2023 WSMO Accuracy Round Problems/Problem 6
Problem
In quadrilateral there exists a point
such that
and
Let
be the foot of the perpendiculars from
to
to
to
and
to
If
find
Solution
The existence of point implies that
is a cyclic quadrilateral. Now, we have
\begin{align*} \angle(AXB) &= 180-\angle(AXD)\\ &= 180-(180-\angle(XAD)-\angle(XDA))\\ &= \angle(XAD)+\angle(XDA) &= \angle(CAD)+\angle(BDA)\\ &= \frac{\overparen{CD}}{2}+\frac{\overparen{AB}}{2}\\ &= \frac{\angle(COD)}{2}+\frac{\angle(AOB)}{2}\\ &= \frac{120}{2} = 60^{\circ}. \end{align*} (Error compiling LaTeX. Unknown error_msg)
So,
In the same manner, we have
We have