Track / Stage 4 / 282 of 340 #542 of 1964

Problem 542

AMC 12 late, AIME early

Geometry Difficulty 4.9 Find the answer

6. Let $a$ , $b$ , $c$ and $m_{a}$ , $m_{b}$ , $m_{c}$ represent the lengths of the three sides and the lengths of the medians to these sides of $\triangle ABC$ , respectively. Given the following conclusions:
(1) $m_{a}+m_{b}+m_{c} \leq \frac{3}{4}(a+b+c)$ ;
(4) $m_{a}^{2}+m_{b}^{2}+m_{c}^{2}=\frac{3}{4}\left(a^{2}+b^{2}+c^{2}\right)$ .
Among these, the correct ones are:

Pick one

Official solution

6.C.

In Figure 6, let the three medians of $\triangle A B C$ be $A D=m_{a}$ ,
$B E=m_{b}, C F=m_{c},$

and they intersect at the centroid $G$ .
Then
$\begin{array}{l} G A+G B>A B, \\ G B+G C>B C, \\ G C+G A>C A . \end{array}$

Adding the three inequalities, we get
$\begin{array}{l} 2(G A+G B+G C) \\ >A B+B C+C A=a+b+c . \end{array}$

It is easy to see that $G A=\frac{2}{3} m_{a}, G B=\frac{2}{3} m_{b}, G C=\frac{2}{3} m_{c}$ .
Therefore, $m_{a}+m_{b}+m_{c}>\frac{3}{4}(a+b+c)$ .
Hence (3) is true.
Noting that $m_{a}^{2}=\frac{1}{2}\left(c^{2}+b^{2}\right)-\frac{a^{2}}{4}$ ,
$\begin{array}{l} m_{b}^{2}=\frac{1}{2}\left(c^{2}+a^{2}\right)-\frac{b^{2}}{4}, \\ m_{c}^{2}=\frac{1}{2}\left(a^{2}+b^{2}\right)-\frac{c^{2}}{4}, \end{array}$

Therefore, $m_{a}^{2}+m_{b}^{2}+m_{c}^{2}=\frac{3}{4}\left(a^{2}+b^{2}+c^{2}\right)$ . Hence (4) is true.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.