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

Unread post by Absur Khan Siam » Tue Jan 17, 2017 11:10 pm

$PQR$ 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

Unread post by Absur Khan Siam » Tue Jan 17, 2017 11:38 pm

My solution:
Let $T'$ is a point on $PQ$ such that $PT' = PT$.
Now,$\angle SPT = \angle SPT' , PT = PT' , PS = PS$
So,$\triangle PST \cong \triangle PST'$.
So,$\angle QTS = 90$ degree.
Because $\frac{QS}{SR} = \frac{9}{7}$.So $\frac{PQ}{PR} = \frac{9}{7}$.
Thus, $PQ = 9q , PR = 7q , q$ is a real number.
$RT = \frac{7q}{2} , QT' = 9q - \frac{7q}{2} = \frac{11q}{2}$
$SR^2 = RT^2 + ST^2 \rightarrow \frac{49q^2}{4} + ST^2 = 49...(i) $
$QS^2 = ST^2 + QT^2 \rightarrow \frac{121q^2}{4} + ST^2 = 81...(ii) $
$(ii)-(i) \rightarrow q = \frac{4}{3}$
$PR = 7 \times \frac{4}{3} = \frac{28}{3}$
$x+y = 31$
"(To Ptolemy I) There is no 'royal road' to geometry." - Euclid

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ahmedittihad
Posts: 181
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Re: Rajshahi '15 \10

Unread post by ahmedittihad » Tue Jan 17, 2017 11:43 pm

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.

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Thanic Nur Samin
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Re: Rajshahi '15 \10

Unread post by Thanic Nur Samin » Wed Jan 18, 2017 12:27 am

Geom.png
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.

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