## Rajshahi '15 \10

For students of class 6-8 (age 12 to 14)
Absur Khan Siam
Posts: 65
Joined: Tue Dec 08, 2015 4:25 pm
Location: Bashaboo , Dhaka

### Rajshahi '15 \10

$PQR$ is a triangle.$SP$ is the angle bisector of $\angle QPR$ and $ST$ is the perpendicular bisector of $PR$.If $QS=9cm$ and $SR=7cm$ then $PR = \frac{x}{y}$ where $x, y$ are coprimes. $x +y = ?$
"(To Ptolemy I) There is no 'royal road' to geometry." - Euclid

Absur Khan Siam
Posts: 65
Joined: Tue Dec 08, 2015 4:25 pm
Location: Bashaboo , Dhaka

### Re: Rajshahi '15 \10

My solution:
"(To Ptolemy I) There is no 'royal road' to geometry." - Euclid

Posts: 181
Joined: Mon Mar 28, 2016 6:21 pm

### Re: Rajshahi '15 \10

Thanic Nur Samin, let us have some glory.
As $ST$ is the perpendicular bisector, $\angle SPT=\angle SRT$. Which is also equal to $\angle QPS$. We get, $\triangle QPS$ is similar to $\triangle QRP$. From length chasing and the similar triangle, we get, $PQ= 9^(1/2)*16^(1/2)= 12$. We have, $PS/PQ= PR/QR$. So, $PR=7*16/12=28/3$ l. So, $x+y=31$.
Frankly, my dear, I don't give a damn.

Thanic Nur Samin
Posts: 176
Joined: Sun Dec 01, 2013 11:02 am

### Re: Rajshahi '15 \10 Geom.png (10.53 KiB) Viewed 918 times
Glory you say?

$SP=SR=7$ for perpendicular biscector reasons. Quickly recalling the angle biscector theorem, $\dfrac{PQ}{PR}=\dfrac{SQ}{SR}=\dfrac{9}{7}$. So we can set $PQ=18t$ and $PR=14t$. Let $Y$ be the foot of perpendicular from $S$ to $PQ$. Now, clearly, $PY=7t$ and $YQ=11t$.

Apply perpendicular lemma to get $9^2-7^2=(11t)^2-(7t)^2$ from which we get $t=\dfrac{2}{3}$. Thus $PR=14t=\dfrac{28}{3}$ and the desired answer is $31$.
Hammer with tact.

Because destroying everything mindlessly isn't cool enough.