Shortest Path #4

Geometry Level 5

The ellipse in the diagram (not drawn to scale) is centered at $O=(0, 0)$ and has major axis with length 20 and minor axis with length 6. The major axis forms an angle of $\theta$ with the $x$ -axis such that $\sin\theta=\frac{3\sqrt{273}}{91}$ and lies in quadrants I and III.

Now, let $P=\big(8,3\sqrt{3}\big), Q=\big(-4, -2\sqrt{3}\big),$ and $R$ be a point on the ellipse. Also, let $m$ and $M$ denote the minimum and maximum of the perimeter of triangle $PQR,$ respectively.

If $m M=a+b\sqrt{c},$ where $a,b,c$ are integers and $c$ is square-free, find the value of $a+b+c.$

This problem is a part of <Shortest Path> series .

The answer is 859.

1 solution

Boi (보이)
Aug 11, 2017

The major axis length is 20, and the minor axis length is 6. We know that the distance between the two foci is $2\sqrt{91}.$ And since $\overline{OP}=\sqrt{91}$ and the sine of the angle that $\overline{OP}$ forms with the $x$ -axis is $\dfrac{3\sqrt{273}}{91},$ point $P$ is one of the foci of this ellipse.

Then consider a point $P',$ which is another focus of this ellipse. $P'(-8,~-3\sqrt{3}).$

Think back to the definition of an ellipse.

$\overline{PR}+\overline{P'R}=\text{major axis length}=20.$

Note that $\overline{QR}+\overline{P'Q}\ge\overline{P'R}$ and $\overline{P'R}+\overline{P'Q}\ge\overline{QR},$ so

$-\overline{P'Q}\le\overline{QR}-\overline{P'R}\le\overline{P'Q}$

$\overline{P'Q}=\sqrt{19}.$

Therefore,

$\begin{aligned} &\overline{PQ}+\overline{QR}+\overline{PR} \\ &=\sqrt{219}+\overline{QR}-\overline{PR'}+20 \\ \end{aligned}$

So, $m=20+\sqrt{219}-\sqrt{19}$ and $M=20+\sqrt{219}+\sqrt{19}.$

$m\cdot M=400+219+40\sqrt{219}-19=600+40\sqrt{219}.$

$\therefore~a+b+c=600+40+219=\boxed{859}.$

Shortest Path #4

The answer is 859.

1 solution

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