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| − | {{duplicate|[[2024 AMC 12A Problems/Problem 9|2024 AMC 12A #9]] and [[2024 AMC 10A Problems/Problem 9|2024 AMC 10A #12]]}}
| + | #redirect[[2024 AMC 10A Problems/Problem 15]] |
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| − | ==Problem==
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| − | Square <math>ABCD</math> has side length <math>6</math> and center <math>O</math>. Points <math>E</math> and <math>F</math> lie in the plane, and <math>AOEF</math> is a rectangle. Suppose that exactly <math>\tfrac{2}{3}</math> of the area of <math>AOEF</math> lies inside square <math>ABCD</math>. What is the area of <math>\triangle CEF</math>?
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| − | <math>\textbf{(A)}~4\qquad\textbf{(B)}~3\sqrt{2}\qquad\textbf{(C)}~6\qquad\textbf{(D)}~4\sqrt{3}\qquad\textbf{(E)}~8</math>
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| − | ==Solution==
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| − | <asy>
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| − | unitsize(20);
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| − | pair A = (6, 6), B = (0, 6), C = (0, 0), D = (6, 0), O = (3, 3), E = (5, 1), F = (8, 4), X = (6, 2);
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| − | draw(A--B--C--D--cycle);
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| − | filldraw(A--O--E--X--cycle, mediumgray);
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| − | label("$A$", A, NE);
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| − | label("$B$", B, NW);
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| − | label("$C$", C, SW);
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| − | label("$D$", D, SE);
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| − | label("$6$", (3, 6), N);
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| − | label("$O$", O, SW);
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| − | dot(A);
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| − | dot(B);
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| − | dot(C);
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| − | dot(D);
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| − | dot(E);
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| − | dot(O);
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| − | dot(F);
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| − | label("$E$", E, S);
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| − | label("$F$", F, NE);
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| − | draw(A--O--E--F--cycle);
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| − | draw(C--E--F--cycle);
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| − | </asy>
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| − | Note that one-third of the area of rectangle <math>AOEF</math> lies outside square <math>ABCD</math>. If <math>X</math> is the intersection of <math>\overline{AD}</math> and <math>\overline{EF}</math>, then the region of the rectangle that lies outside the square is the interior of <math>\triangle AFX</math>. Since <math>\overline{AO} \parallel \overline{EF}</math>, we have <math>\angle AXF = \angle EXD = \angle OAD = 45^{\circ}</math>, and clearly <math>\angle AFX = 90^{\circ}</math>. Thus <math>\triangle AFX</math> is an isosceles right triangle and <math>AF = FX</math>, so its area <math>\tfrac{1}{2}t^{2}</math> where <math>t = AF</math>. The area of <math>AOEF</math> is <math>AF\cdot AO = t \cdot \frac{6}{\sqrt{2}} = t \cdot 3\sqrt{2}</math>. Setting the area of <math>\triangle AFX</math> to one-third of this gives <cmath>\tfrac{1}{2}t^{2} = t\cdot\sqrt{2} \implies t^{2} = t\cdot 2\sqrt{2}\implies t = 0 ~ \operatorname{or} ~ t = 2\sqrt{2}.</cmath> Using <math>t = 0</math> leads to the case of a degenerate rectangle, so we use <math>t = 2\sqrt{2}</math>. The area of <math>\triangle CEF</math> can be computed as <math>\frac{1}{2}\cdot EF\cdot\text{dist}(C,\overleftrightarrow{EF})</math>. Since <math>\overleftrightarrow{AO}\parallel\overleftrightarrow{EF}</math>, all points on <math>\overleftrightarrow{AO}</math> (including <math>C</math>) have the same distance to <math>\overleftrightarrow{EF}</math>, which is precisely <math>t = 2\sqrt{2}</math>, and <math>EF = AO = 3\sqrt{2}</math>. Hence, the area of <math>\triangle CEF</math> is <math>\tfrac{1}{2}\left(2\sqrt{2}\right)\left(3\sqrt{2}\right) = \boxed{\textbf{(C)}~6}</math>.
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| − | ==See also==
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| − | {{AMC12 box|year=2024|ab=A|num-b=8|num-a=10}}
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| − | {{AMC10 box|year=2024|ab=A|num-b=11|num-a=13}}
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| − | {{MAA Notice}}
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