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In a convex quadrilateral $ABCD$ diagonals $AC$ and $BD$ intersect at point $S$. Denote with $P$ the center of circumcircle of triangle $ABS$. Denote with $Q$ the center of circumcircle of triangle $CDS$. Prove that:

$$4 PQ \geq AB + CD.$$

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I was able to calculate the right side based on angle between diagonals and circle radii. I was also able to make some estimate of the distance between circle centers. But I could not make the match between two formulas.

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A graph is drawn below, together with the explanation of the notations. To simplify the explanation, we only restrict to the most non-trivial case: $\angle B \geq \angle A$, $\angle C \geq \angle D$.

two rays and two triangles

It is not hard to see that $\angle PSA = \frac{\theta + \angle A - \angle B}{2}$ and $\angle QSD = \frac{\theta + \angle D - \angle C}{2}$. Therefore

$$ \angle PSQ = \frac{2\theta + \angle A - \angle B + \angle D - \angle C}{2} + 180^\circ - \theta \geq 180^\circ - \frac{\angle B + \angle C}{2} \geq \theta. $$

By law of sines, $\overline{AB} + \overline{CD} = 2(r_1 + r_2)\sin\theta$.

By law of cosines,

$$ \begin{aligned} \overline{PQ} &\geq \sqrt{r_1^2 + r_2^2 - 2r_1r_2\cos\theta} = \sqrt{(r_1+r_2)^2 - 2(1+\cos\theta)r_1r_2}\\ &\geq\sqrt{(r_1+r_2)^2 - \frac{1}{2}(1+\cos\theta)(r_1+r_2)^2}\\ &= (r_1+r_2)\sqrt{\frac{1-\cos\theta}{2}} = (r_1+r_2)\sin\frac{\theta}{2} \geq \frac{1}{2}(r_1+r_2)\sin\theta. \end{aligned} $$

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    $\begingroup$ I have acceppted your solution but I have found a simpler way to prove that $\angle PSQ \ge \theta:$ $\angle PSA = 90 - \angle B$, $\angle DSQ = 90 - \angle C$, $\angle ASD = 180 - \theta$. Therefore: $\angle PSQ = 360 - \angle B - \angle C - \theta \ge \theta$ $\endgroup$ – Oldboy Dec 3 '17 at 9:49
  • $\begingroup$ That's true. Thank you! $\endgroup$ – Hw Chu Dec 3 '17 at 13:21
  • $\begingroup$ You did all the hard work, I have polished just the easies part. :) $\endgroup$ – Oldboy Dec 3 '17 at 15:15

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