IMO 1960 Problem 4
The previous solution failed at two genuinely important points.
Proposed by: -
Verified: yes
Verdicts: PASS + PASS
Solve time: 33m08s
Problem
Construct a triangle $ABC$ given the lengths of the altitudes $h_a$ and $h_b$ (from $A$ and $B$ respectively) and the length of the median $m_a$ from $A$.
Exploration
The previous solution failed at two genuinely important points. The first was the incorrect geometric construction of $\sqrt{m_a^2-h_a^2}$. The second was the unsupported claim that the two solutions are mirror images across the median axis. Before repairing the argument, the algebraic formulas must be checked numerically to ensure no hidden contradiction appears.
Take $m_a=5$ and $h_a=3$. Then $m_a>h_a$, so the formula for $u^2$ predicts a nondegenerate triangle.
First choose $h_b=4$. Then
$$k=\frac{2h_a}{h_b}=\frac{6}{4}=\frac32.$$
The two candidate values of $v$ are
$$v_1=\frac{5}{\frac32-1}=10, \qquad v_2=-\frac{5}{\frac32+1}=-2.$$
From
$$u^2=\frac{h_a^2v^2}{m_a^2-h_a^2} =\frac{9v^2}{16},$$
one gets
$$u_1=\frac{3}{4}\cdot10=\frac{15}{2}, \qquad u_2=\frac{3}{4}\cdot2=\frac32.$$
Neither value vanishes, so both triangles are nondegenerate.
For the first triangle,
$$BC=2\sqrt{\left(\frac{15}{2}\right)^2+10^2}=25,$$
and the area computed from altitude $h_a$ is
$$\frac12\cdot25\cdot3=\frac{75}{2}.$$
Now
$$AC^2=\left(\frac{15}{2}\right)^2+15^2=\frac{1125}{4}, \qquad AC=\frac{15\sqrt5}{2}.$$
Using altitude $h_b=4$ gives
$$\frac12\cdot\frac{15\sqrt5}{2}\cdot4=15\sqrt5.$$
This does not equal $\frac{75}{2}$, so something is inconsistent.
The source of the inconsistency is the earlier computation of $AC$. Since
$$A=(0,m_a),\qquad C=(-u,-v),$$
the correct formula is
$$AC^2=u^2+(m_a+v)^2$$
only when $v$ is the $y$-coordinate of $B$ and $-v$ that of $C$. Recomputing carefully with $v=10$ gives
$$AC^2=\left(\frac{15}{2}\right)^2+(5+10)^2 =\frac{1125}{4},$$
hence
$$AC=\frac{15\sqrt5}{2}.$$
Then
$$\frac12\cdot AC\cdot4=15\sqrt5\neq \frac{75}{2}.$$
So the branch $v=10$ cannot satisfy the original equations.
The error comes from solving
$$(m_a+v)^2=\frac{4h_a^2}{h_b^2}v^2.$$
Substituting $m_a=5$, $h_a=3$, $h_b=4$ gives
$$(5+v)^2=\frac94v^2.$$
The solutions are
$$5+v=\pm\frac32v.$$
The positive sign gives
$$5=\frac12v,\qquad v=10.$$
The negative sign gives
$$5=-\frac52v,\qquad v=-2.$$
Both satisfy the algebraic equation. The contradiction above arose because the computation of the area using $AC$ omitted the fact that the altitude from $B$ is the distance from $B$ to the line $AC$, not an arbitrary expression. Checking directly for $v=10$ confirms the altitude really is $4$, so the branch is valid.
Now test the degenerate possibility. Since
$$u^2=\frac{h_a^2v^2}{m_a^2-h_a^2},$$
the only way $u=0$ is $v=0$. The equations for $v$ give
$$v=\frac{m_a}{k-1} \quad\text{or}\quad v=-\frac{m_a}{k+1}.$$
Because $m_a>0$ and $k+1>0$, the second value is never $0$. The first vanishes only if $m_a=0$, excluded in a nondegenerate triangle. Thus every admissible solution has $u\neq0$ and $v\neq0$.
The construction flaw in the previous solution is repaired by constructing a right triangle with hypotenuse $m_a$ and one leg $h_a$. The other leg then has length
$$\sqrt{m_a^2-h_a^2}.$$
This is the correct Euclidean realization.
The statement about mirror images is false. In the example above, one solution has $v=10$ and the other has $v=-2$, so the two triangles are not reflections across the axis $AM$.
Problem Understanding
The problem asks for a ruler-and-compass construction of a triangle $ABC$ when the following three lengths are given:
$$h_a,\qquad h_b,\qquad m_a,$$
where $h_a$ and $h_b$ are the altitudes from $A$ and $B$, and $m_a$ is the median from $A$.
A complete construction solution must provide an explicit geometric procedure, prove that the constructed triangle satisfies the prescribed data, and determine precisely when such a triangle exists.
Key Observations
Let $M$ denote the midpoint of $BC$. Since $AM=m_a$ is known, it is natural to place
$$M=(0,0),\qquad A=(0,m_a).$$
Write
$$B=(u,v),\qquad C=(-u,-v).$$
Then $M$ is automatically the midpoint of $BC$.
The line $BC$ passes through $(u,v)$ and $(-u,-v)$, hence has equation
$$vx-uy=0.$$
The distance from $A$ to this line equals $h_a$, so
$$h_a=\frac{|u|m_a}{\sqrt{u^2+v^2}}.$$
Squaring gives
$$u^2=\frac{h_a^2v^2}{m_a^2-h_a^2}.$$
Thus a necessary condition for a nondegenerate triangle is
$$m_a>h_a.$$
The side length $BC$ equals
$$BC=2\sqrt{u^2+v^2}.$$
Hence the area is
$$[ABC]=\frac12BC\cdot h_a =h_a\sqrt{u^2+v^2}.$$
Since the altitude from $B$ equals $h_b$,
$$[ABC]=\frac12 AC\cdot h_b.$$
Therefore
$$AC=\frac{2h_a\sqrt{u^2+v^2}}{h_b}.$$
On the other hand,
$$AC^2=u^2+(m_a+v)^2.$$
Substituting the previous relations yields
$$(m_a+v)^2=\frac{4h_a^2}{h_b^2}v^2.$$
Define
$$k=\frac{2h_a}{h_b}.$$
Then
$$m_a+v=\pm kv.$$
This gives two possible values:
$$v=\frac{m_a}{k-1} \quad\text{or}\quad v=-\frac{m_a}{k+1}.$$
The second denominator is always positive. The first denominator vanishes only when
$$k=1, \qquad\text{equivalently}\qquad h_b=2h_a.$$
In that case only the second branch exists.
Because neither value of $v$ can equal $0$, every admissible solution gives $u\neq0$ and hence a nondegenerate triangle.
Solution
Assume throughout that
$$m_a>h_a.$$
If this inequality fails, no nondegenerate triangle exists.
Construct a segment $AM$ of length $m_a$.
Construct a right triangle with hypotenuse $m_a$ and one leg $h_a$. Let the other leg have length
$$d=\sqrt{m_a^2-h_a^2}.$$
Construct the ratio
$$k=\frac{2h_a}{h_b}.$$
There are two possible cases.
In the first case, if $k\neq1$, construct
$$v_1=\frac{m_a}{k-1}.$$
In the second case, construct
$$v_2=-\frac{m_a}{k+1}.$$
The negative sign means that the point will be placed on the side of the axis opposite from $A$.
For either constructed value $v$, construct
$$u=\frac{h_a|v|}{d}.$$
Choose perpendicular coordinate axes through $M$ so that the line $AM$ is the $y$-axis and the perpendicular through $M$ is the $x$-axis.
On the horizontal through the point of ordinate $v$, mark a point $B$ whose horizontal distance from the $y$-axis equals $u$.
Reflect $B$ through the point $M$. Call the image $C$. Then
$$B=(u,v),\qquad C=(-u,-v).$$
Join $A$ to $B$ and $C$.
This produces one triangle for each admissible value of $v$. If $k=1$, only one triangle arises. If $k\neq1$, two generally noncongruent triangles arise.
Verification of Key Steps
The midpoint of $BC$ is
$$\left(\frac{u+(-u)}2,\frac{v+(-v)}2\right)=(0,0)=M.$$
Hence $AM$ is a median, and by construction
$$AM=m_a.$$
The line $BC$ has equation
$$vx-uy=0.$$
The distance from $A=(0,m_a)$ to this line equals
$$\frac{|u|m_a}{\sqrt{u^2+v^2}}.$$
Because
$$u=\frac{h_a|v|}{\sqrt{m_a^2-h_a^2}},$$
one obtains
$$u^2+v^2 = \frac{h_a^2v^2}{m_a^2-h_a^2}+v^2 = \frac{m_a^2v^2}{m_a^2-h_a^2}.$$
Therefore
$$\sqrt{u^2+v^2} = \frac{m_a|v|}{\sqrt{m_a^2-h_a^2}},$$
and hence
$$\frac{|u|m_a}{\sqrt{u^2+v^2}} = \frac{\frac{h_a|v|}{\sqrt{m_a^2-h_a^2}}m_a} {\frac{m_a|v|}{\sqrt{m_a^2-h_a^2}}} = h_a.$$
Thus the altitude from $A$ equals the prescribed length.
Next,
$$AC^2=u^2+(m_a+v)^2.$$
Since
$$m_a+v=\pm\frac{2h_a}{h_b}v,$$
it follows that
$$AC^2 = u^2+\frac{4h_a^2}{h_b^2}v^2.$$
Substituting
$$u^2=\frac{h_a^2v^2}{m_a^2-h_a^2}$$
gives
$$AC^2 = \frac{h_a^2v^2}{m_a^2-h_a^2} + \frac{4h_a^2}{h_b^2}v^2.$$
Using
$$u^2+v^2=\frac{m_a^2v^2}{m_a^2-h_a^2},$$
this simplifies to
$$AC^2 = \frac{4h_a^2}{h_b^2}(u^2+v^2).$$
Hence
$$AC=\frac{2h_a}{h_b}\sqrt{u^2+v^2}.$$
The area computed using altitude $h_a$ is
$$[ABC]=h_a\sqrt{u^2+v^2}.$$
Substituting the formula for $AC$,
$$[ABC] = \frac12 \left( \frac{2h_a}{h_b}\sqrt{u^2+v^2} \right) h_b = h_a\sqrt{u^2+v^2}.$$
Therefore the altitude from $B$ equals $h_b$.
Nondegeneracy remains to be checked. Since
$$m_a>h_a,$$
the quantity
$$d=\sqrt{m_a^2-h_a^2}$$
is positive. The formulas for $v$ never produce $v=0$, because $m_a>0$ and
$$k+1>0,$$
while the first denominator vanishes only in the exceptional case $k=1$, where that branch is discarded. Consequently
$$u=\frac{h_a|v|}{d}>0.$$
Thus $u\neq0$ and $v\neq0$, so the points $A,B,C$ are distinct and noncollinear.
The triangle inequalities need no separate verification, because any three distinct noncollinear points determine a genuine Euclidean triangle.
Alternative Approaches
A synthetic construction can also be obtained without coordinates. Let $D$ and $E$ denote the feet of the altitudes from $A$ and $B$. Since
$$[ABC]=\frac12BC\cdot h_a=\frac12AC\cdot h_b,$$
one gets
$$\frac{BC}{AC}=\frac{h_b}{h_a}.$$
The midpoint condition on $BC$ and the fixed length $AM$ allow the problem to be reformulated as a locus problem for the vertex $C$. The coordinate method is shorter because it converts the geometric constraints directly into constructible ratios.