$ABCD$ is a rectangle with side $AB$ of length 10 and side $BC$ of length 11. An interior point $P$ is chosen such that the sum of angles $APD$ and $BPC$ is $180^\circ$ . All possible points $P$ form a hyperbola that can be represented by $x^2 - y^2 = K,$ where $K$ is a positive real number. Find the value of $K$ .

**
Note:
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The hyperbola assumes that the
$x$
- and
$y$
-axes are parallel to the sides of rectangle
$ABCD$
with the origin at the center of the rectangle.

*
Extension of
this problem
*

The answer is 5.25.

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We have $A = (-5.5, 5 ), B = ( -5.5, -5), C = (5.5, -5), D = (5.5, 5)$ .

If $P$ is a point on the locus, then by translating the image down 10 units, we obtain that $BPCP'$ is a cyclic quadrilateral. Observe that trivially, $ABBC$ and $DCCB$ are cyclic quadrilaterals, which means that $A , D$ lie on the locus. Hence,

with the assumptionthat the locus is a hyperbola of the form $x^2 - y^2 = K$ , we obtain that $K = 5.5^2 - 5^2 = 5.25$ .Now, let's determine the locus properly. Suppose that $P = (X,Y) , P' = (X, Y-10)$ is on the locus. We will use the fact that $BPCP'$ is a cyclic quadrilateral. The center of the circle can be determined by the intersection of perpendicular bisectors. Since the perpendicular bisector of $BC$ is $x = 0$ and the perpendicular bisector of $P P '$ is $y = Y - 5$ , we get that the center of the circle is $E = (0, Y - 5 )$ . The radius of the circle is $EB = (5.5)^2 +(Y-5 + 5)^2$ . Hence, the point $P$ satisfies

$(X - 0)^2 + (Y - (Y-5) ) ^2 = 5.5 ^ 2 + Y^2$

Hence, $X^2 - Y^2 = 5.5^2 - 5^2 = 5.25$ .