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India Math Olympiad Math Olympiad PRMO

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

Check the Answer


But try the problem first…

Answer:$24$

Source
Suggested Reading

PRMO-2018, Problem 10

Pre College Mathematics

Try with Hints


First hint

Medians of triangle ABC

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

Triangle ABC with medians

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

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

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

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3 replies on “Medians of triangle | PRMO-2018 | Problem 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.

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