
97.3. Let $A, B, C$, and $D$ be four different points in the plane. Three of the line segments $A B, A C, A D, B C, B D$, and $C D$ have length a. The other three have length $b$, where $b>a$. Determine all possible values of the quotient $\frac{b}{a}$.


Solution. If the three segments of length $a$ share a common endpoint, say $A$, then the other three points are on a circle of radius $a$, centered at $A$, and they are the vertices of an equilateral triangle of side length $b$. But this means that $A$ is the center of the triangle $B C D$, and

$$
\frac{b}{a}=\frac{b}{\frac{2}{3} \frac{\sqrt{3}}{2} b}=\sqrt{3}
$$

Assume then that of the segments emanating from $A$ at least one has lenght $a$ and at least one has length $b$. We may assume $A B=a$ and $A D=b$. If only one segment of length $a$ would emanate from each of the four poits, then the number of segments of length $a$ would be two, as every segment is counted twice when we count the emanating segments. So we may assume that $A C$ has length $a$, too. If $B C=a$, then $A B C$ would be an equilateral triangle, and the distance of $D$ from each of its vertices would be $b$. This is not possible, since $b>a$. So $B C=b$. Of the segments $C D$ and $B D$ one has length $a$. We may assume $D C=a$. The segments $D C$ and $A B$ are either on one side of thye line $A C$ or on opposite sides of it. In the latter case, $A B C D$ is a parallelogram with a pair of sides of length $a$ and a pair of sides of length $b$, and its diagonals have lengths $a$ and $b$. This is not possible, due to the fact that the sum of the squares of the diagonals of the parallelogram, $a^{2}+b^{2}$, would be equal to the sum of the squares of its sides, i.e. $2 a^{2}+2 b^{2}$. This means that we may assume that $B A C D$ is a convex quadrilateral. Let $\angle A B C=\alpha$ and $\angle A D B=\beta$. From isosceles triangles we obtain for instance $\angle C B D=\beta$, and from the triangle $A B D$ in particular $2 \alpha+2 \beta+\beta=\pi$ as well as $\angle C D A=\alpha, \angle D C B=\frac{1}{2}(\pi-\beta), \angle C A D=\alpha$. The triangle $A D C$ thus yields $\alpha+\alpha+\alpha+\frac{1}{2}(\pi-\beta)=\pi$. From this we solve $\alpha=\frac{1}{5} \pi=36^{\circ}$. The sine theorem applied to $A B C$ gives

$$
\frac{b}{a}=\frac{\sin 108^{\circ}}{\sin 36^{\circ}}=\frac{\sin 72^{\circ}}{\sin 36^{\circ}}=2 \cos 36^{\circ}=\frac{\sqrt{5}+1}{2}
$$

(In fact, $a$ is the side of a regular pentagon, and $b$ is its diagonal.) - Another way of finding the ratio $\frac{b}{a}$ is to consider the trapezium $C D B A$, with $C D \| A B$; if $E$ is the orthogonal projection of $B$ on the segment $C D$, then $C E=b-\frac{1}{2}(b-a)=\frac{1}{2}(b+a)$. The right triangles $B C E$ and $D C E$ yield $C E^{2}=b^{2}-\left(\frac{b+a}{2}\right)^{2}=a^{2}-\left(\frac{b-a}{2}\right)^{2}$, which can be written as $b^{2}-a b-a^{2}=0$. From this we solve $\frac{b}{a}=\frac{\sqrt{5}+1}{2}$.
