Putnam 2004, Problem A-2

For $i = 1, 2$ let $T_i$ be a triangle with side lengths $a_i, b_i, c_i$, and area $A_i$. Suppose that $a_1 \leq a_2, b_1 \leq b_2, c_1 \leq c_2$, and that $T_2$ is an acute triangle. Does it follow that $A_1 \leq A_2$?

Solution:
**First solution:** (partly due to Ravi Vakil) Yes, it does follow. For $i = 1,2$, let $P_i,Q_i,R_i$ be the vertices of $T_i$ opposite the sides of length $a_i,b_i,c_i$, respectively.

We first check the case where $a_1 = a_2$ (or $b_1 = b_2$ or $c_1 = c_2$, by the same argument after relabeling). Imagine $T_2$ as being drawn with the base $Q_2R_2$ horizontal and the point $P_2$ above the line $Q_2R_2$. We may then position $T_1$ so that $Q_1 = Q_2$, $R_1 = R_2$, and $P_1$ lies above the line $Q_1R_1 = Q_2R_2$. Then $P_1$ also lies inside the region bounded by the circles through $P_2$ centered at $Q_2$ and $R_2$. Since $\angle Q_2$ and $\angle R_2$ are acute, the part of this region above the line $Q_2R_2$ lies within $T_2$. In particular, the distance from $P_1$ to the line $Q_2R_2$ is less than or equal to the distance from $P_2$ to the line $Q_2R_2$; hence $A_1 \leq A_2$.

To deduce the general case, put
$$r = \max\{a_1/a_2, b_1/b_2, c_1/c_2\}.$$

Let $T_3$ be the triangle with sides $ra_2, rb_2, rc_2$, which has area $r^2A_2$. Applying the special case to $T_1$ and $T_3$, we deduce that $A_1 \leq r^2A_2$; since $r \leq 1$ by hypothesis, we have $A_1 \leq A_2$ as desired.

**Remark:** Another geometric argument in the case $a_1 = a_2$ is that since angles $\angle Q_2$ and $\angle R_2$ are acute, the perpendicular to $Q_2R_2$ through $P_2$ separates $Q_2$ from $R_2$. If $A_1 > A_2$, then $P_1$ lies above the parallel to $Q_2R_2$ through $P_2$; if then it lies on or to the left of the vertical line through $P_2$, we have $c_1 > c_2$ because the inequality holds for both horizontal and vertical components (possibly with equality for one, but not both). Similarly, if $P_1$ lies to the right of the vertical, then $b_1 > b_2$.

**Second solution:** (attribution unknown) Retain notation as in the first paragraph of the first solution. Since the angle measures in any triangle add up to $\pi$, some angle of $T_1$ must have measure less than or equal to its counterpart in $T_2$. Without loss of generality assume that $\angle P_1 \leq \angle P_2$. Since the latter is acute (because $T_2$ is acute), we have $\sin \angle P_1 \leq \sin \angle P_2$. By the Law of Sines,
$$A_1 = \frac{1}{2}b_1c_1\sin \angle P_1 \leq \frac{1}{2}b_2c_2\sin \angle P_2 = A_2.$$

**Remark:** Many other solutions are possible; for instance, one uses Heron's formula for the area of a triangle in terms of its side lengths.