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 $B=l_1\cap l_3$ .