Squared Medians

We begin by asserting a known, and easy to prove, theorem of parallelograms. The sum of the squares of the sides of a parallelogram is equal to the sum of the squares of the diagonals. The proof follows easily from the use of the law of cosines and the fact that the cosines of adjacent angles of the parallelogram are opposites of each other, one positive and one negative.

For any triangle, $ABC$ , with opposite sides $a$ , $b$ , and $c$ ; and median $AD$ which we will call $m$ . If we make a congruent copy of $ABC$ and rotate it 180 degrees around $D$ , the midpoint of segment $c$ , we get a parallelogram and its two diagonals. We observe that the diagonal $AD$ has length $2m$ , and the other diagonal has length $a$ . From the previously stated theorem it is then clear that : $2b^2+2c^2=a^2+(2m)^2$ , which simply leads to the result $m_a=\frac {(2b^2+2c^2-a^2)^{1/2} }{2}$ . The other two medians can easily be calculated by rotating the variables $a,b$ , and $c$ . Squaring all the 3 equations for the medians, and rearranging , we get $m_a^2+m_b^2+m_c^2=\frac {3}{4} (a^2+b^2+c^2)$ . Therefore, we get our desired relation .

Now we can substitute the values of $a,b$ , and $c$ and our final result will be 303.

Sketch out the triangle ABC.

Then \(\cos A=\frac{AB^{2}+AC^{2}-BC^{2}}{2 \cdot AB \cdot AC} = 29/56 \(\cos B=\frac{AB^{2}+BC^{2}-AC^{2}}{2 \cdot AB \cdot BC} = 1/16 \(\cos C=\frac{AC^{2}+BC^{2}-AB^{2}}{2 \cdot AC \cdot BC} = 1/16

And apply Cosine Rule for each separate triangles ABD, BCE, AFC

\(CF^{2}=AF^{2}+AC^{2} - 2 \cdot AF \cdot AC \cdot cosA \) $AD^{2}=AB^{2}+BD^{2} - 2 \cdot AB \cdot BD \cdot cosB$ $BE^{2}=BF^{2}+EF^{2} - 2 \cdot BF \cdot EF \cdot cosC$

Add the 3 equations up, you'll end up with $303$ .

Hope this helps!

let median from vertex A be represented by m1 let median from vertex B be represented by m2 let median from vertex C be represented by m3 and sides be represented by a,b,c

By Apollonius theorem, the sum of the squares of any two sides of any triangle equals twice the square on half the third side, together with twice the square on the median bisecting the third side"

So, (m1)^2={(2b^2+2c^2-a^2)}/4 (m2)^2={(2a^2+2c^2-b^2)}/4 (m3)^2={(2a^2+2b^2-c^2)}/4

hence required sum={3(a^2+b^2+c^2)}/4

We prove a general result about the length of a median using Stewart's Theorem. Let the sides opposite vertices A,B, and C respectively be a,b, and c.

Then we have c^2(a/2)+b^2(a/2)=(a/2)^2(a)+x^2(a), where x is the length of the median opposite side A. Rearranging this gives x^2=(2b^2+2c^2-a^2)/4. Solving for AD^2, BE^2, and CF^2 gives 55, 94, and 154, for a sum of 303.

We have that AD , BE and CF are medians of the given triangle.

Applying the law of cosines on triangle ADC, we have: AD ^2 + DC ^2 - 2cos(ADC). AD . DC = AC ^2

On triangle ADB, we have: AD ^2 + DB ^2 -2cos(ADB). AD . DB = AB ^2

As ADC + ADB = 180º, cos(ADB) = -cos(ADC) And, as AD is median to BC , BD = CD

Summing the two equations, we have: 2. AD ^2 + 2. DC ^2 + = AB ^2 + AC ^2

We can analogously find that:

2.CF^2 + 2.FB^2 = AE^2 + BA^2 and

2.BE^2 + 2.AE^2 = BA^2 + BC^2

Naming the sum we want S, we have

2S + 2(FB^2 + AE^2 + DC^2) = 2(AC^2 + BC^2 + AB^2)

As E, F and D are midpoints,

FB^2 + AE^2 + DC^2 = 1/4.(AC^2 + BC^2 +AB^2)

Therefore, S = 3/4.(AC^2 + BC^2 + AB^2)

S = 303

$AD=\frac{\sqrt{2\,{b}^{2}+2\,{c}^{2}-{a}^{2}}}{2}$ $BE=\frac{\sqrt{2\,{a}^{2}+2\,{c}^{2}-{b}^{2}}}{2}$ $CF=\frac{\sqrt{2\,{a}^{2}+2\,{b}^{2}-{c}^{2}}}{2}$ {AD}^{2}+{BE}^{2}+{CF}^{2}=\frac{2\,{b}^{2}+2\,{c}^{2}-{a}^{2}}{4}+\frac{2\,{a}^{2}+2\,{c}^{2}-{b}^{2}}{4}+\frac{2\,{a}^{2}+2\,{b}^{2}-{c}^{2}}{4} =\frac{3\,{a}^{2}+3\,{b}^{2}+3\,{c}^{2}}{4} =303

Stewart's Theorem states that for any cevian to the side a in a triangle, its length d relates to the length of the sides of the triangle a , b , c and to the length of the sections m , n of a as \­( d^2 = \frac{b^2m + c^2n - an^2}{a} \­). A median is a special case of a cevian with n = m = a/2 . From here, we simply apply the theorem to every median in our triangle and add the squared lengths up.

SHOULDN'T HAVE BEEN LEVEL 4

Stewart's theorem applied to $AD$ being the median (i.e. Apollonius' theorem) states that $AD^2 = \frac { AB^2 + AC^2 - 2 BD^2} { 2}$ . Adding up the 3 equations ( for $AD, BE, CF$ ) gives us $AD^2+BE^2+CF^2 = \frac {3(AB^2+BC^2+CA^2)} {4}$ .

Hence, the expression is equal to $\frac { 3 ( 8 ^2 + 12 ^2 + 14 ^2) } { 4} = 303$ .