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April 25, 2020

Problem based on Triangles | PRMO-2018 | Problem 12

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


Answer:\(24\)

PRMO-2018, Problem 12

Pre College Mathematics

Try with Hints


Triangle Problem

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 ..........

Figure of the triangle

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

Triangle Problem

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)

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

\(\Rightarrow a=48\)

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

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