# Point $X$ is on the circumference of the circle $PQR$ and $PY$ is a perpendicular on $XR$. Finding the value of $QX + XR$

$$\triangle PQR$$ is an isosceles triangle where $$PQ = PR$$. Point $$X$$ is on the cicumcircle of $$\triangle PQR$$ such that it is in the opposite region of $$P$$ with respect to $$QR$$. $$PY$$ $$\perp$$ $$XR$$ and $$XY$$ = $$12$$. What is the value of $$QX + XR$$?

I am unable to solve the problem because I couldn't use the condition of $$XY$$ = $$12$$ and $$\angle PYX$$ = $$90^\circ$$ $$\triangle PQR$$ being an isosceles triangle. Then how can I suppose to get the value of $$QX + XR$$?

A small hint will be enough for me to proceed.

• What's the source of your problem? Feb 11, 2019 at 9:21
• @Anirban Niloy Is $\Delta PQR$ equilateral? If so, the answer is $24$. Feb 11, 2019 at 9:35
• @Michael Rozenberg No. $\triangle PQR$ is isosceles with having $PQ$ and $PR$ both equal. The diagram really looks like a little bit disturbed. Feb 11, 2019 at 9:41

Choose a point $$R'\in [XY]$$ such that $$\color{red}{YR=YR'}$$ $$\Rightarrow PR'=PR=PQ\Rightarrow \angle PR'Q=\angle R'QP$$.

Since the quadrilateral $$PRXQ$$ is cyclic: $$\angle PRY=\angle YR'P=180°-\angle XQP \Rightarrow \angle PR'X=\angle XQP$$

$$\because \angle PR'Q=\angle R'QP \Rightarrow \angle XQR'=\angle QR'X$$ $$\therefore \color{red}{XR'=XQ}\Rightarrow QX+XR=XR'+XY+R'Y=2XY=24$$

• An alternative construction from which very similar logic follows is PZ perpendicular to XQ. Feb 11, 2019 at 9:53
• Very elegant, doctor! +1 Feb 11, 2019 at 10:01

Let $$\measuredangle PRQ=\alpha$$ and $$\measuredangle XRQ=\beta$$.

Thus, $$PX=\frac{12}{\cos\alpha}$$ and by the law of sines for $$\Delta QPX$$ and for $$\Delta PRX$$ we obtain: $$\frac{QX}{\sin\beta}=\frac{\frac{12}{\cos\alpha}}{\sin(\alpha+\beta)}$$ and $$\frac{RX}{\sin(2\alpha+\beta)}=\frac{\frac{12}{\cos\alpha}}{\sin(\alpha+\beta)}.$$ Id est, $$QX+RX=\frac{12\sin(2\alpha+\beta)+12\sin\beta}{\cos\alpha\sin(\alpha+\beta)}=\frac{24\sin(\alpha+\beta)\cos\alpha}{\cos\alpha\sin(\alpha+\beta)}=24.$$

• I don't know the law of sines that you have applied. So, it is my above my understanding capacity because I'm only in 10th grade. But, well done!!😅😅😅 Feb 11, 2019 at 10:06

Drop A perpendicular from $$P$$ onto $$QR$$ at $$S$$, so $$QR=2\cdot SR$$. And let $$PR=PQ=a$$

Note that $$\angle PXR=\angle PXQ=\angle PRS=\alpha$$

Now from Ptolemy theorem in $$PQXR$$:

$$QX\cdot PR+XR\cdot PQ=PX\cdot QR$$ $$a\cdot(QX+XR)=(12\sec\alpha)\cdot 2a\cos\alpha\\QX+XR=24$$