Question

Point $P$ lies on the diagonal $AC$ of square $ABCD$ with $AP>CP.$ Let $O_1$ and $O_2$ be the circumcentres of $△ABP$ and $△CDP$ respectively. Given that $AB=12$ and $\angle O_{1}P{O}_2={120}^{\circ },$ then $AP=\sqrt{a}+\sqrt{b},$ where $a$ and $b$ are positive integers. Find $a+b.$
(correct answer + 5, wrong answer 0)

Answer 1

Let mid-point of $DC$ be $E$ and mid-point of $AB$ be $F$.
Since $O_1$ and $O_2$ are circumcentres,
therefore both lie on the pependicular bisectors of $DC$ and $AB$ passing through $E$ and $F$.
Now, $O_{1}P=O_{1}B$ and $O_{2}P=O_{2}D$
Also $\angle CAB=\angle ACD={45}^{\circ }$
$\therefore \angle B{O}_{1}P=2\times {45}^{\circ }={90}^{\circ }=\angle D{O}_{2}P$
($\because$ angle subtended by an arc of a circle at its centre $=2\times$ the angle subtended at its circumference)
$\therefore$ $\angle O_{1}PB$ and $\angle O_{2}PD$ are isosceles right triangles.

Using the above information and symmetry,
$\angle DPB={120}^{\circ }$
Also, $△ABP$ is congruent to $△ADP$ (SAS)
and $△CPB$ is congruent to $△CPD$ (SAS)
$\therefore \angle APB=\angle APD={60}^{\circ }$
and $\angle CPB=\angle CPD={120}^{\circ }$

Now, $\angle ABP=(180-60-45{)}^{\circ }={75}^{\circ }$
$\Rightarrow \angle O_{1}BF={30}^{\circ }$
and $\angle PDC=(180-120-45{)}^{\circ }={15}^{\circ }$
$\Rightarrow \angle O_{2}DE={30}^{\circ }$
$\therefore △O_{1}BF$ and $△O_{2}DE$ are right angled triangles.
$BF=DE=\frac{12}{2}=6$
$\Rightarrow O_{1}B=O_{2}D=4\sqrt{3}$
and $PB=PD=\frac{4\sqrt{3}}{\cos {45}^{\circ }}=4\sqrt{6}$

Now, let $x=AP$
Using law of cosine on $△ABP,$ we have
$96=x^2+144-24x\times \frac{1}{\sqrt{2}}$
$\Rightarrow x^2-12\sqrt{2}x+48=0$
$\Rightarrow x=\sqrt{72}\pm \sqrt{24}$
Taking the positive root, $AP=\sqrt{72}+\sqrt{24}$
$\therefore a+b=72+24=96$