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Geometry Difficulty 4.8 AIME Find the answer

Let ABCDA B C D be a rectangle with AB=20A B=20 and AD=23A D=23. Let MM be the midpoint of CDC D, and let XX be the reflection of MM across point AA. Compute the area of triangle XBDX B D.

A number or a short expression. Spacing and $ signs are ignored.

Solution

Observe that [XBD]=[BAD]+[BAX]+[DAX][X B D]=[B A D]+[B A X]+[D A X]. We will find the area of each of these triangles individually. - We have [ABD]=12[ABCD][A B D]=\frac{1}{2}[A B C D]. - Because AM=AX,[BAX]=[BAM]A M=A X,[B A X]=[B A M] as the triangles have the same base and height. Thus, as [BAM][B A M] have the same base and height as ABCD,[BAX]=[BAM]=12[ABCD]A B C D,[B A X]=[B A M]=\frac{1}{2}[A B C D]. - From similar reasoning, we know that [DAX]=[DAM][D A X]=[D A M]. We have that DAMD A M has the same base and half the height of the rectangle. Thus, [DAX]=[DAM]=14[ABCD][D A X]=[D A M]=\frac{1}{4}[A B C D]. Hence, we have [XBD]=[BAD]+[BAX]+[DAX]=12[ABCD]+12[ABCD]+14[ABCD]=54[ABCD]\begin{aligned} {[X B D] } & =[B A D]+[B A X]+[D A X] \\ & =\frac{1}{2}[A B C D]+\frac{1}{2}[A B C D]+\frac{1}{4}[A B C D] \\ & =\frac{5}{4}[A B C D] \end{aligned} Thus, our answer is 54[ABCD]=54(2023)=575\frac{5}{4}[A B C D]=\frac{5}{4}(20 \cdot 23)=575.

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