China Girls' Mathematical Olympiad

Proof I Note that there are $\left(\genfrac{}{}{0ex}{}{6}{2}\right)=15$ possible pairs of ratios among the six given in the problem statement. These pairs are of two types: (i) Three of these pairs are reciprocal pairs involving segments from just one side of triangle $ABC$. (ii) The other 12 pairs involve segments from two sides of the triangle. We first consider the former case.

a. If $\frac{CD}{DB}$ and $\frac{BD}{DC}$ are both integers, then both of these ratios must be 1 and $BD=DC$. Then in triangle $ABC$, $AD$ is the median from $A$ and $D$, because $AD$ contains the circumcenter, is also the perpendicular bisector of segment $BC$. It then follows that $AB=AC$ and the triangle is isosceles. Similarly, if $\frac{CE}{EA}$ and $\frac{AE}{EC}$ are both integers or $\frac{AF}{FB}$ and $\frac{BF}{FA}$ are both integers, then triangle $ABC$ is isosceles.

b. Let $O$ be the circumcenter of triangle $ABC$, and let

$\angle CAB=\alpha$, $\angle ABC=\beta$, and $\angle BCA=\gamma$. We show that any of the ratios can be written in the form $\frac{\sin 2x}{\sin 2y}$ where $x$ and $y$ are two of $\alpha ,\beta ,\gamma$. Since $ABC$ is acute, $0^{\circ }<\alpha ,\beta ,\gamma <90^{\circ }$ and $O$ lies in the interior. Hence $\angle AOB=2\gamma$, $\angle BOC=2\alpha$, and $\angle COA=2\beta$. Applying the sine rule to triangles $BOD$ and $COD$ gives $$\frac{BD}{\sin \angle BOD}=\frac{BO}{\sin \angle BDO}$$ and $$\frac{CD}{\sin \angle COD}=\frac{CO}{\sin \angle CD{O}^{\prime }}$$ Next note that $BO=CO$ and that $$\begin{array}{rl}\angle BDO+\angle CDO& =180^{\circ }\\ & =\angle BOD+\angle AOB\\ & =\angle COD+\angle AOC.\end{array}$$ It follows that $$\frac{BD}{\sin 2\gamma }=\frac{BD}{\sin \angle BOD}=\frac{CD}{\sin \angle COD}=\frac{CD}{\sin 2{\beta }^{\prime }}$$ giving $\frac{BD}{CD}=\frac{\sin 2\gamma }{\sin 2\beta }$. Similarly, $\frac{CE}{EA}=\frac{\sin 2\alpha }{\sin 2\gamma }$ and $\frac{AF}{FB}=\frac{\sin 2\beta }{\sin 2\alpha }$.

Now assume that one of the twelve type (ii) pairs of ratios consists of two integers. Then there are positive integers $m$ and $n$ (with $m\le n$) such that $$\sin 2x=m\sin 2z\text{and}\sin 2y=n\sin 2z$$ or $$\sin 2z=m\sin 2x\text{and}\sin 2z=n\sin 2y$$ for some choice of $x,y,z$ with $\{x,y,z\}=\{\alpha ,\beta ,\gamma \}$.

Without loss of generality we may assume that $$\sin 2\alpha =m\sin 2\gamma \text{and}\sin 2\beta =n\sin 2\gamma$$ or $$\sin 2\gamma =m\sin 2\alpha \text{and}\sin 2\gamma =n\sin 2\beta \stackrel{◯}{1}$$ for some positive integers $m$ and $n$.

Note that there is a triangle with angles $180^{\circ }-2\alpha$, $180^{\circ }-2\beta$, and $180^{\circ }-2\gamma$. (It is easy to check that each of these angles is in the interval $(0^{\circ },180^{\circ })$ and that they sum to $180^{\circ }$.) Furthermore, a triangle with these angles can be constructed by drawing the tangents to the circumcircle of $ABC$ at each of $A$, $B$ and $C$. Denote this triangle by $A_1{B}_1{C}_1$ where $A_1$ is the intersection of the tangents at $B$ and $C$, $B_1$ is the intersection of the tangents at $C$ and $A$, and $C_1$ is the intersection of the tangents at $A$ and $B$. Applying the sine rule to triangle $A_1{B}_1{C}_1$ and by $\stackrel{◯}{1}$ we find $$\begin{array}{rl}{A}_1{B}_1:B_1{C}_1:C_1{A}_1& =\sin \angle C_1:\sin \angle A_1:\sin \angle B_1\\ & =\sin 2\gamma :\sin 2\alpha :\sin 2\beta ,\end{array}$$ that is, $$A_1{B}_1:B_1{C}_1:C_1{A}_1=1:m:n$$ or $$A_1{B}_1:B_1{C}_1:C_1{A}_1=mn:n:m.\stackrel{◯}{2}$$ By the triangle inequality, it follows that $1+m<n$ (that is, $m=n$) or $n+m>nm$ (that is, $(n-1)(m-1)<1$ and $m=1$). We deduce that either $\sin 2\alpha =\sin 2\beta$ or $\sin 2\gamma =\sin 2\alpha$. But then either $\frac{AF}{FB}=\frac{BF}{FA}$ or $\frac{CE}{EA}=\frac{AE}{EC}$, by case (a), triangle $ABC$ is isosceles.

Proof II (We maintain the notations used in Proof I.) We only consider those 12 pairs of ratios of type (ii). Without loss of generality we may assume that each of the following sets $\{\frac{BD}{DC},\frac{CD}{DB}\}$ and $\{\frac{AF}{FB},\frac{BF}{FA}\}$ has an element taking integer values. By symmetry, we consider three cases:

a. In this case, we assume that $\frac{BD}{DC}=m$ and $\frac{BF}{FA}=n$ for some positive integers $m$ and $n$. Applying Menelaus's theorem to line $AOD$ and triangle $BCF$ yields $$\frac{AO}{OD}\cdot \frac{DC}{CB}\cdot \frac{BF}{FA}=1$$ or $$\frac{AO}{OD}=\frac{CB}{DC}\cdot \frac{FA}{BF}=\frac{m+1}{n}.$$ Likewise, applying Menelaus's theorem to line $COF$ and triangle $BAD$ yields $\frac{CO}{OF}=\frac{n+1}{m}$.

Since triangle $ABC$ is acute, $O$ lies on segment $AD$ and $CF$ with $AO>OD$ and $CO>OF$. Hence $m+1>n$ and $n+1>m$, implying that $m-1<n<m+1$. Since $m$ and $n$ are integers, we must have $m=n$. It is then not difficult to see that $A$ and $C$ are symmetric with respect to line $BF$ and triangle $ABC$ is isosceles with $AB=CB$.

b. In this case, we assume that $\frac{CD}{DB}=m$ and $\frac{AF}{FB}=n$ for some positive integers $m$ and $n$. Applying Ceva's theorem gives $$\frac{AF}{FB}\cdot \frac{BD}{DC}\cdot \frac{CE}{EA}=1\text{or}\frac{CE}{EA}=\frac{m}{n}.$$ Applying Menelaus's theorem to line $BOE$ and triangle $ACF$ yields $$\frac{BO}{OE}\cdot \frac{EC}{CA}\cdot \frac{AF}{FB}=1$$ or $$\frac{BO}{OE}=\frac{FB}{AF}\cdot \frac{CA}{EC}=\frac{m+n}{mn}.$$ Since triangle $ABC$ is acute, $O$ lies on segment $BE$ and $EO<BO$. Hence $m+n\ge mn$ or $(m-1)(n-1)\le 1$. Since $m$ and $n$ are positive integers, we deduce that one of $m$ and $n$ is equal to 1; that is, either $AF=FB$ or $BD=DC$. In either case, triangle $ABC$ is isosceles.