AoPS Wiki talk:Problem of the Day/June 18, 2011

Problem

AoPSWiki:Problem of the Day/June 18, 2011

Solution

Let $AD = x$ . Since we know that the perimeter of $ABCD$ is $42$ , $BC = 42 - 9 - 15 - x = 18 - x$ . Construct a perpendicular to $DC$ from point $B$ . Call the intersection of $DC$ and this perpendicular point $E$ . Since angle $A$ is right, $ABED$ is a rectangle, and thus angle $BEC$ is right, and $AB = DE = 9$ . $EC = 15 - 9 = 6$ ; by the Pythagorean Theorem, $BE = x = 8$ , so $AD = 8$ and $BC = 18 - 8 = 10$ . Again by the Pythagorean Theorem, $AC = 17$ . Since $AB$ and $DC$ are bases, $AB$ is parallel to $DC$ , and so angle $BAX$ = angle $DCX$ ; by vertical angles, angle $AXB$ = angle $DXC$ . By AA, triangle $DXC$ ~ triangle $BXA$ in a ratio of $\frac{9}{15}$ . Thus, $\frac{AX}{XC} = \frac{9}{15}$ , so $XC$ is $\frac{15}{24}$ of $AC$ . Note that the distance from $X$ to $DC$ is really the length of the perpendicular to $DC$ drawn from $X$ by definition. Drawing this perpendicular and letting the point of intersection of $DC$ and $X$ be point $F$ , we have that triangle $XFC$ is similar to triangle $ADC$ by AA in a ratio of $\frac{15}{24}$ . Thus $\frac{XF}{AD} = \frac{15}{24}$ , so $\frac{XF}{8} = \frac{15}{24}$ . Solving, we get $XF = \boxed{5}$ .

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AoPS Wiki talk:Problem of the Day/June 18, 2011

Problem

Solution