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India Math Olympiad Math Olympiad PRMO

Medians | Geometry | PRMO-2018 | Problem 13

Try this beautiful problem from PRMO, 2018 based on Geometry. You may use sequential hints to solve the problem.

Try this beautiful Geometry problem from PRMO, 2018 based on Medians.

Medians | PRMO | Problem-13


In a triangle ABC, right-angled at A, the altitude through A and the internal bisector of \(\angle A \) have lengths \(3\) and \(4\), respectively. Find the length of the median through \(A\).

  • $20$
  • $24$
  • $13$

Key Concepts


Geometry

Medians

Triangle

Check the Answer


Answer:\(24\)

PRMO-2018, Problem 13

Pre College Mathematics

Try with Hints


medians of a triangle

Now in the Right angle Triangle ABC ,\(\angle A=90\).\(AD=3\) is Altitude and \(AE=4\) is the internal Bisector and \(AF\) is the median.Now we have to find out the length of \(AF\)

Now \(\angle CAE = 45^{\circ }= \angle BAE\).

Let\( BC = a\), \(CA = b\), \(AB = c\)
so \(\frac{bc}{2}=\frac{3a}{2}\) \(\Rightarrow bc=3a\)

Can you now finish the problem ……….

a triangle with medians

Therefore,

\(\frac{2bc}{b+c} cos\frac{A}{2}=4\)

\(\Rightarrow \frac{6a}{b+c} .\frac{1}{\sqrt 2}=4\)

\(\Rightarrow 2\sqrt 2 (b+c)=3a\)

\(\Rightarrow 8(b^2 + c^2 + 2bc) = 9a^2\)

\(\Rightarrow 8(a^2 + 6a) = 9a^2\)

\(\Rightarrow 48a=a^2\)

\(\Rightarrow a=48\)

Can you finish the problem……..

Now \(AF\) is the median , \(AF=\frac{a}{2}=24\)

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