In an equilateral triangle $PQR$, let the vertex $P$ be at $(3, 5)$ and the side $QR$ be along the line $x + y = 4$. If the orthocentre of the triangle $PQR$ is $(\alpha, \beta)$, then $9(\alpha + \beta)$ is equal to:

Let $P$ be an interior point of a convex quadrilateral $ABCD$ and $K, L, M, N$ be the mid-points of $AB, BC, CD, DA$ respectively. If $\text{Area}(PKAN) = 25$,$\text{Area}(PLBK) = 36$,and $\text{Area}(PMDN) = 41$,then $\text{Area}(PLCM)$ is

Consider four triangles with side lengths $(5, 12, 9)$,$(5, 12, 11)$,$(5, 12, 13)$,and $(5, 12, 15)$. Among these,the triangle with the maximum area has sides:

Let $ABC$ be a triangle with $A(-3, 1)$ and $\angle ACB = \theta$,where $0 < \theta < \frac{\pi}{2}$. If the equation of the median through $B$ is $2x + y - 3 = 0$ and the equation of the angle bisector of $C$ is $7x - 4y - 1 = 0$,then $\tan \theta$ is equal to:

In a triangle $ABC$,$\angle BAC = 90^{\circ}$; $AD$ is the altitude from $A$ onto $BC$. Draw $DE$ perpendicular to $AC$ and $DF$ perpendicular to $AB$. Suppose $AB = 15$ and $BC = 25$. Then the length of $EF$ is

In a $\triangle ABC$,points $X$ and $Y$ are on $AB$ and $AC$,respectively,such that $XY$ is parallel to $BC$. Which of the two following equalities always hold? (Here $[PQR]$ denotes the area of $\triangle PQR$).
$I$. $[BCX] = [BCY]$
$II$. $[ACX] \cdot [ABY] = [AXY] \cdot [ABC]$