Let $ABC$ be a triangle with $AB > AC$. Let $P$ be a point on the line $AB$ beyond $A$ such that $AP +P C = AB$. Let $M$ be the mid-point of $BC$ and let $Q$ be the point on the side $AB$ such that $CQ \perp AM$. Prove that $BQ = 2AP.$
Problem
Source: RMO 2014
Tags: geometry
Mathphile01
13.08.2018 20:35
Looks like I can manage geometry even now Solution: Let $X$ be a point on the line $AB$, above $P$ such that $PX$=$PC$ Hence using the sum of sides condition We observe That $A$ is the midpoint of side $XB$ Thus in triangle $XCB$,$AM$ is parallel to $XC$ (By midpoint theorem's converse) Now,denoting Angle $PXC$ as $\alpha$ and angle chasing Gives Angle $PCQ$=$PQC$=$90-\alpha$ Implying triangle $PCQ$ is isosceles Thus $PC=PQ$ Or $AB-AP=AB-BQ+AP$ which gives $BQ=2AP$ Notice That $AB$>$AC$ allows us to construct the desired line-segment and finish the problem
guptaamitu1
09.10.2020 00:14
file:///C:/Users/arjun/OneDrive/0_Arjun/IMO/RMO%20past%20year%20papers/my%20solution,%20RMO%202014-5%20(paper%202).pdf Go to the above link to check my solution to the problem. I used similarity and manipulated some ratios to prove the desired result (Though, I agree that the official solution is much more elegant)
franzliszt
09.10.2020 00:20
I get an error when viewing the file. Please attach since it appears to be stored on your local.
Jishnu4414l
13.12.2023 19:32
Note that the problem statement is equivalent to saying that $\triangle PQC$ is isosceles.
Proof: Extend $BP$ to $F$ beyond $P$ such that $PC=PF$.
Now, we have $AF=AP+PF=AP+PC=AB$.
This leads us to the conclusion that $A$ is the midpoint of $BF$.
Now by the converse of Midpoint-Theorem, we have that $AM \parallel FC$.
Now let $\angle AFC=\angle ACF=\theta$.
$\angle APC=2\theta$ (exterior angle).
also as $AM \parallel CF$, we must have $CF\perp QC$.
Now a very easy and trivial angle chase gives us $\angle PCQ=90^{\circ}-\theta$.
This gives $\triangle PQC$ is isosceles.
Our proof is thus complete.