What is the NO-SHORTCUT approach for learning great Mathematics?

# How to Pursue Mathematics after High School?

For Students who are passionate for Mathematics and want to pursue it for higher studies in India and abroad.

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

## What to do to shape your Career in Mathematics after 12th?

From the video below, let's learn from Dr. Ashani Dasgupta (a Ph.D. in Mathematics from the University of Milwaukee-Wisconsin and Founder-Faculty of Cheenta) how you can shape your career in Mathematics and pursue it after 12th in India and Abroad. These are some of the key questions that we are discussing here:

• What are some of the best colleges for Mathematics that you can aim to apply for after high school?
• How can you strategically opt for less known colleges and prepare yourself for the best universities in India or Abroad for your Masters or Ph.D. Programs?
• What are the best universities for MS, MMath, and Ph.D. Programs in India?
• What topics in Mathematics are really needed to crack some great Masters or Ph.D. level entrances?
• How can you pursue a Ph.D. in Mathematics outside India?
• What are the 5 ways Cheenta can help you to pursue Higher Mathematics in India and abroad?

## Want to Explore Advanced Mathematics at Cheenta?

Cheenta has taken an initiative of helping College and High School Passout Students with its "Open Seminars" and "Open for all Math Camps". These events are extremely useful for students who are really passionate for Mathematic and want to pursue their career in it.

To Explore and Experience Advanced Mathematics at Cheenta

### 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...

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