Triangle of Medians

The image below is a "proof without words" that a triangle formed by the medians of another triangle has an area that is $\frac{3}{4}$ that of the other triangle.

Heron's formula gives the area of $\Delta ABC$ as 336, so the area of $\Delta DEF$ would be $\frac{3}{4} \cdot 336 = \boxed{252}$ .

First, hats off to Matt!! Here is a vector solution -

Imgur

Let the triangle be formed by three vectors $\vec{a}, \vec{b}, \vec{c}$ such that $\vec{a} + \vec{b} + \vec{c} = \vec{0}$

Its area will be half the cross product of any two vectors like $\frac{\vec{a} \times \vec{b}}{2}$

Denoting median on vector $\vec{c} = \vec{C} = \frac{\vec{a}-\vec{b}}{2}$

By symmetry, $\vec{A} + \vec{B} + \vec{C} = \vec{0}$ . Thus medians must form a closed polygon (triangle) when placed cyclically!

Area of such a triangle = $\frac{\vec{A} \times \vec{B}}{2} =\frac{1}{4}(\vec{b}-\vec{c})\times(\vec{c}-\vec{a}) = \frac{1}{4}(\vec{b}\times \vec{c} - \vec{c}\times \vec{c} - \vec{b} \times \vec{a} + \vec{c} \times \vec{a})$

Expanding and plugging $\vec{c}\times \vec{c} = \vec{0}$ and $\vec{b}\times\vec{c} = \vec{c}\times\vec{a}=\vec{a}\times \vec{b} = - \vec{b}\times\vec{a}$

Thus the area of median triangle is 3/4 area of origina, which is 336 using Heron's formulal

Thus the answer is $\boxed{252}$

I used Stewart theorem

$m_a^2=\frac 1 2*(b^2+c^2)-\frac 1 4a^2=\frac 1 2*(a^2+b^2+c^2)-\frac 3 4 a^2=1873.\\ m_b^2=\frac 1 2*(a^2+b^2+c^2)-\frac 3 4 b^2=1386.25.\\ m_c^2=\frac 1 2*(a^2+b^2+c^2)-\frac 3 4 c^2=198.25.\\ 2s=m_a+m_b+m_c= 94.5906,\\ Area^2=s(s-m_a)(s-m_b)(s-m_c)=63504.\\ Area=\Large\ \ \ \color{#D61F06}{252}\$

I used Apollonius Theorem on the three sides and calculated the the three sides and then by appyling heron's theorem I calculated the answer. It came nearly about 252.02

It’s kind tricky, here’s how I understand it, The sides of ∆DEF r d red lines (medians) in ∆ABC,

To find the medians: Assume: AB = 24, AC = 35, BC = 53 Median1: 4m^2 = [2x(35)^2] + [2x(53)^2] – 24^2 m = 43.278 Median2: 4m^2 = [2x(24)^2] + [2x(35)^2] – 53^2 m = 14.0801 Median3: 4m^2 = [2x(24)^2] + [2x(53)^2] – 35^2 m = 37.2324

area (∆DEF) = square-root of[s(s-a)(s-b)(s-c)] --->(I think d s d hero’s formula, search it) (s = (a+b+c)/2)

>AREA = 252.00

…hope I calc. it correctly (◠‿ ◠)

In an equilateral triangle side 1, all medians are $\frac{\sqrt{3}}{2}$ long. Triangle made of medians in place of the 1's has therefore area 3/4 that of the area of the original triangle. Triangle ABC in the problem can be obtained from a unit equilateral triangle by a linear transformation. Applying this transformation to the entire plane containing both triangles will change the areas of $\triangle ABC$ and $\triangle DEF$ , but it will not change the ratio of these areas. Therefore area of $\triangle DEF$ will still be 3/4 of the area of $\triangle ABC$ , and 3/4 of 336 is 252.

Mr Enlow's demonstration is, of course, very nice. For those of us who don't mind horrific series of calculations there's Appolonius' theorem, mentioned in https://en.wikipedia.org/wiki/Median_(geometry).