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

Revision as of 19:13, 27 December 2020

Problem

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 1

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 $6$ .

$[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), S); label("$3$", (0,1.5), W); label("$2$", (.8,1), E); label("$S$", (0,0), NE); draw((0.3,0.3)--(1.4,1.9), dashed); draw((0.3,0.3)--(4,0), dashed); draw((0.3,0.3)--(0,3), dashed); label("$\small{x}$", (0.15,0.3), N); label("$\small{x}$", (0.3,0.15), E); [/asy]$

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 $\dfrac{6-\dfrac{4}{49}}{6}=\boxed{\dfrac{145}{147}}$ .

Alternatively, once you get $x=\frac{2}{7}$ , you can avoid computation by noticing that there is a denominator of $7$ , so the answer must have a factor of $7$ in the denominator, which only $\boxed{\dfrac{145}{147}}$ does.

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 larger similar triangle is $\frac{5}{3}(2)=\frac{10}{3}$ . Now, let's extend this larger 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 $\boxed{\dfrac{145}{147}}$ .

Solution 5

$[asy] draw((0,0)--(4,0)--(0,3)--(0,0)); draw((0,0)--(0.3,0)--(0.3,0.3)--(0,0.3)--(0,0)); fill(origin--(0.3,0)--(0.3,0.3)--(0,0.3)--cycle, gray); draw((0.3,0.3)--(3.6,0.3), dashed); draw((0.3,2.7)--(0.3,0.3), dashed); label("$S$", (-0.05,-0.05), NE); draw((0.3,0.3)--(1.41,1.91)); draw((1.63,1.78)--(1.48,1.56)); draw((1.28,1.70)--(1.48,1.56)); label("$4$", (2,0), S); label("$3$", (0,1.5), W); label("$\frac{10}{3}$", (2,0.3), N); label("$\frac{5}{2}$", (0.3,1.5), E); label("$2$", (1,1.2), E); draw((3.6,0)--(3.6,0.3), dashed); draw((0,2.7)--(0.3,2.7), dashed); label("$\small{l}$", (3.6,0.15), W); label("$\small{l}$", (0.15,2.7), S); label("$\small{l}$", (0.3,0.15), E); label("$\small{l}$", (0.15,0.3), N); [/asy]$

On the diagram above, find two smaller triangles similar to the large one with side lengths $3$ , $4$ , and $5$ ; consequently, the segments with length $\frac{5}{2}$ and $\frac{10}{3}$ .

With $l$ being the side length of the square, we need to find an expression for $l$ . Using the hypotenuse, we can see that $\frac{3}{2}+\frac{8}{3}+\frac{5}{4}l+\frac{5}{3}l=5$ . Simplifying, $\frac{35}{12}l=\frac{5}{6}$ , or $l=2/7$ .

A different calculation would yield $l+\frac{3}{4}l+\frac{5}{2}=3$ , so $\frac{7}{4}l=\frac{1}{2}$ . In other words, $l=\frac{2}{7}$ , while to check, $l+\frac{4}{3}l+\frac{10}{3}=4$ . As such, $\frac{7}{3}l=\frac{2}{3}$ , and $l=\frac{2}{7}$ .

Finally, we get $A(\Square S)=l^2=\frac{4}{49}$ , to finish. As a proportion of the triangle with area $6$ , the answer would be $1-\frac{4}{49\cdot6}=1-\frac{2}{147}=\frac{145}{147}$ , so $\boxed{\textit D}$ is correct.

Solution 6: Also Coordinate Geo

Let the right angle be at $(0,0)$ , the point $(x,x)$ be the far edge of the unplanted square and the hypotenuse be the line $y=-\frac{3}{4}x+3$ . Since the line from $(x,x)$ to the hypotenuse is the shortest possible distance, we know this line, call it line $\l$ , is perpendicular to the hypotenuse and therefore has a slope of $\frac{4}{3}$ .

Since we know $m=\frac{4}{3}$ , we can see that the line rises by $\frac{8}{5}$ and moves to the right by $\frac{6}{5}$ to meet the hypotenuse. (Let $2 = 5x$ and the rise be $4x$ and the run be $3x$ and then solve.) Therefore, line $\l$ intersects the hypotenuse at the point $(x+\frac{6}{5}, x+\frac{8}{5})$ . Plugging into the equation for the hypotenuse we have $x=\frac{2}{7}$ , and after a bit of computation we get $\boxed{\textbf{(D) } \frac{145}{147}}$

Video Solution by Richard Rusczyk

https://www.youtube.com/watch?v=p9npzq4FY_Y

~ dolphin7

Solution 7(Slightly different from first solution):

Same drawing as before:

$[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), S); label("$3$", (0,1.5), W); label("$2$", (.8,1), E); label("$S$", (0,0), NE); draw((0.3,0.3)--(1.4,1.9), dashed); draw((0.3,0.3)--(4,0), dashed); draw((0.3,0.3)--(0,3), dashed); label("$\small{a}$", (0.15,0.3), N); label("$\small{a}$", (0.3,0.15), E); [/asy]$

Let's assign $a$ as the side length of box S. We then get each of the smaller triangle areas. The sum of all the triangular areas(not including the box) is equal to $\frac{(3-a) \cdot a}{2} + \frac{(4-a) \cdot a}{2} + \frac{5 \cdot 2}{2} = \frac{3 \cdot 4}{2} - a^2$

You can solve for $a=\frac{2}{7}$

Then, the ratio would be $1-\dfrac{{\frac{2}{7}}^2}{6}$ which is equal to $\boxed{\textbf{(D) } \frac{145}{147}}$

~Starshooter11

Solution 8(Similar but cleaner than first solution):

Instead of dividing the large triangle into three right triangles plus a square, simply draw one diagonal of the square (can someone asymptote this for me I don't know how) divide the large triangle into three triangles with the sides $3,4,$ and $5$ as their bases. The total area is $6$ as found above, and the square's side length is $x$ . The area equation is: $\dfrac{3x}{2}+\dfrac{4x}{2}+\dfrac{5\cdot}{2}=6,$ which solves to $x=\dfrac{4}{7}$ , the same as the first solution. The final answer is $\boxed \dfrac{145}{147}$ (Error compiling LaTeX. Unknown error_msg).

@@ Line 124: / Line 124: @@
 ~Starshooter11
+===Solution 8(Similar but cleaner than first solution):===
+Instead of dividing the large triangle into three right triangles plus a square, simply draw one diagonal of the square (can someone asymptote this for me I don't know how) divide the large triangle into three triangles with the sides <math>3,4,</math> and <math>5</math> as their bases. The total area is <math>6</math> as found above, and the square's side length is <math>x</math>. The area equation is:
+<cmath>\dfrac{3x}{2}+\dfrac{4x}{2}+\dfrac{5\cdot}{2}=6,</cmath> which solves to <cmath>x=\dfrac{4}{7}</cmath>, the same as the first solution. The final answer is <math>\boxed \dfrac{145}{147}</math>.
 ==See Also==

2018 AMC 10A (Problems • Answer Key • Resources)
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 (Problems • Answer Key • Resources)
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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