Side Median Ratio

Geometry Level 3

A D AD , B E BE and C F CF are the medians of the triangle A B C ABC . If A D B E AD \perp BE , then A B C F \dfrac{AB}{CF} can be expressed as p q \dfrac{p}{q} where p p and q q are co-prime positive integers. What is p + q p+q ?


Details and assumptions : The picture isn't drawn accurately.


The answer is 5.

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3 solutions

Mursalin Habib
Apr 19, 2014

We'll need to know the following things for this solution:

1 1 . The Pythagorean Theorem;

2 2 . The fact that the medians of a triangle trisect one another;

3 3 . Apollonius' theorem.


We're going to call the centroid [this is the point where the medians intersect one another] G G for this problem. Also let A B = c AB=c , B C = a BC=a and C A = b CA=b for the sake of convenience.

Now notice that if A D = 3 x AD=3x , then A G = 2 x AG=2x and D G = x DG=x .

Similarly, if B E = 3 y BE=3y , then B G = 2 y BG=2y and E G = y EG=y .

Now apply the Pythagorean Theorem on triangles A G B AGB , B D G BDG and A G E AGE to get the following equations respectively:

c 2 = 4 x 2 + 4 y 2 ( 1 ) c^2=4x^2+4y^2 \cdots (1)

a 2 4 = 4 y 2 + x 2 ( 2 ) \frac{a^2}{4}=4y^2+x^2 \cdots (2)

b 2 4 = 4 x 2 + y 2 ( 3 ) \frac{b^2}{4}=4x^2+y^2\cdots (3)

Add ( 2 ) (2) and ( 3 ) (3) together to get

a 2 + b 2 4 = 5 x 2 + 5 y 2 \frac{a^2+b^2}{4}=5x^2+5y^2

So, a 2 + b 2 = 5 ( 4 x 2 + 4 y 2 ) = 5 c 2 a^2+b^2=5(4x^2+4y^2)=5c^2 [See ( 1 ) (1) ].

Now use Apollonius' theorem on A B C \triangle ABC to get

a 2 + b 2 = 2 ( c 2 4 + C F 2 ) a^2+b^2=2(\dfrac{c^2}{4}+CF^2) .

That means 5 c 2 = c 2 2 + 2 C F 2 5c^2=\dfrac{c^2}{2}+2CF^2

And from that, c C F 2 = 4 9 = 2 3 \dfrac{c}{CF^2}=\sqrt{\dfrac{4}{9}}=\dfrac{2}{3} .

So, p + q = 2 + 3 = 5 p+q=2+3=\boxed{5} and that is our answer.

What is. Apollonius' theorem

Mardokay Mosazghi - 7 years, 1 month ago

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Apollonius' theorem states that if A D AD is a median of A B C \triangle ABC ,

Something Something

The area of the green+ The area of the blue= The area of the red

In other words, 2 B D 2 + 2 A D 2 = A B 2 + A C 2 2BD^2+2AD^2=AB^2+AC^2 .

Hope this helps! More info here .

EDIT: Did you have a solution that doesn't use Apollonius' theorem? I'd like to see it.

Mursalin Habib - 7 years, 1 month ago

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Consider the triangle defined by the coordinates ( 2 , 0 ) (2,0) , ( 0 , 2 ) (0,2) , and ( 2 , 2 ) (-2,-2) . Easy solve? Perhaps. Cheating? Definitely ¨ \ddot\smile

Daniel Liu - 7 years, 1 month ago

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@Daniel Liu I agree.

Mardokay Mosazghi - 7 years, 1 month ago

@Daniel Liu Did the same. Take the axes as the medians and take the triangle isosceles.

Himanshu Arora - 6 years, 11 months ago

I used same method but I never new it was called Apollonius' theorem but thanks now that I have read the link you gave me.

Mardokay Mosazghi - 7 years, 1 month ago

I did it in exactly the same way but made that formula for median on my own and I didn't know that it was called Appolonius Theorem.

Kushagra Sahni - 6 years, 8 months ago
Richard Rodriguez
Apr 30, 2014

Let G be the point where AD meets BE. Then, Triangle AGB is a rectangular triangle. Then, as F is midpoint of AB, GF=AF=BF. By means, AB = 2GF.

On the other hand, CG=2GF (it's a properte of the Medians), then CF = 3GF.

Therefore, AB/CF = 2GF/3GF = 2/3 = p/q. Then, p = 2 and q = 3 ->> p + q = 5

same method mine :)

Nihar Mahajan - 6 years, 7 months ago

Decent one

Satyam Tripathi - 4 years, 6 months ago
Jack D'Aurizio
Apr 22, 2014

If D D is the midpoint of A B AB and G G is the centroid of A B C ABC , since B G A BGA is a right triangle, G D A B = 1 2 \frac{GD}{AB}=\frac{1}{2} . Since C D CD is a median, G D C D = 1 3 \frac{GD}{CD}=\frac{1}{3} , hence the ratio is two thirds.

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