2018 AMC 10A Problems/Problem 23

Revision as of 20:46, 8 February 2018 by Rekt4 (talk | contribs) (Solution 5)


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]

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


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: $\frac{3\cdot 4 \cdot \frac{1}{2} -\frac{4}{49}}{3\cdot 4 \cdot \frac{1}{2}}=\frac{6-\frac{4}{49}}{6}=\frac{294-4}{294}=\frac{290}{294}=\boxed{\frac{145}{147}.}$

Solution 3

We use coordinate geometry. Let the right angle be at $(0,0)$ and the hypotenuse be the line $3x+4y = 12$ for $0\le x\le 3$. Denote the position of $S$ as $(s,s)$, and by the point to line distance formula, we know that \[\frac{|3s+4s-12|}{5} = 2\] \[\Rightarrow |7s-12| = 10\] Obviously $s<\frac{22}{7}$, so $s = \frac{2}{7}$, and from here the rest of the solution follows to get $\boxed{\frac{145}{147}}$.

Solution 4

Let the side length of the square be $x$. First off, let us make a similar triangle with the segment of length $2$ and the top-right corner of $S$. Therefore, the longest side of the smaller triangle must be $2 \cdot \frac54 = \frac52$. We then do operations with that side in terms of $x$. We subtract $x$ from the bottom, and $\frac{3x}{4}$ from the top. That gives us the equation of $3-\frac{7x}{4} = \frac{5}{2}$. Solving, \[12-7x = 10 \implies x = \frac{2}{7}.\]

Thus, $x^2 = \frac{4}{49}$, so the fraction of the triangle (area $6$) covered by the square is $\frac{2}{147}$. The answer is then $\boxed{\dfrac{145}{147}}$.

See Also

2018 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
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
2018 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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

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