2024 AIME I Problems/Problem 5
We use simple geometry to solve this problem.
We are given that ,
,
, and
are concyclic; call the circle that they all pass through circle
with center
. We know that, given any chord on a circle, the perpendicular bisector to the chord passes through the center; thus, given two chords, taking the intersection of their perpendicular bisectors gives the center. We therefore consider chords
and
and take the midpoints of
and
to be
and
, respectively.
We could draw the circumcircle, but actually it does not matter for our solution; all that matters is that , where
is the circumradius.
By the Pythagorean Theorem, . Also,
. We know that
, and
;
;
; and finally,
. Let
. We now know that
and
. Recall that
; thus,
. We solve for
:
\begin{align*} (x+92)^2+8^2&=25^2+92^2 \\ (x+92)^2&=625+(100-8)^2-8^2 \\ &=625+10000-1600+64-64 \\ &=9025 \\ x+92&=95 \\ x&=3. \\ \end{align*}
The question asks for , which is
.
~Technodoggo