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

## Check the Answer

But try the problem first…

Answer:\(71\)

PRMO-2016, Problem 10

Pre College Mathematics

## Try with Hints

First hint

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 ……….

Second Step

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 \(\triangle ABC \sim \triangle ODC\)?

Second Step

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 \((m+n)=63+8=71\)

## Other useful links

- https://www.youtube.com/watch?v=PfRqs9W8nPQ
- https://www.cheenta.com/problem-based-on-integer-prmo-2018-problem-4/

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