JHMT 2021 Geometry Problem 6

2021 JHMT Geometry Problem 6

JHMT is a convex quadrilateral with perimeter $68$ and satisfies $\angle HJT = 120^\circ$, $HM = 20$, and $JH + JT = JM > HM$. Furthermore, ray $\overrightarrow{JM}$ bisects $\angle HJT$. Compute the length of $JM$.

Diagram

There’s no diagram given in this problem, so our first step is to make our own. This was my first pass (digitized from paper, not to scale):

It seemed like JT needed to look fairly “small” in order for $JH+JT=JM$, so this is how I decided to visualize it.

Construction

I’m reminded of some advice I saw in Carl Joshua Quines’ excellent handout Constructions¹.

If you have a condition like $AB + CD = EF$, you almost always want to “unroll” or “open the gates”.

The first is constructive: turning the sum of two segments into a single segment. If $AX + AY$ appears on one side, construct a point $Y'$ on ray $AX$ such that $XY' = AY$. Then $AX + AY = AY'$, and $AY'$ will hopefully form an isosceles triangle, parallelogram, isosceles trapezoid, or cyclic quadrilateral.

Note that you can try constructing on ray $AY$ instead. Constructing in the opposite direction sometimes helps as well.

This problem follows that advice perfectly. Start by extending ray $\overrightarrow{JH}$ to the point $T'$ such that $HT'=JT$:

Beautiful! Not only is $\triangle JMT'$ isoceles, it’s equilateral since the vertex angle is $60^\circ$. From this we know three pieces of information:

• $\angle HT'M=60^\circ$,
• $MT'=MJ$, and
• $HT'=JT$.

By SAS, $\triangle HMT'\cong \triangle TMJ$! This congruence gives us the critical fact that $MT=MH$. Since the perimeter of the whole quadrilateral is $68$ and two of the sides are $20$, the remaining segments $JH$ and $JT$ must add up to $28$. But we’re given that $JM$ is the sum of those two segments so our final answer is $\boxed{28}$.

Alternative Construction

Try rotating the entire diagram $60^\circ$ CCW around point $J$.

Footnotes

1. https://cjquines.com/files/constructions.pdf ↩