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# Inequality for sides of a triangles

2017-12-03 03:06:08

I recently encountered this geometric problem (I could not find a better title).

Let $ABC$ a triangle and $P$ a point on the segment $AB$. Let $Q$ be

the intersection of the line $CP$ with the circumscribed circle of the

triangle different from $C$. Show the inequality

$$\frac{\overline{PQ}}{\overline{CQ}} \le \Big(\frac {\overline{AB}}{\overline{AC}+\overline{CB}}\Big)^2 .$$ Further,

show that equality holds if and only if $CP$ is the angle bisector of

the angle $\angle ACB$.

As in most problems of this kind, finding the solution is a matter drawing the right lines. I already come up with a solution. I wanted to share the problem as there are probably many different ways to solve it.

To be clear: I don't need help with it, I just thought maybe there are some people here who like to do this kind of problems and want to try themselves on it.

First, let $CPQ$ bisect $\angle ACB$, and join $AQ$, $BQ$. Then $AQ=BQ$, and by Ptolemy's Theorem $$AB\ • First, let CPQ bisect \angle ACB, and join AQ, BQ. Then AQ=BQ, and by Ptolemy's Theorem$$AB\times CQ=AQ\times(CA+CB)$$Dividing both sides by AC\times AQ, inverting and squaring$$\frac{AQ^2}{CQ^2}=\frac{AB^2}{(AC+CB)^2}$$Thus we must show that$$\frac{PQ}{CQ}=\frac{AQ^2}{CQ^2}=\frac{AB^2}{(AC+CB)^2}$$or$$PQ\times CQ=AQ^2$$But since \angle ACB is bisected, \angle ACQ=\angle QAB, making triangles ACQ and QAB similar. Hence$$\frac{PQ}{AQ}=\frac{AQ}{CQ}$$or$$PQ\times CQ=AQ^2$$Thus the ratios are equal when \angle ACB is bisected. Now let R be any other point on AB, and draw CR thru to S on the circumference, and join QS. Then since arcs AQ, BQ are equal, the tangent at Q is parallel to AB, and therefore chord QS is inclined toward PR, making$$\frac{SR}{RC}<\frac{QP}{PC}$$Hence$$\frac{SR}{SR+RC}<\frac{QP}{QP+PC}$$or$$\frac{SR}{SC}<\frac{QP}{QC}$$And since \frac{AB^2}{(AC+CB)^2} is fixed$$\frac{SR}{SC}<\frac{AB^2}{(AC+CB)^2}

2017-12-03 05:37:36