#21 of Sept 2017 Grade 10 CSAT (Korean SAT) Mock test

$\overline{AC}=2,$ and therefore $\triangle ABC$ is an isosceles triangle, leading to $\overline{BE}=\overline{AE}=\sqrt{3}$ and $\overline{CE}=1.$

Since $\overline{CE}$ and $\overline{AD}$ are both midlines of $\triangle ABC,$ $\overline{PE}=\dfrac{1}{3},$ $\overline{PC}=\dfrac{2}{3},$ $\overline{PA}=\dfrac{2\sqrt{7}}{3},$ and $\overline{PD}=\dfrac{\sqrt{7}}{3}.$ $P$ is the centroid of $\triangle ABC.$

$\overline{QR}$ bisects $\angle APE$ and $\angle CPD,$ so we know that

$\overline{RA}:\overline{RE}=\overline{PA}:\overline{PE}=2\sqrt{7}:1,$ and $\overline{QD}:\overline{QC}=\overline{PD}:\overline{PC}=\sqrt{7}:2.$

Then, $\triangle PRE=\dfrac{1}{2\sqrt{7}+1}\triangle APE,$ and $\triangle PQC=\dfrac{2}{2+\sqrt{7}}\triangle CPD.$

$\triangle APE=\triangle CPD$ from $P$ being the centroid, and therefore,

$\begin{aligned} a+b\sqrt{7} &=\dfrac{\dfrac{2}{2+\sqrt{7}}}{\dfrac{1}{2\sqrt{7}+1}} \\ \\ &=\dfrac{2(2\sqrt{7}+1)}{\sqrt{7}+2} \\ \\ &=\dfrac{2(2\sqrt{7}+1)(\sqrt{7}-2)}{3} \\ \\ &=\dfrac{2(12-3\sqrt{7})}{3} \\ \\ &=8-2\sqrt{7}. \end{aligned}$

$\therefore \, ab=8\cdot(-2)=\boxed{-16}.$

Here's my horribly messy solution.

First, let us find AC. I haven't heard of Apollonius' theorem before checking H.M.'s solution, so I did something else.

Notice how $S_{\Delta ABD}=S_{\Delta ACD}$ (as AD is a median they have the same base, and the height from A is the same for both). Thus, we can use Heron's formula to deduce that:

$(1+12+7)^2-2(1+49+144)=(1+7+AC^2)^2-2(1+49+AC^4)$

$AC^4-16AC^2+48=0$

$(AC^2-8)^2=16$

$AC^2=8\pm 4$

$AC=2\ or\ 2\sqrt{3}$

The second solution doesn't work as the length of AD should be $\sqrt{11}$ by the Pythagorean theorem . Thus, $AC=2$ and the triangle is isosceles, so CE is also a median and a height and by the Pythagorean theorem $CE=1$ . By P being the centroid , $CP=\frac{2}{3},\ EP=\frac{1}{3},\ DP=\frac{\sqrt{7}}{3}\,\ AP=\frac{2\sqrt{7}}{3}$ .

Lets denote $\angle DPC=\angle APE=\theta$ . Now see that $S_{\Delta DPC}=\frac{DP·CP·sin\theta}{2}=\frac{\sqrt{7} sin\theta}{9}=\frac{EP·AP·sin\theta}{2}=S_{\Delta APE}$ , which we'll denote $S$ .

By the angle bisector theorem $CQ=\frac{CP}{CP+DP}CD=\frac{2}{2+\sqrt{7}}CD$ , and similarly $ER=\frac{1}{1+2\sqrt{7}}$ .

Now $S_{\Delta QPC}=S_{\Delta DPC} \frac{2}{2+\sqrt{7}} = S_{\Delta EPA} \frac{2}{2+\sqrt{7}} = S_{\Delta EPR} \frac{2}{2+\sqrt{7}}(1+2\sqrt{7})= S_{\Delta EPR}(8-2\sqrt{7})$ .

Thus, $ab=8·(-2)=-16$ .

▄

P.S. I think this is horribly hard for 10th graders. Ah well, maybe korean 10th graders are tougher than this.

Did same as me myself.But this has much toughness in it compared to class 10 graders.This maybe for the maths Olympiads enthusiasts instead.