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(Tags: New redirect, Rollback) |
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| − | ==Problem==
| + | #redirect[[2024 AMC 10A Problems/Problem 7]] |
| − | Equilateral triangle <math>ABC</math> is partitioned into six smaller equilateral triangles and one smaller regular hexagon, as shown below. If the regular hexagon has area <math>12</math>, what is the area of <math>\triangle ABC</math>?
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| − | | |
| − | <asy>
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| − | import graph; size(4.5cm);
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| − | real labelscalefactor = 0.5; /* changes label-to-point distance */
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| − | pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
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| − | pen dotstyle = black; /* point style */
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| − | fill((-3.46, 6)--(-2.6, 6.5)--(-1.73, 6)--(-1.73, 5)--(-2.6, 4.5)--(-3.46, 5)--cycle, lightgrey);
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| − | /* draw figures */
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| − | draw((-3.4641016151377544,6)--(-4.330127018922193,5.5));
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| − | draw((-4.330127018922193,5.5)--(-3.4641016151377544,5));
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| − | draw((-3.4641016151377544,5)--(-3.4641016151377544,6));
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| − | draw((-3.4641016151377544,5)--(-2.598076211353316,4.5));
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| − | draw((-2.598076211353316,4.5)--(-1.7320508075688772,5));
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| − | draw((-1.7320508075688772,5)--(-1.7320508075688772,6));
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| − | draw((-1.7320508075688772,6)--(-2.598076211353316,6.5));
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| − | draw((-2.598076211353316,6.5)--(-3.4641016151377544,6));
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| − | draw((-4.330127018922193,5.5)--(-4.330127018922193,3.5));
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| − | draw((-4.330127018922193,3.5)--(-2.598076211353316,4.5));
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| − | draw((-1.7320508075688772,5)--(-1.7320508075688772,2));
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| − | draw((-1.7320508075688772,2)--(-4.330127018922193,3.5));
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| − | draw((-1.7320508075688772,6)--(1.7320508075688772,4));
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| − | draw((1.7320508075688772,4)--(-1.7320508075688772,2));
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| − | draw((-4.330127018922193,5.5)--(-4.330127018922193,7.5));
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| − | draw((-4.330127018922193,7.5)--(-2.598076211353316,6.5));
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| − | draw((-4.330127018922193,3.5)--(-4.330127018922193,0.5));
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| − | draw((-4.330127018922193,0.5)--(-1.7320508075688772,2));
| |
| − | /*
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| − | dot((-3.4641016151377544,6),dotstyle);
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| − | dot((-4.330127018922193,5.5),dotstyle);
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| − | dot((-3.4641016151377544,5),dotstyle);
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| − | dot((-2.598076211353316,4.5),dotstyle);
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| − | dot((-1.7320508075688772,5),dotstyle);
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| − | dot((-1.7320508075688772,6),dotstyle);
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| − | dot((-2.598076211353316,6.5),dotstyle);
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| − | dot((-4.330127018922193,3.5),dotstyle);
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| − | dot((-1.7320508075688772,2),dotstyle);
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| − | */
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| − | dot((1.7320508075688772,4),dotstyle);
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| − | label("$A$", (1.8087225843418686,4), E);
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| − | dot((-4.330127018922193,7.5),dotstyle);
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| − | label("$B$", (-4.266904757984109,7.678590535956637), NW);
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| − | dot((-4.330127018922193,0.5),dotstyle);
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| − | label("$C$", (-4.266904757984109,0.6655806893860435), SW * 2.5);
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| − | label("$12$", (-2.6, 5.5));
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| − | </asy>
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| − | | |
| − | <math>\textbf{(A)}~ 72 \qquad \textbf{(B)}~ 84 \qquad \textbf{(C)}~ 98 \qquad \textbf{(D)}~ 128 \qquad \textbf{(E)}~ 147</math>
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| − | | |
| − | ==Solution==
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| − | Let the side length of the regular hexagon be <math>s</math>. There are the equivalent of seven equilateral triangles of side length <math>s</math>, two of side <math>2s</math>, two of side <math>3s</math>, and one of side <math>4s</math>. Let <math>X</math> be the area of the entire figure so that <cmath>\frac{X}{12} = \frac{7s^{2} + 2(2s)^{2} + 2(3s)^{2} + (4s)^{2}}{6s^{2}} =\frac{7 + 8 + 18 + 16}{6} = \frac{49}{6}.</cmath> Solving gives <math>X = \boxed{\textbf{(C)}~98}</math>.
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| − | | |
| − | ==See also==
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| − | {{AMC12 box|year=2024|ab=A|num-b=5|num-a=7}}
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| − | {{MAA Notice}}
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