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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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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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