Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus
During AMC testing, the AoPS Wiki is in read-only mode and no edits can be made.

Difference between revisions of "2024 AMC 12A Problems/Problem 6"

(Redirected page to 2024 AMC 10A Problems/Problem 7)
(Tag: New redirect)
 
(Removed redirect to 2024 AMC 10A Problems/Problem 7)
(Tag: Removed redirect)
Line 1: Line 1:
−
#redirect[[2024 AMC 10A Problems/Problem 7]]
+
==Problem==
 +
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>?
 +
 
 +
<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>
 +
 
 +
<math>\textbf{(A)}~ 72 \qquad \textbf{(B)}~ 84 \qquad \textbf{(C)}~ 98 \qquad \textbf{(D)}~ 128 \qquad \textbf{(E)}~ 147</math>
 +
 
 +
==Solution==
 +
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>.
 +
 
 +
==See also==
 +
{{AMC12 box|year=2024|ab=A|num-b=5|num-a=7}}
 +
{{MAA Notice}}

Revision as of 20:41, 20 March 2025

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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_AMC_12A_Problems/Problem_6&oldid=245635"