# 1960 IMO Problems/Problem 4

## Problem

Construct triangle $ABC$, given $h_a$, $h_b$ (the altitudes from $A$ and $B$), and $m_a$, the median from vertex $A$.

## Solution

Let $M_a$, $M_b$, and $M_c$ be the midpoints of sides $\overline{BC}$, $\overline{CA}$, and $\overline{AB}$, respectively. Let $H_a$, $H_b$, and $H_c$ be the feet of the altitudes from $A$, $B$, and $C$ to their opposite sides, respectively. Since $\triangle ABC\sim\triangle M_bM_aC$, with $M_bM_a=\frac12 AB$, the distance from $M_a$ to side $\overline{AC}$ is $\frac{h_b}{2}$.

Construct $AM_a$ with length $m_a$. Draw a circle centered at $A$ with radius $h_a$. Construct the tangent $l_1$ to this circle through $M_a$. $\overline{BC}$ lies on $l_1$.

Draw a circle centered at $M_a$ with radius $\frac{h_b}{2}$. Construct the tangent $l_2$ to this circle through $A$. $\overline{AC}$ lies on $l_2$. Then $C=l_1\cap l_2$.

Construct the line $l_3$ parallel to $l_2$ so that the distance between $l_2$ and $l_3$ is $h_b$ and $M_a$ lies between these lines. $B$ lies on $l_3$. Then $B=l_1\cap l_3$.