IMO 1974 Problem 2
In triangle $ABC$, we seek a point $D$ on the side $AB$ such that
Proposed by: -
Verified: yes
Verdicts: PASS + PASS
Solve time: 18m19s
Problem
In the triangle $ABC$, prove that there is a point $D$ on side $AB$ such that $CD$ is the geometric mean of $AD$ and $DB$ if and only if $$ \sin{A} \sin{B} \leq \sin^2 \frac{C}{2}. $$
Problem Understanding
In triangle $ABC$, we seek a point $D$ on the side $AB$ such that
$$CD^{2}=AD\cdot DB.$$
The goal is to determine exactly when such a point exists and to prove that this happens if and only if
$$\sin A,\sin B\le \sin^{2}\frac C2.$$
The strategy is to express $AD$ and $DB$ in terms of $CD$ and the angles at $C$, reducing the condition $CD^{2}=AD\cdot DB$ to a trigonometric equation involving a single parameter $x=\angle ACD$. The crucial point is then to justify rigorously that every $x\in[0,C]$ arises from a unique point $D\in AB$.
Key Observations
For an interior point $D\in AB$, let
$$x=\angle ACD.$$
Since $D$ lies on the segment $AB$, the ray $CD$ lies inside the angle $\angle ACB$, so
$$\angle BCD=C-x.$$
Applying the law of sines in triangles $ACD$ and $BCD$ gives expressions for $AD$ and $DB$ in terms of $CD$, $x$, and $C-x$. Multiplying them will convert the condition $CD^{2}=AD\cdot DB$ into
$$\sin A,\sin B=\sin x,\sin(C-x).$$
Thus the problem becomes determining whether the quantity $\sin A,\sin B$ belongs to the range of the function
$$g(x)=\sin x,\sin(C-x), \qquad 0\le x\le C.$$
Solution
Let $D$ be an interior point of $AB$. In triangle $ACD$, the law of sines gives
$$\frac{AD}{\sin\angle ACD} = \frac{CD}{\sin A},$$
hence
$$AD = CD\cdot\frac{\sin(\angle ACD)}{\sin A}.$$
Similarly, in triangle $BCD$,
$$\frac{DB}{\sin\angle BCD} = \frac{CD}{\sin B},$$
hence
$$DB = CD\cdot\frac{\sin(\angle BCD)}{\sin B}.$$
Let
$$x=\angle ACD.$$
Since $\angle ACB=C$,
$$\angle BCD=C-x.$$
Therefore
$$AD\cdot DB = CD^{2}, \frac{\sin x,\sin(C-x)} {\sin A,\sin B}.$$
The condition
$$CD^{2}=AD\cdot DB$$
is therefore equivalent to
$$\sin A,\sin B = \sin x,\sin(C-x).$$
We now justify the parameterization by $x$.
Consider the map that sends a point $D\in AB$ to the angle
$$x=\angle ACD.$$
The segment $AB$ is contained in the interior of the angle $\angle ACB$. For every ray from $C$ making an angle $x\in(0,C)$ with $CA$, that ray lies strictly inside $\angle ACB$. Since $AB$ joins a point on one side of the angle to a point on the other side, the ray intersects the segment $AB$ in exactly one point. Thus for every $x\in(0,C)$ there exists a unique point $D\in AB$ with $\angle ACD=x$.
Conversely, every interior point $D\in AB$ determines a unique ray $CD$, hence a unique angle $x\in(0,C)$. Therefore the correspondence
$$D\longleftrightarrow x$$
is a bijection between the interior of $AB$ and the interval $(0,C)$. Extending to the endpoints gives $x=0$ at $D=A$ and $x=C$ at $D=B$. Hence the possible values of $x$ are exactly the whole interval $[0,C]$.
Consequently, the existence of a point $D$ satisfying $CD^{2}=AD\cdot DB$ is equivalent to the existence of $x\in[0,C]$ such that
$$\sin A,\sin B = \sin x,\sin(C-x).$$
Now define
$$g(x)=\sin x,\sin(C-x).$$
Using
$$\sin u,\sin v = \frac{\cos(u-v)-\cos(u+v)}2,$$
we obtain
$$g(x) = \frac{\cos(2x-C)-\cos C}{2}.$$
Since $-1\le \cos(2x-C)\le 1$,
$$g(x)\le \frac{1-\cos C}{2} = \sin^{2}\frac C2.$$
Equality is attained when $2x-C=0$, that is, at
$$x=\frac C2.$$
Also,
$$g(0)=g(C)=0.$$
Because $g$ is continuous on $[0,C]$, its image is the entire interval
$$\left[0,\sin^{2}\frac C2\right].$$
Therefore there exists $x\in[0,C]$ satisfying
$$g(x)=\sin A,\sin B$$
if and only if
$$\sin A,\sin B \le \sin^{2}\frac C2.$$
Since such an $x$ corresponds to a point $D\in AB$, we conclude that a point $D$ satisfying
$$CD^{2}=AD\cdot DB$$
exists if and only if
$$\boxed{\sin A,\sin B\le \sin^{2}\frac C2}.$$
This completes the proof.
$\square$
Verification of Key Steps
The law-of-sines identities are valid for every interior point $D$, where triangles $ACD$ and $BCD$ are nondegenerate. Since $CD>0$, dividing by $CD^{2}$ introduces no extraneous solutions.
The crucial parameterization is justified geometrically. Every ray from $C$ making an angle $x\in(0,C)$ with $CA$ lies inside $\angle ACB$, and such a ray meets the segment $AB$ exactly once. Hence every $x\in(0,C)$ corresponds to exactly one point $D\in AB$, and every interior point $D$ yields exactly one such $x$. Thus the parameter $x$ runs through the entire interval $[0,C]$.
The function
$$g(x)=\sin x,\sin(C-x)$$
is continuous on $[0,C]$, satisfies
$$g(0)=g(C)=0,$$
and attains its maximum
$$\sin^{2}\frac C2$$
at $x=C/2$. Therefore its range is exactly
$$\left[0,\sin^{2}\frac C2\right].$$
This establishes both directions of the equivalence. If a suitable $D$ exists, then $g(x)=\sin A\sin B$, so necessarily $\sin A\sin B\le \sin^{2}(C/2)$. Conversely, if the inequality holds, then $\sin A\sin B$ lies in the range of $g$, so some $x\in[0,C]$ satisfies $g(x)=\sin A\sin B$, yielding the required point $D$.
Alternative Approaches
A different solution can be obtained by introducing a coordinate system with $AB$ as the $x$-axis and writing $D$ as a variable point on that axis. The condition $CD^{2}=AD\cdot DB$ becomes a quadratic equation in the coordinate of $D$. An analysis of the discriminant and the requirement that the root lie on the segment $AB$ eventually reduces to the same inequality
$$\sin A,\sin B\le \sin^{2}\frac C2.$$
The trigonometric method above is more transparent because it converts the existence question directly into the range of a simple one-variable function.