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# Medians of triangle | PRMO-2018 | Problem 10

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

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

## 3 replies on “Medians of triangle | PRMO-2018 | Problem 10” Arnab Chattopadhyay.says:

How do you find that $BC = 20$ so that $BD = DC = 10.$ Arnab Chattopadhyay.says:

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. KOUSHIK SOMsays:

yes…

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