From Putnam 1985 A2:
The problem is attached above. I'm not sure my solution is entirely rigorous, since I solve it by defining a function which I claim is continuous. Would a solution like this earn full marks?
We will let $b_s, b_r, b_t$ denote the bases of $S, R,$ and $T$ respectively. Similarly, $h_s, h_r, h_t$ denote the heights of $S, R, $ and $T$ respectively.
By similar triangles, $$b_s = b_t\left(\frac{h_t-h_s-h_r}{h_t}\right), b_r = b_t\left(\frac{h_t-h_r}{h_t}\right).$$
Then, \begin{align*} \frac{A(R)+A(S)}{A(t)} &= 2\left(\frac{b_rh_r + b_sh_s}{b_th_t}\right) \\ &= \frac{2}{h_tb_t}\left(b_th_r\left(\frac{h_t-h_r}{h_t}\right) + b_th_s\left(\frac{h_t-h_s-h_r}{h_t}\right)\right) \\ &= \frac{2}{h_t^2}\left(h_rh_t - h_r^2 +h_sh_t -h_s^2-h_sh_r \right)\\ &= 2\left(\frac{h_r}{h_t} - \left(\frac{h_r}{h_t}\right)^2 + \frac{h_s}{h_t} - \left(\frac{h_s}{h_t}\right)^2 - \frac{h_sh_r}{h_t^2}\right). \\ \end{align*} Now, define $\gamma_r = \frac{h_r}{h_t}, \gamma_h = \frac{h_s}{h_t}$. Note that the values $\gamma_r, \gamma_h$ run continuously in the interval $[0, 1]$, under the condition that $\gamma_r + \gamma_h \le 1$. Thus, we can define the continuous function $\gamma : R^2 \rightarrow R$ such that $$\gamma(\gamma_s, \gamma_r) = 2(\gamma_s - \gamma_s^2 + \gamma_r-\gamma_r^2 - \gamma_s\gamma_r),$$ corresponding to the ratio $\frac{A(R)+A(S)}{A(T)}$ found above. It suffices to use the multivariate second derivative test for finding maxima/minima.
First, $\nabla \gamma = \vec{0}$ when $\gamma_r = \gamma_s = 1/3$(note that this satisfies the condition that $\gamma_r + \gamma_s < 1$).
Since $$\det H(\gamma) = \frac{\partial^2\gamma}{\partial\gamma_s^2} \frac{\partial^2\gamma}{\partial\gamma_r^2} - \left(\frac{\partial^2\gamma}{\partial\gamma_s\gamma_r}\right)^2 = (-4)^2 - (-2)^2 = 12 > 0,$$ the critical point $(1/3, 1/3)$ corresponds to a relative maximum. This corresponding to the value $\gamma(1/3, 1/3) = 2/3$. Note that this maximum is achievable. Consider an equilateral triangle with height 3, and $h_r = h_s = 1$. In this case, $A(T) = \frac{9}{\sqrt{3}}, A(R) = \frac{4}{\sqrt{3}}, A(S) = \frac{2}{\sqrt{3}}$, which gives $\frac{A(R) + A(S)}{A(T)} = \frac{2}{3},$ as desired.