-1
$\begingroup$

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 $m$?

  • 0
    This is from the **ongoing** [USAMTS competition (problem 4)](http://www.usamts.org/Tests/Problems_23_2.pdf). Of course, Alan, I'm sure you are aware that it would against [the rules of the competition](http://www.usamts.org/Rules/U_RulesComplete.php) to submit this as an answer - so you're just asking out of curiosity. Right?2011-11-05

1 Answers 1

1

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 $P$).