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2024 AMC 12A Problems/Problem 6

Revision as of 21:41, 20 March 2025 by Maa is stupid (talk | contribs) (Removed redirect to 2024 AMC 10A Problems/Problem 7)

Problem

Equilateral triangle $ABC$ is partitioned into six smaller equilateral triangles and one smaller regular hexagon, as shown below. If the regular hexagon has area $12$, what is the area of $\triangle ABC$?

[asy] import graph; size(4.5cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ fill((-3.46, 6)--(-2.6, 6.5)--(-1.73, 6)--(-1.73, 5)--(-2.6, 4.5)--(-3.46, 5)--cycle, lightgrey); /* draw figures */ draw((-3.4641016151377544,6)--(-4.330127018922193,5.5)); draw((-4.330127018922193,5.5)--(-3.4641016151377544,5)); draw((-3.4641016151377544,5)--(-3.4641016151377544,6)); draw((-3.4641016151377544,5)--(-2.598076211353316,4.5)); draw((-2.598076211353316,4.5)--(-1.7320508075688772,5)); draw((-1.7320508075688772,5)--(-1.7320508075688772,6)); draw((-1.7320508075688772,6)--(-2.598076211353316,6.5)); draw((-2.598076211353316,6.5)--(-3.4641016151377544,6)); draw((-4.330127018922193,5.5)--(-4.330127018922193,3.5)); draw((-4.330127018922193,3.5)--(-2.598076211353316,4.5)); draw((-1.7320508075688772,5)--(-1.7320508075688772,2)); draw((-1.7320508075688772,2)--(-4.330127018922193,3.5)); draw((-1.7320508075688772,6)--(1.7320508075688772,4)); draw((1.7320508075688772,4)--(-1.7320508075688772,2)); draw((-4.330127018922193,5.5)--(-4.330127018922193,7.5)); draw((-4.330127018922193,7.5)--(-2.598076211353316,6.5)); draw((-4.330127018922193,3.5)--(-4.330127018922193,0.5)); draw((-4.330127018922193,0.5)--(-1.7320508075688772,2)); /* dot((-3.4641016151377544,6),dotstyle); dot((-4.330127018922193,5.5),dotstyle); dot((-3.4641016151377544,5),dotstyle); dot((-2.598076211353316,4.5),dotstyle); dot((-1.7320508075688772,5),dotstyle); dot((-1.7320508075688772,6),dotstyle); dot((-2.598076211353316,6.5),dotstyle); dot((-4.330127018922193,3.5),dotstyle); dot((-1.7320508075688772,2),dotstyle); */ dot((1.7320508075688772,4),dotstyle); label("$A$", (1.8087225843418686,4), E); dot((-4.330127018922193,7.5),dotstyle); label("$B$", (-4.266904757984109,7.678590535956637), NW); dot((-4.330127018922193,0.5),dotstyle); label("$C$", (-4.266904757984109,0.6655806893860435), SW * 2.5); label("$12$", (-2.6, 5.5)); [/asy]

$\textbf{(A)}~ 72 \qquad \textbf{(B)}~ 84 \qquad \textbf{(C)}~ 98 \qquad \textbf{(D)}~ 128 \qquad \textbf{(E)}~ 147$

Solution

Let the side length of the regular hexagon be $s$. There are the equivalent of seven equilateral triangles of side length $s$, two of side $2s$, two of side $3s$, and one of side $4s$. Let $X$ be the area of the entire figure so that \[\frac{X}{12} = \frac{7s^{2} + 2(2s)^{2} + 2(3s)^{2} + (4s)^{2}}{6s^{2}} =\frac{7 + 8 + 18 + 16}{6} = \frac{49}{6}.\] Solving gives $X = \boxed{\textbf{(C)}~98}$.

See also

2024 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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

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