United States competition problems

Solution:

Let equal medians $AD$ and $BE$ in triangle $ABC$ meet at $F$. It is well known that
$$AF : FD = BF : FE = 2 : 1.$$
Triangle $ABC$ is isosceles if we can show $AC = BC$ or equivalently, that $AE = BD$. But triangles $AFE$ and $BFD$ are congruent with vertical angles plus sides that are $1/3$ and $2/3$ of the equal median length.

Solution:

Let medians $AM = BN$ in $\triangle ABC$. Extend each median to $AM_1$ and $BN_1$ so that $M$ and $N$ are the midpoints of $AM_1$ and $BN_1$, respectively. By the property of bisecting diagonals, $ABM_1C$ and $ABC N_1$ are parallelograms. Hence $CM_1$ and $CN_1$ are each parallel and equal to $AB$. We conclude that $C$ lies on $N_1M_1$, $C$ is the midpoint of $N_1M_1$, and $AM_1 = BN_1$ as they are twice the lengths of the original medians $AM$ and $BN$. Summarizing, $ABM_1N_1$ is a trapezoid with equal diagonals.

It is easy to see that such a trapezoid is isosceles. One way to see this is to draw a line through $A$ parallel to diagonal $BN_1$, until it intersects line $N_1M_1$ in point $L$. Thus, $ABN_1L$ is a parallelogram, so $\angle ALN_1 = \angle ABN_1$. On the other hand, $\triangle AM_1L$ is isosceles since $AL = BN_1 = AM_1$; hence, $\angle ALN_1 = \angle AM_1N_1$. Finally, $AB \parallel N_1M_1$ implies $\angle AM_1N_1 = \angle BAM_1$. We conclude that $\angle BAM_1 = \angle ABN_1$, and $\triangle ABN_1$ and $\triangle BAM_1$ are congruent by two equal sides and angles between these sides. Therefore, $BM_1 = AN_1$ and our trapezoid is isosceles. Hence $\angle AN_1C = \angle BM_1C$.

Finally, $\triangle ACN_1$ and $\triangle BCM_1$ are congruent by $AN_1 = BM_1$, $CN_1 = CM_1$ and $\angle AN_1C = \angle BM_1C$. We conclude that $AC = BC$ and our original $\triangle ABC$ is also isosceles.

Solution:

As a variation of the above solution, note that $NM$ is the midsegment of $\triangle ABC$, and as such it is parallel to $AB$. Thus $ABMN$ is a trapezoid with equal diagonals, which by a similar argument as in Solution 1 is isosceles. Therefore, $\angle BAC = \angle ABC$ and $AC = BC$.

Solution:

A well-known formula for a parallelogram $ABM_1C$ says: $2(AB^2 + AC^2) = AM_1^2 + BC^2$ (it can be easily proved with vectors for example). From here one derives a formula for the median $AM$ of a triangle $\triangle ABC$:
$$AM^2 = \frac{1}{2}(AB^2 + AC^2) - \frac{1}{4} BC^2.$$
Similarly, the other median $BN$ in $\triangle ABC$ satisfies:
$$BN^2 = \frac{1}{2}(AB^2 + BC^2) - \frac{1}{4} AC^2.$$
Since $AM = BN$, easy algebraic cancellations lead to $AC^2 = BC^2$, i.e. $AC = BC$ and our triangle is isosceles.