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Difference between revisions of "2018 AMC 10A Problems/Problem 23"

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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>.
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==Solution 2==
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Let the square have side length <math>s</math>. If we were to extend the sides of the square further into the triangle until they intersect on point on the hypotenuse, we'd have a similar right triangle formed between the hypotenuse and the two new lines, and 2 smaller similar triangles that share a side of length 2. Using the side-to-side ratios of these triangles, we can find that the length of the big similar triangle is <math>\frac{5}{3}(2)=\frac{10}{3}</math>. Now, let's extend this big similar right triangle to the left until it hits the side of length 3. Now, the length is <math>\frac{10}{3}+s</math>, and using the ratios of the side lengths, the height is <math>\frac{3}{4}(\frac{10}{3}+s)=\frac{5}{2}+\frac{3s}{4}</math>. Looking at the diagram, if we add the height of this triangle to the side length of the square, we'd get 3, so <cmath>\frac{5}{2}+\frac{3s}{4}+s=\frac{5}{2}+\frac{7s}{4}=3 \\ \frac{7s}{4}=\frac{1}{2} \\ s=\frac{2}{7} \implies \textrm{ area of square is } (\frac{2}{7})^2=\frac{4}{49}</cmath>
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Now comes the easy part: finding the ratio of the areas: <math>\frac{3\cdot 4 \cdot \frac{1}{2} -\frac{4}{49}}{3\cdot 4 \cdot \frac{1}{2}}=\frac{6-\frac{4}{49}}{6}=\frac{294-4}{294}=\frac{290}{294}=\boxed{\frac{145}{147}}</math>

Revision as of 17:17, 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 $\dfrac{6-\dfrac{4}{49}}{6}=\boxed{\dfrac{145}{147}}$.

Solution 2

Let the square have side length $s$. If we were to extend the sides of the square further into the triangle until they intersect on point on the hypotenuse, we'd have a similar right triangle formed between the hypotenuse and the two new lines, and 2 smaller similar triangles that share a side of length 2. Using the side-to-side ratios of these triangles, we can find that the length of the big similar triangle is $\frac{5}{3}(2)=\frac{10}{3}$. Now, let's extend this big similar right triangle to the left until it hits the side of length 3. Now, the length is $\frac{10}{3}+s$, and using the ratios of the side lengths, the height is $\frac{3}{4}(\frac{10}{3}+s)=\frac{5}{2}+\frac{3s}{4}$. Looking at the diagram, if we add the height of this triangle to the side length of the square, we'd get 3, so \[\frac{5}{2}+\frac{3s}{4}+s=\frac{5}{2}+\frac{7s}{4}=3 \\ \frac{7s}{4}=\frac{1}{2} \\ s=\frac{2}{7} \implies \textrm{ area of square is } (\frac{2}{7})^2=\frac{4}{49}\]

Now comes the easy part: finding the ratio of the areas: $\frac{3\cdot 4 \cdot \frac{1}{2} -\frac{4}{49}}{3\cdot 4 \cdot \frac{1}{2}}=\frac{6-\frac{4}{49}}{6}=\frac{294-4}{294}=\frac{290}{294}=\boxed{\frac{145}{147}}$

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