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

Difference between revisions of "2024 AMC 12A Problems/Problem 14"

(Redirected page to 2024 AMC 10A Problems/Problem 21)
(Tag: New redirect)
 
(Removed redirect to 2024 AMC 10A Problems/Problem 21)
(Tag: Removed redirect)
Line 1: Line 1:
−
#redirect[[2024 AMC 10A Problems/Problem 21]]
+
{{duplicate|[[2024 AMC 12A Problems/Problem 14|2024 AMC 12A #14]] and [[2024 AMC 10A Problems/Problem 18|2024 AMC 10A #18]]}}
 +
 
 +
==Problem==
 +
Points <math>X</math> and <math>Y</math> lie on sides <math>\overline{BC}</math> and <math>\overline{CD}</math>, respectively, of parallelogram <math>ABCD</math> such that <math>\angle AXC = \angle AYC = 90^{\circ}</math>. Suppose <math>BX = 5</math> and <math>DY = 3</math>, as shown. If <math>ABCD</math> has perimeter <math>48</math>, what is its area?
 +
 
 +
<asy>
 +
import olympiad; import graph;
 +
size(8cm);
 +
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 */
 +
pair A = (0, 0), B = (15, 0), C = (12, -6 * sqrt(2)), D = (-3, -6 * sqrt(2));
 +
pair X = (15 - 3 * 5/9, -6 * sqrt(2) * 5 / 9);
 +
pair Y = (0, -6 * sqrt(2));
 +
dot(A); dot(B); dot(C); dot(D); dot(X); dot(Y);
 +
draw(A--B--C--D--cycle);
 +
draw(A--X); draw(A--Y);
 +
draw(rightanglemark(A,X,C,15)); draw(rightanglemark(A,Y,C,15));
 +
label("$A$", A, N * 1.5);
 +
label("$B$", B, N * 1.5);
 +
label("$C$", C, S * 1.5);
 +
label("$D$", D, S * 1.5);
 +
label("$X$", X, E * 1.5);
 +
label("$Y$", Y, S * 1.5);
 +
label("$3$", midpoint(D--Y), S * 1.5);
 +
label("$5$", midpoint(B--X), E * 1.5);
 +
</asy>
 +
 
 +
<math>\textbf{(A)}~40\sqrt{5}\qquad\textbf{(B)}~56\sqrt{3}\qquad\textbf{(C)}~48\sqrt{7}\qquad\textbf{(D)}~90\sqrt{2}\qquad\textbf{(E)}~60\sqrt{5}</math>
 +
 
 +
==Solution==
 +
Note that opposite angles in a parallelogram are equal, so <math>\angle ADY = \angle ABX</math>. Also, <math>\angle AYD = \angle AXB = 90^{\circ}</math>, so <math>\triangle AYD \sim \triangle AXB</math>. The ratio of similarity of these triangles is <math>\tfrac{AB}{AD} = \tfrac{BX}{DY} = \tfrac{5}{3}</math>, so let <math>AB = 5x</math> and <math>AD = 3x</math>. The perimeter of <math>ABCD</math> is <math>5x + 3x + 5x + 3x = 16x = 48</math>, so <math>x = 3</math>. Therefore <math>AD = 3 \cdot 3 = 9</math>, <math>AB = 5 \cdot 3 = 15</math>, and the area of <math>ABCD</math> is <math>AY \cdot AB = \sqrt{9^{2} - 3^{2}} \cdot 15 = \sqrt{72} \cdot 15 = 6\sqrt{2} \cdot 15 = \boxed{\textbf{(D)}~90\sqrt{2}}</math>.
 +
 
 +
==See also==
 +
{{AMC12 box|year=2024|ab=A|num-b=13|num-a=15}}
 +
{{AMC10 box|year=2024|ab=A|num-b=17|num-a=19}}
 +
{{MAA Notice}}

Revision as of 21:43, 20 March 2025

The following problem is from both the 2024 AMC 12A #14 and 2024 AMC 10A #18, so both problems redirect to this page.

Problem

Points $X$ and $Y$ lie on sides $\overline{BC}$ and $\overline{CD}$, respectively, of parallelogram $ABCD$ such that $\angle AXC = \angle AYC = 90^{\circ}$. Suppose $BX = 5$ and $DY = 3$, as shown. If $ABCD$ has perimeter $48$, what is its area?

[asy] import olympiad; import graph; size(8cm); 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 */ pair A = (0, 0), B = (15, 0), C = (12, -6 * sqrt(2)), D = (-3, -6 * sqrt(2)); pair X = (15 - 3 * 5/9, -6 * sqrt(2) * 5 / 9); pair Y = (0, -6 * sqrt(2)); dot(A); dot(B); dot(C); dot(D); dot(X); dot(Y); draw(A--B--C--D--cycle); draw(A--X); draw(A--Y); draw(rightanglemark(A,X,C,15)); draw(rightanglemark(A,Y,C,15)); label("$A$", A, N * 1.5); label("$B$", B, N * 1.5); label("$C$", C, S * 1.5); label("$D$", D, S * 1.5); label("$X$", X, E * 1.5); label("$Y$", Y, S * 1.5); label("$3$", midpoint(D--Y), S * 1.5); label("$5$", midpoint(B--X), E * 1.5); [/asy]

$\textbf{(A)}~40\sqrt{5}\qquad\textbf{(B)}~56\sqrt{3}\qquad\textbf{(C)}~48\sqrt{7}\qquad\textbf{(D)}~90\sqrt{2}\qquad\textbf{(E)}~60\sqrt{5}$

Solution

Note that opposite angles in a parallelogram are equal, so $\angle ADY = \angle ABX$. Also, $\angle AYD = \angle AXB = 90^{\circ}$, so $\triangle AYD \sim \triangle AXB$. The ratio of similarity of these triangles is $\tfrac{AB}{AD} = \tfrac{BX}{DY} = \tfrac{5}{3}$, so let $AB = 5x$ and $AD = 3x$. The perimeter of $ABCD$ is $5x + 3x + 5x + 3x = 16x = 48$, so $x = 3$. Therefore $AD = 3 \cdot 3 = 9$, $AB = 5 \cdot 3 = 15$, and the area of $ABCD$ is $AY \cdot AB = \sqrt{9^{2} - 3^{2}} \cdot 15 = \sqrt{72} \cdot 15 = 6\sqrt{2} \cdot 15 = \boxed{\textbf{(D)}~90\sqrt{2}}$.

See also

2024 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
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 17 		Followed by
Problem 19
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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_AMC_12A_Problems/Problem_14&oldid=245643"