[u]Set 3[/u]


[b]p7.[/b] On unit square $ABCD$, a point $P$ is selected on segment $CD$ such that $DP =\frac14$ . The segment $BP$ is drawn and its intersection with diagonal $AC$ is marked as $E$. What is the area of triangle $AEP$?


[b]p8.[/b] Five distinct points are arranged on a plane, creating ten pairs of distinct points. Seven pairs of points are distance $1$ apart, two pairs of points are distance $\sqrt3$ apart, and one pair of points is distance $2$ apart. Draw a line segment from one of these points to the midpoint of a pair of these points. What is the longest this line segment can be?


[b]p9.[/b] The inhabitants of Mars use a base $8$ system. Mandrew Mellon is competing in the annual Martian College Interesting Competition of Math (MCICM). The first question asks to compute the product of the base $8$ numerals $1245415_8$, $7563265_8$, and $ 6321473_8$. Mandrew correctly computed the product in his scratch work, but when he looked back he realized he smudged the middle digit. He knows that the product is $1014133027\blacksquare 27662041138$. What is the missing digit?


PS. You should use hide for answers.
1. Place the unit square \(ABCD\) onto the Cartesian Plane such that:
   \[
   A = (0, 1), \quad B = (1, 1), \quad C = (1, 0), \quad D = (0, 0)
   \]
2. Point \(P\) is selected on segment \(CD\) such that \(DP = \frac{1}{4}\). Therefore, the coordinates of \(P\) are:
   \[
   P = \left(0, \frac{1}{4}\right)
   \]
3. The equation of line \(BP\) can be found using the slope formula. The slope \(m\) of \(BP\) is:
   \[
   m = \frac{\frac{1}{4} - 1}{0 - 1} = \frac{-\frac{3}{4}}{-1} = \frac{3}{4}
   \]
   Thus, the equation of line \(BP\) is:
   \[
   y = \frac{3}{4}x + \frac{1}{4}
   \]
4. The equation of the diagonal \(AC\) is:
   \[
   y = x
   \]
5. To find the intersection point \(E\) of lines \(BP\) and \(AC\), set the equations equal to each other:
   \[
   \frac{3}{4}x + \frac{1}{4} = x
   \]
   Solving for \(x\):
   \[
   \frac{3}{4}x + \frac{1}{4} = x \implies \frac{1}{4} = \frac{1}{4}x \implies x = \frac{1}{3}
   \]
   Therefore, the coordinates of \(E\) are:
   \[
   E = \left(\frac{1}{3}, \frac{1}{3}\right)
   \]
6. To find the area of triangle \(AEP\), we use the Shoelace Formula. The vertices of triangle \(AEP\) are \(A(0, 1)\), \(E\left(\frac{1}{3}, \frac{1}{3}\right)\), and \(P\left(0, \frac{1}{4}\right)\). The Shoelace Formula is given by:
   \[
   \text{Area} = \frac{1}{2} \left| x_1y_2 + x_2y_3 + x_3y_1 - (y_1x_2 + y_2x_3 + y_3x_1) \right|
   \]
   Substituting the coordinates:
   \[
   \text{Area} = \frac{1}{2} \left| 0 \cdot \frac{1}{3} + \frac{1}{3} \cdot \frac{1}{4} + 0 \cdot 1 - (1 \cdot \frac{1}{3} + \frac{1}{3} \cdot 0 + \frac{1}{4} \cdot 0) \right|
   \]
   Simplifying:
   \[
   \text{Area} = \frac{1}{2} \left| 0 + \frac{1}{12} + 0 - \left(\frac{1}{3} + 0 + 0\right) \right| = \frac{1}{2} \left| \frac{1}{12} - \frac{1}{3} \right| = \frac{1}{2} \left| \frac{1}{12} - \frac{4}{12} \right| = \frac{1}{2} \left| -\frac{1}{4} \right| = \frac{1}{2} \cdot \frac{1}{4} = \frac{1}{8}
   \]

The final answer is \(\boxed{\frac{1}{8}}\)