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Difference between revisions of "2018 AMC 10A Problems/Problem 23"
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==Solution==
 
==Solution==
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Let the square have sidel length <math>x</math>. Connect the upper-right vertex of square <math>S</math> with the two vertices of the triangle's hypotenuse. This divides the triangle in several regions whose areas must add up to the area of the whole triangle, which is <math>6</math>.
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Let the square have side length <math>x</math>. Connect the upper-right vertex of square <math>S</math> with the two vertices of the triangle's hypotenuse. This divides the triangle in several regions whose areas must add up to the area of the whole triangle, which is <math>6</math>.
  
 
Square <math>S</math> has area <math>x^2</math>, and the two thin triangle regions have area <math>\dfrac{x(3-x)}{2}</math> and <math>\dfrac{x(4-x)}{2}</math>. The final triangular region with the hypotenuse as its base and height <math>2</math> has area <math>5</math>. Thus, we have <cmath>x^2+\dfrac{x(3-x)}{2}+\dfrac{x(4-x)}{2}+5=6</cmath>
 
Square <math>S</math> has area <math>x^2</math>, and the two thin triangle regions have area <math>\dfrac{x(3-x)}{2}</math> and <math>\dfrac{x(4-x)}{2}</math>. The final triangular region with the hypotenuse as its base and height <math>2</math> has area <math>5</math>. Thus, we have <cmath>x^2+\dfrac{x(3-x)}{2}+\dfrac{x(4-x)}{2}+5=6</cmath>
  
 
Solving gives <math>x=\dfrac{2}{7}</math>. The area of <math>S</math> is <math>\dfrac{4}{49}</math> and the desired ratio is <math>\dfrac{6-\dfrac{4}{49}}{6}=\boxed{\dfrac{145}{147}}</math>.
 
Solving gives <math>x=\dfrac{2}{7}</math>. The area of <math>S</math> is <math>\dfrac{4}{49}</math> and the desired ratio is <math>\dfrac{6-\dfrac{4}{49}}{6}=\boxed{\dfrac{145}{147}}</math>.

Revision as of 14:49, 8 February 2018

Farmer Pythagoras has a field in the shape of a right triangle. The right triangle's legs have lengths 3 and 4 units. In the corner where those sides meet at a right angle, he leaves a small unplanted square $S$ so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortest distance from $S$ to the hypotenuse is 2 units. What fraction of the field is planted?

[asy] draw((0,0)--(4,0)--(0,3)--(0,0)); draw((0,0)--(0.3,0)--(0.3,0.3)--(0,0.3)--(0,0)); fill(origin--(0.3,0)--(0.3,0.3)--(0,0.3)--cycle, gray); label("$4$", (2,0), N); label("$3$", (0,1.5), E); label("$2$", (.8,1), E); label("$S$", (0,0), NE); draw((0.3,0.3)--(1.4,1.9), dashed); [/asy]

$\textbf{(A) } \frac{25}{27} \qquad \textbf{(B) } \frac{26}{27} \qquad \textbf{(C) } \frac{73}{75} \qquad \textbf{(D) } \frac{145}{147} \qquad \textbf{(E) } \frac{74}{75}$

Solution

Let the square have side length $x$. Connect the upper-right vertex of square $S$ with the two vertices of the triangle's hypotenuse. This divides the triangle in several regions whose areas must add up to the area of the whole triangle, which is $6$.

Square $S$ has area $x^2$, and the two thin triangle regions have area $\dfrac{x(3-x)}{2}$ and $\dfrac{x(4-x)}{2}$. The final triangular region with the hypotenuse as its base and height $2$ has area $5$. Thus, we have \[x^2+\dfrac{x(3-x)}{2}+\dfrac{x(4-x)}{2}+5=6\]

Solving gives $x=\dfrac{2}{7}$. The area of $S$ is $\dfrac{4}{49}$ and the desired ratio is $\dfrac{6-\dfrac{4}{49}}{6}=\boxed{\dfrac{145}{147}}$.

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