# Difficult Geometry Problem--Circular Arcs

Good Afternoon,

I have spent a couple of hours trying to solve this problem, but it appears that I have not gotten anywhere.

Let $R$,$P$,$Q$ be points such that $\angle RQP=75^\circ$. Also we are given that $RQ=\sqrt{2}$ and $PQ=\sqrt{3}$. A lens is a region bounded by two circular arcs meeting at the endpoints. Let $m$ be the area of the largest lens that does not intersect lines $QR$ and $QP$, with endpoints at $R$ and $P$. How would we find $m$?

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This is from the ongoing USAMTS competition (problem 4). Of course, Alan, I'm sure you are aware that it would against the rules of the competition to submit this as an answer - so you're just asking out of curiosity. Right? –  Zev Chonoles Nov 5 '11 at 19:54

The max luns with that requirements will have one arc being tangent to a side by $R$. The other one will be tangent to the other side by $P$. It depends on the length of the lines wich one will be the outer one, and wich one the inner one.

How to find it?

Find the middle point of $RP$ (call it $X$) and the perpendicular to $RP$ by $X$ (call it center-line).

The inner part of the luns will emerge from a circle having its center on this line (center-line). Find its center intersecting the center-line with the perpendicular to $QP$ by $P$ (longest segment).

The outer part will emerge from a circle having its center on center-line, but intersecting it with the perpendicular to $QR$ by $R$ (shortest segment).

Proof: by finding center-line you are finding all points equidistant to $R$ and $P$. So any circle starting and that line and passing by $R$ will also pass by $P$ (and viceversa). By finding the perpendicular to $QR$ passing by $R$ you find a line containg the center of all circles being tangent to $QR$ by $R$ (and the same goes for $QP$ by $P$).

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