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