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In △PQR, ∠PRQ=60° and PS=SR. If ∠PQS=x°, compare x with 60 (i.e, greater, equal or the relation can't be determined)

To keep ∠PRQ fixed, I first thought of an equilateral triangle, so that x=30°. Now if I increase angle ∠QPR, angle ∠PQR will decrease. Though I don't understand here whether ∠PQS will be greater than 30° or less than 30°. But, ∠PQS will definitely be less than 60°. Now if I extend RP to the left, let upto T point, so that ∠QPR decreases, while ∠PQR will increase, in which case I think ∠PQS (=x) can be greater than 60°.

But, can't grasp the whole picture here. Can anyone clarify, how to deal with it?

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  • $\begingroup$ In fact, the largest possible value of $ x^{\circ} $ is $ \cos^{-1} \! \left( \dfrac{9 \sqrt{2} - 1}{14} \right) = 33.10104 \ldots^{\circ} $. $\endgroup$ – Transcendental Sep 29 '16 at 9:56
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Consider a circle $\omega$ passing through points $P$ and $S$ that is tangent to line $RQ$ at point $X$. Then $Q$ lies outside the circle or coincides with $X$. Therefore $$\angle PQS \le \angle PXS.$$

enter image description here

The problem reduces to showing that $\angle PXS < 60^\circ$.

Since $\angle RPX = \angle RXS$ and $\angle XRP = \angle SRX$, triangles $RPX$ is similar to triangle $RXS$. Therefore $$\frac{PR}{RX}=\frac{RX}{RS}$$ which implies $XR^2 = PR \cdot SR = 2SR^2$. Denoting $SR=x$ we find $PR=2x$ and $XR = \sqrt 2 x$.

Let $PX = y$. Using law of cosines in triangle $PXR$ we calculate $$y^2 = (2x^2) + (\sqrt 2 x)^2 - 2 \cdot 2x \cdot \sqrt 2 x \cdot \cos 60^\circ = (6-2\sqrt 2)x^2.$$

Using similarity of $RPX, RXS$ again we get $\frac{XS}{PX} = \frac{XR}{PR} = \frac 1{\sqrt 2}$.

Finally, law of cosines in triangle $PXS$ yields $$\cos \angle PXS = \frac{y^2 + \left(\frac{y}{\sqrt 2}\right)^2-x^2}{2\cdot y \cdot \frac{y}{\sqrt 2}} = \ldots = \frac{9\sqrt 2-1}{14}.$$

$\cos$ is a decreasing function on $[0,\pi]$ so you only need to check that $\cos \angle PXS > \cos 60^\circ = \frac 12$. I'll leave it to you.

More geometric way: As we calculated above, $y^2 = (6-2\sqrt 2)x^2 > 2x^2$, therefore $XP>XR$. Therefore $\angle PXS < \frac 12 \angle PXR$. This is because the angle bisector of $PXR$ cuts $PR$ in a point $Y$ such that $$\frac{PY}{YR} = \frac{PX}{XR} >1$$

enter image description here

(this is called angle bisector theorem). So $$\angle PXS < \frac 12 \angle PXR = \frac 12 (180^\circ - \angle XRP - \angle RPX) = \frac 12(120^\circ - \angle RPX) < \frac 12 \cdot 120^\circ = 60^\circ.$$

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  • $\begingroup$ Thank you for replying, but I din't understand! can you please clarify a bit? $\endgroup$ – Mahmudul Hasan Sep 29 '16 at 8:29
  • $\begingroup$ I finished the proof. What exactly don't you understand? $\endgroup$ – timon92 Sep 29 '16 at 8:32
  • $\begingroup$ I missed it from the beginning-- "Consider a circle ω passing through points PP and SS that is tangent to line RQ at point X. Then Q lies outside the circle or coincides with X" $\endgroup$ – Mahmudul Hasan Sep 29 '16 at 8:38
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    $\begingroup$ I'll attach a picture, maybe it will be clear then $\endgroup$ – timon92 Sep 29 '16 at 8:45
  • $\begingroup$ Using coordinate geometry, one can show that the largest possible value of the angle is $ \cos^{-1} \! \left( \dfrac{9 \sqrt{2} - 1}{14} \right) = 33.10104 \ldots^{\circ} $. $\endgroup$ – Transcendental Sep 29 '16 at 10:04

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