Dhaka secondary '16 \6

For students of class 9-10 (age 14-16)
Absur Khan Siam
Posts:65
Joined:Tue Dec 08, 2015 4:25 pm
Location:Bashaboo , Dhaka
Dhaka secondary '16 \6

Unread post by Absur Khan Siam » Fri Jan 27, 2017 5:26 pm

Two tangent $PQ$ and $PR$ are drawn from external point $P$ to a circle with center $O$; where $Q, R$ are not the point of tangency. $Q, R$ are two points such that $PQ=PR$ and $O$ is the midpoint of the line $QR$. $X, Y$ are two points situated on $PQ$ and $PR$ respectively in such a way so that $XY$ is a tangent to the circle. If $QR=10$. Then find the value of $QX \times RY$ = ?
"(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: Dhaka secondary '16 \6

Unread post by Absur Khan Siam » Fri Jan 27, 2017 5:46 pm

My solution:
Let the point of tangency of $PQ$ is $T$ and the point of tangency of $PQ$ is $T'$.
Note that $\triangle PQO \cong \triangle PRO$.Thus $\angle POQ = \angle POR = 90$ degree.
Thus $QR || XY$ , $\triangle PXY$and $\triangle PQR$ similar , $PX = PY$ and $QX = YR$.Moreover $XT = XT'$.
Thus $\triangle XTO \cong \triangle XT'O$ , $\angle TXO = \angle T'XO$ and $\angle PXT' = \angle PQO$.
Therefore $\angle QXO = \angle XOQ = 90-\frac{\angle PQO}{2}$ and $QO = QX = 5$.
And we get,$QX \times RY = QX^2 = 5^2 = 25$ :mrgreen:
"(To Ptolemy I) There is no 'royal road' to geometry." - Euclid

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