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Difference between revisions of "2024 AMC 12A Problems/Problem 9"

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#redirect[[2024 AMC 10A Problems/Problem 15]]
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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]]}}
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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}}

Revision as of 21:42, 20 March 2025

The following problem is from both the 2024 AMC 12A #9 and 2024 AMC 10A #12, so both problems redirect to this page.

Problem

Square $ABCD$ has side length $6$ and center $O$. Points $E$ and $F$ lie in the plane, and $AOEF$ is a rectangle. Suppose that exactly $\tfrac{2}{3}$ of the area of $AOEF$ lies inside square $ABCD$. What is the area of $\triangle CEF$?

$\textbf{(A)}~4\qquad\textbf{(B)}~3\sqrt{2}\qquad\textbf{(C)}~6\qquad\textbf{(D)}~4\sqrt{3}\qquad\textbf{(E)}~8$

Solution

[asy] unitsize(20); 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); draw(A--B--C--D--cycle); filldraw(A--O--E--X--cycle, mediumgray); label("$A$", A, NE); label("$B$", B, NW); label("$C$", C, SW); label("$D$", D, SE); label("$6$", (3, 6), N); label("$O$", O, SW); dot(A); dot(B); dot(C); dot(D); dot(E); dot(O); dot(F); label("$E$", E, S); label("$F$", F, NE); draw(A--O--E--F--cycle); draw(C--E--F--cycle); [/asy] Note that one-third of the area of rectangle $AOEF$ lies outside square $ABCD$. If $X$ is the intersection of $\overline{AD}$ and $\overline{EF}$, then the region of the rectangle that lies outside the square is the interior of $\triangle AFX$. Since $\overline{AO} \parallel \overline{EF}$, we have $\angle AXF = \angle EXD = \angle OAD = 45^{\circ}$, and clearly $\angle AFX = 90^{\circ}$. Thus $\triangle AFX$ is an isosceles right triangle and $AF = FX$, so its area $\tfrac{1}{2}t^{2}$ where $t = AF$. The area of $AOEF$ is $AF\cdot AO = t \cdot \frac{6}{\sqrt{2}} = t \cdot 3\sqrt{2}$. Setting the area of $\triangle AFX$ to one-third of this gives \[\tfrac{1}{2}t^{2} = t\cdot\sqrt{2} \implies t^{2} = t\cdot 2\sqrt{2}\implies t = 0 ~ \operatorname{or} ~ t = 2\sqrt{2}.\] Using $t = 0$ leads to the case of a degenerate rectangle, so we use $t = 2\sqrt{2}$. The area of $\triangle CEF$ can be computed as $\frac{1}{2}\cdot EF\cdot\text{dist}(C,\overleftrightarrow{EF})$. Since $\overleftrightarrow{AO}\parallel\overleftrightarrow{EF}$, all points on $\overleftrightarrow{AO}$ (including $C$) have the same distance to $\overleftrightarrow{EF}$, which is precisely $t = 2\sqrt{2}$, and $EF = AO = 3\sqrt{2}$. Hence, the area of $\triangle CEF$ is $\tfrac{1}{2}\left(2\sqrt{2}\right)\left(3\sqrt{2}\right) = \boxed{\textbf{(C)}~6}$.

See also

2024 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions
2024 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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