Try this beautiful problem from PRMO, 2018 based on Triangles.

## Problem based on Triangles | PRMO | Problem 12

In the triangle, ABC, the right angle at A. The altitude through A and the internal bisector of \(\angle A\) have lengths \(3\) and \(4\), respectively. Find the length of the medium through A?

- \(24\)
- \(30\)
- \(22\)
- \(18\)

**Key Concepts**

Geometry

Triangle

Pythagoras

## Check the Answer

But try the problem first…

Answer:\(24\)

PRMO-2018, Problem 12

Pre College Mathematics

## Try with Hints

First hint

We have to find out the value of \(AM\) .Now we can find out the area of Triangle dividing two parts , area of \(\triangle AMC\) + area of \(\triangle ABM\) ( as M is the mid point of BC)

Can you now finish the problem ……….

Second Step

Now \(AN\) is the internal bisector…Therefore \(\angle NAB=\angle NAC= 45^{\circ}\).Let \(AC=b\) ,\(AB=c\) and \(BC=c\).using this values find out the area of triangles AMC and Triangle ABM

Final Step

Area of \(\triangle ABC\)=\(\frac{1}{2} bc=\frac{1}{2} \times a \times 3\)

\(\Rightarrow bc=3a\)………………….(1)

Now Area of \(\triangle ABN\) + Area of \(\triangle ANC\)=Area of \(\triangle ABC\)

\(\Rightarrow \frac{1}{2} c 4 sin 45^{\circ} +\frac{1}{2} b 4 sin 45^{\circ}=\frac{1}{2} bc\)

\(\Rightarrow b+c =\frac{1}{2\sqrt 2} bc\)

(squarring both sides we get……….)

\(\Rightarrow b^2 +c^2 +2bc=\frac{1}{8} b^2 c^2\)

\(\Rightarrow a^2 +6a=\frac{9}{8} a^2\) (from 1)

\(\Rightarrow a +6 =\frac{9}{8} a\) \((as a \neq 0)\)

\(\Rightarrow a=48\)

\(\Rightarrow AM=BM=MC=\frac{a}{2}=24\)

## Other useful links

- https://www.cheenta.com/roots-of-equations-prmo-2016-problem-8/
- https://www.youtube.com/watch?v=XOrePzJWFiE

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