Subfield: Geometry


Question: Cyclic quadrilateral $A B C D$ has $A B=A D=1, C D=\cos \angle A B C$, and $\cos \angle B A D=-\frac{1}{3}$. Prove that $B C$ is a diameter of the circumscribed circle.

Solution:
Consider $\triangle B A D$. Since we know the lengths of sides $B A$ and $A D$ and the cosine of the angle between them, we can calculate the length of $B D$ using the cosine law:

$$
\begin{aligned}
B D & =\sqrt{B A^{2}+A D^{2}-2(B A)(A D) \cos \angle B A D} \\
& =\sqrt{2-2\left(-\frac{1}{3}\right)} \\
& =\sqrt{\frac{8}{3}}
\end{aligned}
$$

<img_3167>

Next, let $x=\cos \angle A B C$. Note that $D C=x$.



Since $A B C D$ is a cyclic quadrilateral, then $\angle A D C=180^{\circ}-\angle A B C$, and so $\cos \angle A D C=-\cos \angle A B C=-x$.

Similarly, $\cos \angle B C D=-\cos \angle B A D=\frac{1}{3}$ (since $A B C D$ is a cyclic quadrilateral).

So we can now use the cosine law simultaneously in $\triangle A D C$ and $\triangle A B C$ (since side $A C$ is common) in order to try to solve for $B C$ :

$$
\begin{aligned}
1^{2}+x^{2}-2(1)(x) \cos \angle A D C & =1^{2}+B C^{2}-2(1)(B C) \cos \angle A B C \\
1^{2}+x^{2}-2(1)(x)(-x) & =1^{2}+B C^{2}-2(1)(B C)(x) \\
0 & =B C^{2}-2(B C) x-3 x^{2} \\
0 & =(B C-3 x)(B C+x)
\end{aligned}
$$

Since $x$ is already a side length, then $x$ must be positive (ie. $\angle A B C$ is acute), so $B C=3 x$.

Since $\cos \angle B C D=\frac{1}{3}$ and sides $D C$ and $B C$ are in the ratio $1: 3$, then $\triangle B C D$ must indeed be right-angled at $D$. (We could prove this by using the cosine law to calculate $B D^{2}=8 x^{2}$ and then noticing that $D C^{2}+B D^{2}=B C^{2}$.)

Since $\triangle B C D$ is right-angled at $D$, then $B C$ is a diameter of the circle.
Let $x=\cos \angle A B C=C D$, and let $B C=y$.

Since the opposite angles in a cyclic quadrilateral are supplementary, their cosines are negatives of each other. Thus, $\cos \angle A D C=-x$ and $\cos \angle B C D=\frac{1}{3}$.

<img_3217>

Next, we use the cosine law four times: twice to calculate $A C^{2}$ in the two triangles $A B C$ and $A D C$, and then twice to calculate $B D^{2}$ in the triangles $A D B$ and $C D B$ to obtain:

$$
\begin{aligned}
1^{2}+x^{2}-2(1)(x) \cos \angle A D C & =1^{2}+y^{2}-2(1)(y) \cos \angle A B C \\
1+x^{2}-2 x(-x) & =1+y^{2}-2 y(x) \\
0 & =y^{2}-2 x y-3 x^{2} \\
0 & =(y-3 x)(y+x)
\end{aligned}
$$

and

$$
\begin{aligned}
1^{2}+1^{2}-2(1)(1) \cos \angle B A D & =x^{2}+y^{2}-2 x y \cos \angle B C D \\
2-2\left(-\frac{1}{3}\right) & =x^{2}+y^{2}-2 x y\left(\frac{1}{3}\right) \\
\frac{8}{3} & =x^{2}+y^{2}-\frac{2}{3} x y
\end{aligned}
$$

From the first equation, since $x$ is already a side length and so is positive, we must have that $y=3 x$.

Substituting into the second equation, we obtain



$$
\begin{aligned}
& \frac{8}{3}=x^{2}+(3 x)^{2}-\frac{2}{3} x(3 x) \\
& \frac{8}{3}=8 x^{2} \\
& x=\frac{1}{\sqrt{3}}
\end{aligned}
$$

since $x$ must be positive. Thus, since $y=3 x$, then $y=\sqrt{3}$.

Looking then at $\triangle B D C$, we have side lengths $B C=\sqrt{3}, C D=\frac{1}{\sqrt{3}}$ and $B D=\sqrt{\frac{8}{3}}$. (The last is from the left side of the second cosine law equation.) Thus, $B C^{2}=C D^{2}+B D^{2}$, and so $\triangle B D C$ is right-angled at $D$, whence $B C$ is a diameter of the circle.