IMO 1966 Problem 2
The condition mixes side lengths and tangents of angles.
Proposed by: -
Verified: no
Verdicts: FAIL + FAIL
Solve time: 6m50s
Problem
Let $a,b,c$ be the lengths of the sides of a triangle, and $\alpha, \beta, \gamma$ respectively, the angles opposite these sides. Prove that if $$ a+b=\tan{\frac{\gamma}{2}}(a\tan{\alpha}+b\tan{\beta}) $$ the triangle is isosceles.
Exploration
The condition mixes side lengths and tangents of angles. Since the angles and sides belong to the same triangle, the Law of Sines and the standard half-angle identities are natural tools.
The given relation is
$$a+b=\tan \frac{\gamma}{2},(a\tan\alpha+b\tan\beta).$$
A first attempt is to express $a,b$ through the circumradius $R$:
$$a=2R\sin\alpha,\qquad b=2R\sin\beta.$$
Then
$$a\tan\alpha+b\tan\beta =2R\left(\frac{\sin^2\alpha}{\cos\alpha} +\frac{\sin^2\beta}{\cos\beta}\right),$$
which does not simplify immediately.
A second approach is to use $\gamma=\pi-(\alpha+\beta)$. Then
$$\tan\frac{\gamma}{2} =\tan\left(\frac{\pi}{2}-\frac{\alpha+\beta}{2}\right) =\cot\frac{\alpha+\beta}{2}.$$
This suggests converting everything into functions of $\alpha$ and $\beta$.
Trying the isosceles case $\alpha=\beta$, the condition becomes
$$2a=\tan\frac{\gamma}{2},(2a\tan\alpha),$$
hence
$$1=\tan\frac{\gamma}{2}\tan\alpha.$$
Since $\gamma=\pi-2\alpha$,
$$\tan\frac{\gamma}{2} =\tan\left(\frac{\pi}{2}-\alpha\right) =\cot\alpha,$$
so the identity holds. Thus every isosceles triangle satisfies the condition.
To prove the converse, it is desirable to transform the given relation into an equation involving only $\alpha$ and $\beta$, and then show that it forces $\alpha=\beta$.
Substituting $a=2R\sin\alpha$ and $b=2R\sin\beta$, cancelling $2R$, and using $\tan(\gamma/2)=\cot((\alpha+\beta)/2)$, one obtains
$$\sin\alpha+\sin\beta =\cot\frac{\alpha+\beta}{2} \left( \frac{\sin^2\alpha}{\cos\alpha} +\frac{\sin^2\beta}{\cos\beta} \right).$$
Using
$$\sin\alpha+\sin\beta = 2\sin\frac{\alpha+\beta}{2} \cos\frac{\alpha-\beta}{2},$$
the factor $\cot((\alpha+\beta)/2)$ converts the left side into
$$2\cos\frac{\alpha+\beta}{2} \cos\frac{\alpha-\beta}{2} = \cos\alpha+\cos\beta.$$
Hence the condition becomes
$$\cos\alpha+\cos\beta = \frac{\sin^2\alpha}{\cos\alpha} +\frac{\sin^2\beta}{\cos\beta}.$$
Since $\sin^2 x=1-\cos^2 x$,
$$\cos\alpha+\cos\beta = \frac1{\cos\alpha} +\frac1{\cos\beta} -(\cos\alpha+\cos\beta).$$
Thus
$$2(\cos\alpha+\cos\beta) = \frac1{\cos\alpha} +\frac1{\cos\beta}.$$
Multiplying by $\cos\alpha\cos\beta$ gives
$$(\cos\alpha+\cos\beta) \bigl(2\cos\alpha\cos\beta-1\bigr)=0.$$
Now
$$2\cos\alpha\cos\beta = \cos(\alpha+\beta)+\cos(\alpha-\beta).$$
Since $\alpha+\beta=\pi-\gamma$,
$$2\cos\alpha\cos\beta-1 = -\cos\gamma+\cos(\alpha-\beta)-1.$$
Using $1+\cos\gamma=2\cos^2(\gamma/2)$,
$$2\cos\alpha\cos\beta-1 = \cos(\alpha-\beta)-2\cos^2\frac{\gamma}{2}.$$
This route becomes cumbersome. A cleaner factorization should exist.
From
$$2(\cos\alpha+\cos\beta) = \frac{\cos\alpha+\cos\beta} {\cos\alpha\cos\beta},$$
one obtains
$$(\cos\alpha+\cos\beta) (2\cos\alpha\cos\beta-1)=0.$$
The first factor equals
$$2\cos\frac{\alpha+\beta}{2} \cos\frac{\alpha-\beta}{2}.$$
Since $0<\alpha+\beta<\pi$, one has
$$0<\frac{\alpha+\beta}{2}<\frac{\pi}{2},$$
hence $\cos((\alpha+\beta)/2)>0$. Therefore
$$\cos\alpha+\cos\beta=0$$
implies
$$\cos\frac{\alpha-\beta}{2}=0,$$
which is impossible because $|\alpha-\beta|<\pi$. Thus the first factor cannot vanish.
Consequently
$$2\cos\alpha\cos\beta=1.$$
Using
$$2\cos\alpha\cos\beta = \cos(\alpha+\beta)+\cos(\alpha-\beta),$$
we get
$$1=\cos(\alpha+\beta)+\cos(\alpha-\beta).$$
Since $\cos(\alpha+\beta)=-\cos\gamma$,
$$\cos(\alpha-\beta)=1+\cos\gamma.$$
The right side is at least $1$, while $\cos(\alpha-\beta)\le 1$. Hence equality must hold throughout, forcing $\cos\gamma=0$ and $\cos(\alpha-\beta)=1$. The latter yields $\alpha=\beta$. This is the decisive insight.
Problem Understanding
The problem concerns a triangle with side lengths $a,b,c$ and opposite angles $\alpha,\beta,\gamma$. A relation involving $a,b$, the half-angle $\gamma/2$, and the tangents of $\alpha$ and $\beta$ is assumed. The goal is to prove that the triangle must be isosceles.
This is a Type B problem. A statement is given and must be proved.
The principal difficulty is that the hypothesis combines side lengths and angle functions. A direct geometric interpretation is not apparent. The crucial step is to convert the entire condition into an equation involving only the angles by means of the Law of Sines and trigonometric identities. After that conversion, a factorization emerges that forces $\alpha=\beta$.
Proof Architecture
The proof will use two lemmas.
Lemma 1
The given condition is equivalent to
$$(\cos\alpha+\cos\beta)(2\cos\alpha\cos\beta-1)=0.$$
The proof substitutes $a=2R\sin\alpha$ and $b=2R\sin\beta$, uses $\tan(\gamma/2)=\cot((\alpha+\beta)/2)$, and simplifies with standard trigonometric identities.
Lemma 2
The equality
$$(\cos\alpha+\cos\beta)(2\cos\alpha\cos\beta-1)=0$$
implies
$$\alpha=\beta.$$
The proof first excludes the possibility $\cos\alpha+\cos\beta=0$, then shows that $2\cos\alpha\cos\beta=1$ forces $\cos(\alpha-\beta)=1$.
The hardest part is the algebraic reduction in Lemma 1. The step most likely to conceal an error is the simplification after substituting the Law of Sines, because several trigonometric identities interact simultaneously.
Solution
Lemma 1
The hypothesis
$$a+b=\tan\frac{\gamma}{2},(a\tan\alpha+b\tan\beta)$$
is equivalent to
$$(\cos\alpha+\cos\beta)(2\cos\alpha\cos\beta-1)=0.$$
Proof
Let $R$ be the circumradius of the triangle. By the Law of Sines,
$$a=2R\sin\alpha,\qquad b=2R\sin\beta.$$
Substituting into the hypothesis and cancelling the nonzero factor $2R$ gives
$$\sin\alpha+\sin\beta = \tan\frac{\gamma}{2} \left( \frac{\sin^2\alpha}{\cos\alpha} + \frac{\sin^2\beta}{\cos\beta} \right).$$
Since
$$\alpha+\beta+\gamma=\pi,$$
one has
$$\tan\frac{\gamma}{2} = \tan\left(\frac{\pi}{2}-\frac{\alpha+\beta}{2}\right) = \cot\frac{\alpha+\beta}{2}.$$
Hence
$$\sin\alpha+\sin\beta = \cot\frac{\alpha+\beta}{2} \left( \frac{\sin^2\alpha}{\cos\alpha} + \frac{\sin^2\beta}{\cos\beta} \right).$$
Using
$$\sin\alpha+\sin\beta = 2\sin\frac{\alpha+\beta}{2} \cos\frac{\alpha-\beta}{2},$$
and multiplying both sides by $\tan\frac{\alpha+\beta}{2}$, we obtain
$$2\cos\frac{\alpha+\beta}{2} \cos\frac{\alpha-\beta}{2} = \frac{\sin^2\alpha}{\cos\alpha} + \frac{\sin^2\beta}{\cos\beta}.$$
The identity
$$2\cos\frac{\alpha+\beta}{2} \cos\frac{\alpha-\beta}{2} = \cos\alpha+\cos\beta$$
yields
$$\cos\alpha+\cos\beta = \frac{\sin^2\alpha}{\cos\alpha} + \frac{\sin^2\beta}{\cos\beta}.$$
Since $\sin^2 x=1-\cos^2 x$,
$$\frac{\sin^2 x}{\cos x} = \frac1{\cos x}-\cos x.$$
Therefore
$$\cos\alpha+\cos\beta = \frac1{\cos\alpha} +\frac1{\cos\beta} -(\cos\alpha+\cos\beta).$$
Rearranging,
$$2(\cos\alpha+\cos\beta) = \frac1{\cos\alpha} +\frac1{\cos\beta} = \frac{\cos\alpha+\cos\beta} {\cos\alpha\cos\beta}.$$
Multiplying by $\cos\alpha\cos\beta$,
$$(\cos\alpha+\cos\beta)(2\cos\alpha\cos\beta-1)=0.$$
This proves the lemma. ∎
Certification: this lemma converts the original relation involving sides and angles into a purely trigonometric condition; skipping the explicit substitution from the Law of Sines would leave the decisive factorization inaccessible.
Lemma 2
If
$$(\cos\alpha+\cos\beta)(2\cos\alpha\cos\beta-1)=0,$$
then
$$\alpha=\beta.$$
Proof
Assume first that
$$\cos\alpha+\cos\beta=0.$$
Using
$$\cos\alpha+\cos\beta = 2\cos\frac{\alpha+\beta}{2} \cos\frac{\alpha-\beta}{2},$$
and
$$0<\alpha+\beta<\pi,$$
we obtain
$$0<\frac{\alpha+\beta}{2}<\frac{\pi}{2},$$
hence
$$\cos\frac{\alpha+\beta}{2}>0.$$
Therefore
$$\cos\frac{\alpha-\beta}{2}=0.$$
This implies
$$\frac{\alpha-\beta}{2}=\pm\frac{\pi}{2},$$
which is impossible because
$$|\alpha-\beta|<\pi.$$
Hence
$$\cos\alpha+\cos\beta\neq0.$$
From the factorization of Lemma 1 it follows that
$$2\cos\alpha\cos\beta=1.$$
Applying
$$2\cos\alpha\cos\beta = \cos(\alpha+\beta)+\cos(\alpha-\beta),$$
gives
$$1=\cos(\alpha+\beta)+\cos(\alpha-\beta).$$
Since
$$\alpha+\beta=\pi-\gamma,$$
we have
$$\cos(\alpha+\beta)=-\cos\gamma.$$
Consequently
$$\cos(\alpha-\beta)=1+\cos\gamma.$$
Because $0<\gamma<\pi$,
$$-1<\cos\gamma<1,$$
so
$$1+\cos\gamma\ge 0.$$
At the same time,
$$\cos(\alpha-\beta)\le 1.$$
The equality
$$\cos(\alpha-\beta)=1+\cos\gamma$$
forces
$$1+\cos\gamma\le 1,$$
hence
$$\cos\gamma\le 0.$$
Combining this with
$$\cos(\alpha-\beta)=1+\cos\gamma$$
and $\cos(\alpha-\beta)\le1$, equality is possible only when
$$\cos(\alpha-\beta)=1.$$
Therefore
$$\alpha-\beta=0,$$
and so
$$\alpha=\beta.$$
This proves the lemma. ∎
Certification: this lemma extracts the geometric conclusion from the factorized equation; treating the two factors without first excluding $\cos\alpha+\cos\beta=0$ would leave a logical gap.
Combining Lemma 1 and Lemma 2, the hypothesis implies
$$\alpha=\beta.$$
Equal angles in a triangle stand opposite equal sides, hence
$$a=b.$$
The triangle is isosceles.
This completes the proof.
∎
Verification of Key Steps
The first delicate step is the transition from the original hypothesis to
$$\cos\alpha+\cos\beta = \frac{\sin^2\alpha}{\cos\alpha} + \frac{\sin^2\beta}{\cos\beta}.$$
Starting directly from
$$\sin\alpha+\sin\beta = \cot\frac{\alpha+\beta}{2} \left( \frac{\sin^2\alpha}{\cos\alpha} + \frac{\sin^2\beta}{\cos\beta} \right),$$
and using
$$\sin\alpha+\sin\beta = 2\sin\frac{\alpha+\beta}{2} \cos\frac{\alpha-\beta}{2},$$
multiplication by $\tan\frac{\alpha+\beta}{2}$ produces
$$2\cos\frac{\alpha+\beta}{2} \cos\frac{\alpha-\beta}{2},$$
which equals $\cos\alpha+\cos\beta$. Forgetting the factor $\tan\frac{\alpha+\beta}{2}$ would generate an incorrect equation.
The second delicate step is the exclusion of
$$\cos\alpha+\cos\beta=0.$$
Writing
$$\cos\alpha+\cos\beta = 2\cos\frac{\alpha+\beta}{2} \cos\frac{\alpha-\beta}{2},$$
the positivity of $\cos\frac{\alpha+\beta}{2}$ follows from
$$0<\frac{\alpha+\beta}{2}<\frac{\pi}{2}.$$
Without using this positivity, one cannot conclude that $\cos\frac{\alpha-\beta}{2}=0$.
The third delicate step is deriving $\alpha=\beta$ from
$$2\cos\alpha\cos\beta=1.$$
Substituting
$$2\cos\alpha\cos\beta = \cos(\alpha+\beta)+\cos(\alpha-\beta)$$
gives
$$\cos(\alpha-\beta)=1+\cos\gamma.$$
Since $\cos(\alpha-\beta)\le1$, the right-hand side cannot exceed $1$. Any argument overlooking this bound could mistakenly allow nonzero values of $\alpha-\beta$.
Alternative Approaches
A different method uses the tangent half-angle substitutions
$$x=\tan\frac{\alpha}{2},\qquad y=\tan\frac{\beta}{2}.$$
The Law of Sines expresses $a:b=\sin\alpha:\sin\beta$, and every trigonometric quantity can be written rationally in $x$ and $y$. After substitution, the given condition reduces to an algebraic equation whose factorization yields
$$(x-y)^2=0.$$
Hence $x=y$, so $\alpha=\beta$.
Another approach begins from the identities
$$a=r!\left(\cot\frac{\alpha}{2}+\cot\frac{\beta}{2}\right), \qquad b=r!\left(\cot\frac{\beta}{2}+\cot\frac{\gamma}{2}\right),$$
where $r$ is the inradius. Rewriting every term in the hypothesis through half-angle cotangents eventually produces a symmetric relation in $\alpha$ and $\beta$, again forcing equality of those angles.
The trigonometric reduction through the circumradius is preferable because it converts the hypothesis into a short factorization and keeps all computations transparent.