IMO 1960 Problem 3
The reviewers identified two genuine problems in the previous proof.
Proposed by: -
Verified: yes
Verdicts: PASS + PASS
Solve time: 17m36s
Problem
A right-angled triangle $ABC$ is given for which the hypotenuse $BC$ has length $a$ and is divided into $n$ equal segments ($n$ odd). Let $\alpha$ be the angle subtended at $A$ by the segment containing the midpoint of $BC$. If $h$ denotes the altitude from $A$, prove that $$\tan\alpha = \frac{4nh}{(n^2-1)a}.$$
Exploration
The reviewers identified two genuine problems in the previous proof. The first was a justification gap: the identity
$$(x_1-p)(x_2-p)+p(a-p)=\frac{(n^2-1)a^2}{4n^2}$$
was asserted without calculation. The second was critical: the proof stopped before deriving the target formula.
Before rebuilding the argument, the small cases must be checked carefully.
For $n=1$, the “middle segment” is the entire hypotenuse $BC$. The claimed formula becomes
$$\tan\alpha=\frac{4h}{0},$$
which is meaningless. Hence the theorem cannot include $n=1$. The valid range is odd $n\ge3$.
For $n=3$, the middle segment is
$$\left[\frac a3,\frac{2a}3\right].$$
Take a concrete triangle with $a=10$ and $p=2$. Then
$$h^2=p(a-p)=2\cdot8=16, \qquad h=4.$$
The claimed formula gives
$$\tan\alpha=\frac{4\cdot3\cdot4}{(9-1)\cdot10} =\frac{48}{80} =\frac35.$$
Using coordinates directly,
$$x_1=\frac{10}3,\qquad x_2=\frac{20}3,$$
and the slopes are
$$m_1=-\frac4{10/3-2}=-3, \qquad m_2=-\frac4{20/3-2}=-\frac67.$$
Thus
$$\tan\alpha = \left| \frac{m_2-m_1}{1+m_1m_2} \right| = \left| \frac{-6/7+3}{1+18/7} \right| = \frac{15/7}{25/7} = \frac35,$$
which agrees with the formula.
For $n=5$, the formula predicts
$$\tan\alpha=\frac{20h}{24a}=\frac{5h}{6a}.$$
Using the same triangle with $a=10$, $h=4$, this gives
$$\tan\alpha=\frac13.$$
A direct slope computation confirms the same value.
These tests support the corrected coordinate method. The earlier invalid assumption $p=a/2$ is completely unnecessary, since the right angle condition only gives
$$p(a-p)=h^2.$$
Problem Understanding
A right triangle $ABC$ has hypotenuse $BC$ of length $a$. The segment $BC$ is divided into $n$ equal parts, where $n$ is odd and at least $3$. The segment containing the midpoint of $BC$ determines an angle $\alpha$ at the vertex $A$. If $h$ is the altitude from $A$ to $BC$, the task is to prove
$$\tan\alpha=\frac{4nh}{(n^2-1)a}.$$
The expression depends only on $a$, $h$, and $n$, so any auxiliary coordinate parameter must disappear during the computation.
Key Observations
Place the triangle in the coordinate plane with
$$B=(0,0),\qquad C=(a,0),\qquad A=(p,h),$$
where $0<p<a$.
Since the angle at $A$ is right,
$$(B-A)\cdot(C-A)=0.$$
This gives
$$(-p,-h)\cdot(a-p,-h)=0,$$
hence
$$-p(a-p)+h^2=0,$$
so
$$p(a-p)=h^2.$$
Because $n$ is odd, the segment containing the midpoint of $BC$ has endpoints
$$x_1=\frac{(n-1)a}{2n}, \qquad x_2=\frac{(n+1)a}{2n}.$$
The two rays from $A$ to these endpoints have slopes
$$m_1=-\frac{h}{x_1-p}, \qquad m_2=-\frac{h}{x_2-p}.$$
The tangent of the angle between two lines with slopes $m_1,m_2$ is
$$\tan\alpha= \left| \frac{m_2-m_1}{1+m_1m_2} \right|.$$
Solution
Using the coordinate setup above, the slopes of the two rays are
$$m_1=-\frac{h}{x_1-p}, \qquad m_2=-\frac{h}{x_2-p}.$$
Therefore
$$\tan\alpha = \left| \frac{m_2-m_1}{1+m_1m_2} \right|.$$
Substituting the expressions for $m_1,m_2$ gives
$$\tan\alpha = \left| \frac{ -\frac{h}{x_2-p} +\frac{h}{x_1-p} }{ 1+\frac{h^2}{(x_1-p)(x_2-p)} } \right|.$$
The numerator simplifies to
$$h\left( \frac1{x_1-p}-\frac1{x_2-p} \right) = h\cdot \frac{x_2-x_1}{(x_1-p)(x_2-p)}.$$
Since
$$x_2-x_1=\frac an,$$
we obtain
$$\tan\alpha = \frac{h(a/n)} {(x_1-p)(x_2-p)+h^2}.$$
Using
$$h^2=p(a-p),$$
the denominator becomes
$$(x_1-p)(x_2-p)+p(a-p).$$
Now compute this expression explicitly. Since
$$x_1=\frac{a(n-1)}{2n}, \qquad x_2=\frac{a(n+1)}{2n},$$
we have
$$x_1+x_2=a$$
and
$$x_1x_2 = \frac{a^2(n^2-1)}{4n^2}.$$
Hence
$$(x_1-p)(x_2-p) = x_1x_2-p(x_1+x_2)+p^2 = \frac{a^2(n^2-1)}{4n^2}-ap+p^2.$$
Adding
$$p(a-p)=ap-p^2$$
gives
$$(x_1-p)(x_2-p)+p(a-p) = \frac{a^2(n^2-1)}{4n^2}.$$
Substituting this into the formula for $\tan\alpha$ yields
$$\tan\alpha = \frac{h(a/n)} {\frac{a^2(n^2-1)}{4n^2}}.$$
Simplifying,
$$\tan\alpha = \frac{ha}{n}\cdot \frac{4n^2}{a^2(n^2-1)} = \frac{4nh}{(n^2-1)a}.$$
This completes the proof.
∎
Verification of Key Steps
The identity
$$p(a-p)=h^2$$
must hold for every right triangle in the coordinate setup. Testing the earlier numerical example with $a=10$ and $p=2$ gives
$$h^2=2\cdot8=16,$$
so $h=4$, which is consistent.
The algebraic identity
$$(x_1-p)(x_2-p)+p(a-p) = \frac{a^2(n^2-1)}{4n^2}$$
is independent of $p$. Testing $n=3$ gives
$$\frac{a^2(9-1)}{36} = \frac{2a^2}{9}.$$
Using
$$x_1=\frac a3, \qquad x_2=\frac{2a}3,$$
the left side becomes
$$\left(\frac a3-p\right)\left(\frac{2a}3-p\right)+p(a-p),$$
which expands to
$$\frac{2a^2}{9}-ap+p^2+ap-p^2 = \frac{2a^2}{9}.$$
The formula is correct.
The theorem excludes $n=1$, since the denominator $n^2-1$ vanishes. The statement is valid for all odd integers $n\ge3$.
Alternative Approaches
A trigonometric approach avoids slope formulas entirely. If the endpoints of the middle segment are denoted by $P$ and $Q$, then one may express
$$\angle PAQ=\angle PAC-\angle QAC$$
and use tangent subtraction identities together with similar triangles arising from the altitude to the hypotenuse. This eventually leads to the same expression for $\tan\alpha$.
A vector approach is also possible. Writing vectors from $A$ to the two endpoints of the middle segment, the tangent of the angle between them can be computed from
$$\tan\alpha = \frac{\lVert u\times v\rVert}{u\cdot v}.$$
The resulting algebra simplifies using the relation
$$p(a-p)=h^2.$$