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Difference between revisions of "1987 AHSME Problems/Problem 14"

(Created page with "==Problem== <math>ABCD</math> is a square and <math>M</math> and <math>N</math> are the midpoints of <math>BC</math> and <math>CD</math> respectively. Then <math>\sin \theta=</...")
 
(→‎Solution)
 
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\textbf{(C)}\ \frac{\sqrt{10}}{5} \qquad
 
\textbf{(C)}\ \frac{\sqrt{10}}{5} \qquad
 
\textbf{(D)}\ \frac{4}{5}\qquad
 
\textbf{(D)}\ \frac{4}{5}\qquad
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\textbf{(E)}\ \text{none of these} </math>  
+
\textbf{(E)}\ \text{none of these} </math>
  
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==Solution 1==
 +
Use the Sine Area Formula. We can isolate the triangle for which the angle <math>\theta</math> is contained in. WLOG, denote the side length of a triangle as <math>2</math>. Our midpoints are then <math>1</math>. Subtract the areas of the triangles that don't include the area of our desired triangle: <math>4-1-1-\frac{1}{2} = \frac{3}{2}.</math> The Sine Area Formula tells us <math>\frac{1}{2}(\sqrt{5})^2\sin\theta=\frac{3}{2}.</math> Solving this equation, we get <math>\sin\theta=\textbf{(B)}\ \frac{3}{5} \qquad</math>
  
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Solution: Everyoneintexas
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 +
==Solution 2==
 +
Let <math>\alpha = \angle MAB = \angle DAN</math>, and let <math>s</math> be the side length of <math>ABCD</math>.  Then <math>BM = \frac{s}{2}</math> and <math>AM = \frac{s\sqrt{5}}{2}</math>, so <math>\sin \alpha = \frac{1}{\sqrt{5}}</math>.  Therefore, <math>\sin \theta = \sin \left(\frac{\pi}{2} - 2\alpha \right) = \cos 2\alpha = 1 - 2 \sin^{2}\alpha = 1 - 2\cdot \frac{1}{5} = \boxed{\textbf{(B)}\ \frac{3}{5}}</math>.
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-j314andrews
  
 
== See also ==
 
== See also ==

Latest revision as of 23:15, 7 July 2025

Problem

$ABCD$ is a square and $M$ and $N$ are the midpoints of $BC$ and $CD$ respectively. Then $\sin \theta=$

[asy] draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); draw((0,0)--(2,1)); draw((0,0)--(1,2)); label("A", (0,0), SW); label("B", (0,2), NW); label("C", (2,2), NE); label("D", (2,0), SE); label("M", (1,2), N); label("N", (2,1), E); label("$\theta$", (.5,.5), SW); [/asy]

$\textbf{(A)}\ \frac{\sqrt{5}}{5} \qquad \textbf{(B)}\ \frac{3}{5} \qquad \textbf{(C)}\ \frac{\sqrt{10}}{5} \qquad \textbf{(D)}\ \frac{4}{5}\qquad \textbf{(E)}\ \text{none of these}$

Solution 1

Use the Sine Area Formula. We can isolate the triangle for which the angle $\theta$ is contained in. WLOG, denote the side length of a triangle as $2$. Our midpoints are then $1$. Subtract the areas of the triangles that don't include the area of our desired triangle: $4-1-1-\frac{1}{2} = \frac{3}{2}.$ The Sine Area Formula tells us $\frac{1}{2}(\sqrt{5})^2\sin\theta=\frac{3}{2}.$ Solving this equation, we get $\sin\theta=\textbf{(B)}\ \frac{3}{5} \qquad$

Solution: Everyoneintexas

Solution 2

Let $\alpha = \angle MAB = \angle DAN$, and let $s$ be the side length of $ABCD$. Then $BM = \frac{s}{2}$ and $AM = \frac{s\sqrt{5}}{2}$, so $\sin \alpha = \frac{1}{\sqrt{5}}$. Therefore, $\sin \theta = \sin \left(\frac{\pi}{2} - 2\alpha \right) = \cos 2\alpha = 1 - 2 \sin^{2}\alpha = 1 - 2\cdot \frac{1}{5} = \boxed{\textbf{(B)}\ \frac{3}{5}}$.

-j314andrews

See also

1987 AHSME (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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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