Try this beautiful problem from Geometry based on medians of triangle
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 \((BC^2 +AC^2+AB^2)/100\)?
Geometry
Medians
Centroid
Answer:$24$
PRMO-2018, Problem 10
Pre College Mathematics
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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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...