Try this beautiful problem from Geometry based on medians of triangle

## Medians of triangle | PRMO | Problem 10

In a triangle ABC, the medians from B to CA is perpendicular to the median from C to AB. If the median from A to BC is 30,determine \((BC^2 +AC^2+AB^2)/100\)?

- $56$
- $24$
- $34$

**Key Concepts**

Geometry

Medians

Centroid

## Check the Answer

But try the problem first…

Answer:$24$

PRMO-2018, Problem 10

Pre College Mathematics

## Try with Hints

First hint

We have to find out \((BC^2 +AC^2+AB^2)/100\). So, we have to find out \(AB, BC, CA\) at first. Now given that, the medians from B to CA is perpendicular to the median from C to AB and the median from A to BC is 30,

So clearly \(\triangle BGC\),\(\triangle BGF\),\(\triangle EGC\) are right angle triangle.Let \(CF=3x\) & \(BE =3y\) then clearly \(CG=2x\) & \(BG= 2y\) given that \(AD=30\) SO \(AG=20\) & \(DG =10\) (as \(G\) is centroid, medians intersects at 2:1). Therefore from pythagoras theorem we can find out \(BC,BF,CE\) i.e we can find out the value \(AB,BC,CA\)

Can you now finish the problem ……….

Second Hint

\(CE^2=(2x)^2+y^2=4x^2 +y^2\)

\(BF^2=(2y)^2+x^2=4y^2+x^2\)

Also, \(CG^2+BG^2=BC^2\) \(\Rightarrow 4x^2 + 4y^2={20}^2\) \(\Rightarrow x^2+y^2=100\)

\(AC^2=(2CE)^2=4(4x^2+y^2)\)

\(AB^2=(2BF)^2=4(4y^2+x^2)\)

Can you finish the problem……..

Final Step

\((BC^2 +AC^2+AB^2)=20(x^2+y^2)+20^2=2400\)

so, \((BC^2 +AC^2+AB^2)/100\)=24

## Other useful links

- https://www.cheenta.com/problem-on-semicircle-amc-8-2013-problem-20/
- https://www.youtube.com/watch?v=7AlfBAPWEMg

## 3 replies on “Medians of triangle | PRMO-2018 | Problem 10”

How do you find that $BC = 20$ so that $BD = DC = 10.$

Oh! I got it. Because median of right triangle drawn from the vertex of right angle to the opposite side is half the length of it’s hypotenuse.

yes…

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