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Medians of triangle | PRMO-2018 | Problem 10

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

Answer:$24$

PRMO-2018, Problem 10

Pre College Mathematics

Try with Hints

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 ..........

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

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

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........

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

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

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3 comments 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.

1. KOUSHIK SOM says:

yes...