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Difference between revisions of "1975 AHSME Problems/Problem 23"

(Created page with "== Problem 23 == In the adjoining figure <math>AB</math> and <math>BC</math> are adjacent sides of square <math>ABCD</math>; <math>M</math> is the midpoint of <math>AB</math>...")
 
(→‎Solution)
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== Solution ==
 
== Solution ==
  
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First, let's draw a few auxiliary lines. Drop altitudes from <math>O</math> to <math>AB</math> and from <math>O</math> to <math>BC</math>. We can label them as <math>J</math> and <math>K</math>, respectively. This forms square <math>OKBJ</math>. Connect <math>OB</math>
+
First, let's draw a few auxiliary lines. Drop altitudes from <math>O</math> to <math>AB</math> and from <math>O</math> to <math>BC</math>. We can label the points as <math>J</math> and <math>K</math>, respectively. This forms square <math>OKBJ</math>. Connect <math>OB</math>.
  
 
Without loss of generality, set the side length of the square equal to <math>1</math>. Let <math>NK=x</math>, and since <math>N</math> is the midpoint of <math>CB</math>, <math>KB</math> would be <math>\frac{1}{2} - x</math>. With the same reasoning, <math>MJ=x</math> and <math>JB = \frac{1}{2} - x</math>
 
Without loss of generality, set the side length of the square equal to <math>1</math>. Let <math>NK=x</math>, and since <math>N</math> is the midpoint of <math>CB</math>, <math>KB</math> would be <math>\frac{1}{2} - x</math>. With the same reasoning, <math>MJ=x</math> and <math>JB = \frac{1}{2} - x</math>
  
 
<asy>
 
<asy>
−
draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)--(2,1)--(2,2)--(1,0)--(4/3, 0)--(4/3, 2/3)--(2, 2/3)--(2,0)--(4/3, 2/3));
+
draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)--(2,1)--(2,2)--(1,0));
 +
draw((4/3, 0)--(4/3, 2/3)--(2, 2/3)--(2,0)--(4/3, 2/3), dashed);
 
label("A", (0,0), S);
 
label("A", (0,0), S);
 
label("B", (2,0), S);
 
label("B", (2,0), S);

Revision as of 17:24, 27 December 2020

Problem 23

In the adjoining figure $AB$ and $BC$ are adjacent sides of square $ABCD$; $M$ is the midpoint of $AB$; $N$ is the midpoint of $BC$; and $AN$ and $CM$ intersect at $O$. The ratio of the area of $AOCD$ to the area of $ABCD$ is

[asy] draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)--(2,1)--(2,2)--(1,0)); label("A", (0,0), S); label("B", (2,0), S); label("C", (2,2), N); label("D", (0,2), N); label("M", (1,0), S); label("N", (2,1), E); label("O", (1.2, .8)); [/asy]

$\textbf{(A)}\ \frac{5}{6}\qquad \textbf{(B)}\ \frac{3}{4}\qquad \textbf{(C)}\ \frac{2}{3}\qquad \textbf{(D)}\ \frac{\sqrt{3}}{2}\qquad \textbf{(E)}\ \frac{(\sqrt{3}-1)}{2}$

Solution

First, let's draw a few auxiliary lines. Drop altitudes from $O$ to $AB$ and from $O$ to $BC$. We can label the points as $J$ and $K$, respectively. This forms square $OKBJ$. Connect $OB$.

Without loss of generality, set the side length of the square equal to $1$. Let $NK=x$, and since $N$ is the midpoint of $CB$, $KB$ would be $\frac{1}{2} - x$. With the same reasoning, $MJ=x$ and $JB = \frac{1}{2} - x$

[asy] draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)--(2,1)--(2,2)--(1,0)); draw((4/3, 0)--(4/3, 2/3)--(2, 2/3)--(2,0)--(4/3, 2/3), dashed); label("A", (0,0), S); label("B", (2,0), S); label("C", (2,2), N); label("D", (0,2), N); label("M", (1,0), S); label("N", (2,1), E); label("O", (1.2, .8)); label("J", (4/3, 0), S); label("K", (2,2/3), E); [/asy]

We can also see that $\triangle COK$ is similar to $\triangle CMB$. That means $\frac{CK}{OK} = \frac{CB}{MB}$.

Plugging in the values, we get: $\frac{\frac{1}{2} + x}{\frac{1}{2} - x} = \frac{1}{\frac{1}{2}}$. Solving, we find that $x=\frac{1}{6}$. Then, $OK=\frac{1}{2} - \frac{1}{6} = \frac{1}{3}$. The area of $\triangle COB$ and $\triangle AOB$ together would be $2 (\frac{1}{2} \cdot 1 \cdot \frac{1}{3}) = \frac{1}{3}$. Subtract this area from the total area of $ABCD$ to get the area of $AOCD$.

So, $AOCD = 1 - \frac{1}{3} = \frac{2}{3}$. The question asks for the ratio of the area of $AOCD$ to the area of $ABCD$, which is $\frac{\frac{2}{3}}{1} = \frac{2}{3}$.

The answer is $\boxed{\textbf{(C)}\ \frac{2}{3}}$. ~jiang147369

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