O is choosen on the side BC such that the circle Y centerer at O of radius OB touches the side AC.Let AB =63 and BC=16 and the radius of Y be the form mn wherw m,n are relatively prime positive integers,find the value of m+n?

• $75$
• $80$
• $71$

Key Concepts

Geometry

Triangle

Circle

Answer:$71$

PRMO-2016, Problem 10

Pre College Mathematics

Try with Hints

Given that ABC is a triangle where AB=63 and BC=16 and $\angle ABC=90^{\circ}$. O is any point on BC.Let OB=$r$=radius of the circle.we have to find out the value of $r$ for $m+n$.

Now $\triangle ABC$ is aright angle triangle at$B$ So using pythagoras theorm $AC=\sqrt {(AB)^2+(BC)^2}=\sqrt{(63)^2+(16)^2}=65$.now $\triangle ODC$ is also a right angle triangle as OD radius and AC tangent,can you find out the value of $r$?

Can you now finish the problem ..........

Given bc=16,Let OB=$r$ then OC=$(16- r)$.we have to find out the value of $r$.if we can show that $\triangle ABC \sim \triangle ODC$ then we can find out the value of r.can you proof $\triangle ABC \sim \triangle ODC$?

In $\triangle ABC$ & $\triangle ODC$,

$\angle ABC=\angle ODC$

$\angle C= \angle C$

Therefore $\triangle ABC \sim \triangle ODC$

so we have ....

$\frac{AC}{OC}=\frac{AB}{OD}=\frac{BC}{AC}$ $\Rightarrow \frac{65}{16-r}=\frac{63}{r}=\frac{16}{65}$

Therefore,$\frac{65}{16-r}=\frac{63}{r}\Rightarrow r=\frac{63}{8}$=$\frac{m}{n}$

Therefore $(m+n)=63+8=71$

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