IMO 1961 Problem 5
We are asked to construct a triangle $ABC$ when the side lengths
Proposed by: -
Verified: yes
Verdicts: PASS + PASS
Solve time: 39m11s
Problem
Construct a triangle ABC if the following elements are given: $AC = b, AB = c$, and $\angle AMB = \omega \left(\omega < 90^{\circ}\right)$ where M is the midpoint of BC. Prove that the construction has a solution if and only if
$b \tan{\frac{\omega}{2}} \le c < b$
In what case does equality hold?
Problem Understanding
We are asked to construct a triangle $ABC$ when the side lengths
$AC = b, \qquad AB = c,$
and the angle
$\angle AMB = \omega, \qquad \omega<90^\circ,$
are prescribed, where $M$ denotes the midpoint of $BC$. Furthermore, we must prove that the construction is possible if and only if
$b\tan\frac{\omega}{2} \le c < b,$
and determine the equality case.
The central difficulty is that both the location of $A$ and the unknown side $BC$ are constrained through the midpoint condition and the prescribed angle at $M$. We therefore introduce suitable auxiliary quantities and derive algebraic conditions equivalent to the existence of the triangle.
Key Observations
Let
$m=AM, \qquad x=MB=MC.$
Since $M$ is the midpoint of $BC$, the points $B,M,C$ are collinear and
$\angle AMC = 180^\circ-\omega.$
Applying the law of cosines in $\triangle AMB$ and $\triangle AMC$ gives
$c^2=m^2+x^2-2mx\cos\omega,$
and
$b^2=m^2+x^2+2mx\cos\omega.$
Adding and subtracting these relations yields
$m^2+x^2=\frac{b^2+c^2}{2},$
and
$b^2-c^2=4mx\cos\omega.$
Since $\omega<90^\circ$, we have $\cos\omega>0$, and therefore
$b^2-c^2>0,$
so necessarily
$c<b.$
The lower bound on $c$ comes from the condition
$(m-x)^2\ge0.$
The crucial point is to show rigorously that this condition is equivalent to
$c\ge b\tan\frac{\omega}{2}.$
This equivalence is also exactly what guarantees the existence of positive real numbers $m,x$ in the sufficiency argument.
Solution
Set
$S=\frac{b^2+c^2}{2}, \qquad P=\frac{b^2-c^2}{4\cos\omega}.$
Then the relations derived above become
$m^2+x^2=S, \qquad mx=P.$
We first derive the necessary inequalities.
From
$(m-x)^2=m^2+x^2-2mx,$
we obtain
$$= S-2P = \frac{b^2+c^2}{2} - \frac{b^2-c^2}{2\cos\omega}.$$
Since $(m-x)^2\ge0$, we have
- \frac{b^2-c^2}{2\cos\omega} \ge0.$$Multiplying by $2\cos\omega>0$ gives$$(b^2+c^2)\cos\omega-(b^2-c^2)\ge0.$$Rearranging,$$c^2(1+\cos\omega)\ge b^2(1-\cos\omega).$$Using the half-angle identity$$\tan^2\frac{\omega}{2} = \frac{1-\cos\omega}{1+\cos\omega},
we obtain
$c^2\ge b^2\tan^2\frac{\omega}{2}.$
Since $b,c>0$,
$c\ge b\tan\frac{\omega}{2}.$
Thus necessity gives
$b\tan\frac{\omega}{2}\le c<b.$
We now prove sufficiency.
Assume
$b\tan\frac{\omega}{2}\le c<b.$
The inequality $c<b$ implies
$P=\frac{b^2-c^2}{4\cos\omega}>0,$
because $\cos\omega>0$.
Next we show that
$S-2P\ge0.$
Indeed,
$$= \frac{b^2+c^2}{2} - \frac{b^2-c^2}{2\cos\omega}.$$
Therefore
$S-2P\ge0$
is equivalent, after multiplying by $2\cos\omega>0$, to
$(b^2+c^2)\cos\omega-(b^2-c^2)\ge0,$
that is,
$c^2(1+\cos\omega)\ge b^2(1-\cos\omega).$
Using again
$$= \frac{1-\cos\omega}{1+\cos\omega},$$
this becomes
$c^2\ge b^2\tan^2\frac{\omega}{2},$
which holds by assumption. Hence
$S-2P\ge0.$
Now compute
$(m-x)^2=S-2P\ge0,$
and
$(m+x)^2=S+2P.$
Since $S>0$ and $P>0$, we have
$S+2P>0.$
Define
$m+x=\sqrt{S+2P}, \qquad m-x=\sqrt{S-2P}.$
Then
$m=\frac{\sqrt{S+2P}+\sqrt{S-2P}}{2},$
and
$x=\frac{\sqrt{S+2P}-\sqrt{S-2P}}{2}.$
These are real numbers because $S+2P>0$ and $S-2P\ge0$.
Moreover,
$S+2P>S-2P,$
because $P>0$. Hence
$\sqrt{S+2P}>\sqrt{S-2P},$
so
$x>0.$
Also,
$m>0.$
Thus positive real numbers $m,x$ satisfying
$m^2+x^2=S, \qquad mx=P$
exist.
We can now construct the triangle.
Draw a line $\ell$ and choose a point $M$ on it. Draw a ray from $M$ making an angle $\omega$ with $\ell$. On this ray mark a point $A$ such that
$AM=m.$
On the line $\ell$ place points $B$ and $C$ on opposite sides of $M$ such that
$MB=MC=x.$
Then $M$ is the midpoint of $BC$, and by construction
$\angle AMB=\omega.$
Finally, by the law of cosines,
$$= m^2+x^2-2mx\cos\omega = c^2,$$
and
$$= m^2+x^2+2mx\cos\omega = b^2.$$
Hence
$AB=c, \qquad AC=b.$
Therefore the construction is possible exactly when
$\boxed{b\tan\frac{\omega}{2}\le c<b}.$
For the equality case,
$c=b\tan\frac{\omega}{2}$
is equivalent to
$S-2P=0,$
which is equivalent to
$(m-x)^2=0.$
Hence
$m=x,$
that is,
$AM=MB=MC.$
Thus equality holds precisely when the three segments $AM$, $MB$, and $MC$ are equal.
Verification of Key Steps
The cosine-law identities are correct because
$\angle AMC=180^\circ-\omega,$
and
$\cos(180^\circ-\omega)=-\cos\omega.$
The condition $c<b$ follows from
$b^2-c^2=4mx\cos\omega,$
since $m>0$, $x>0$, and $\cos\omega>0$.
The derivation of the lower bound is rigorous because
$(m-x)^2\ge0$
is transformed explicitly into
$c^2(1+\cos\omega)\ge b^2(1-\cos\omega),$
and then into
$c\ge b\tan\frac{\omega}{2}.$
Conversely, the assumption
$c\ge b\tan\frac{\omega}{2}$
was shown explicitly to imply
$S-2P\ge0,$
which is exactly the condition needed for the existence of real numbers $m,x$.
The formulas
x=\frac{\sqrt{S+2P}-\sqrt{S-2P}}{2}$$indeed produce positive real numbers because$$P>0.$$The construction itself is explicit, and the verification by the law of cosines confirms that all required conditions are satisfied. Finally, equality occurs precisely when$$(m-x)^2=0,$$that is,$$AM=MB=MC.$$## Alternative Approaches A coordinate method leads to the same relations. Place$$M=(0,0), \qquad B=(x,0), \qquad C=(-x,0).$$Let$$A=(m\cos\omega,m\sin\omega).$$Then$$AB^2=(m\cos\omega-x)^2+(m\sin\omega)^2 =m^2+x^2-2mx\cos\omega,$$and$$AC^2=(m\cos\omega+x)^2+(m\sin\omega)^2 =m^2+x^2+2mx\cos\omega.$$Thus$$m^2+x^2=\frac{b^2+c^2}{2}, \qquad mx=\frac{b^2-c^2}{4\cos\omega},$$and the same argument yields$$b\tan\frac{\omega}{2}\le c<b.$$The synthetic method, however, makes the geometric role of the midpoint and the equality case more transparent.