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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

But try the problem first…

Answer:$24$

Source

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