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

Revision as of 12:31, 7 July 2022

The following problem is from both the 2018 AMC 10A #23 and 2018 AMC 12A #17, so both problems redirect to this page.

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] /* Edited by MRENTHUSIASM */ size(160); pair A, B, C, D, F; A = origin; B = (4,0); C = (0,3); D = (2/7,2/7); F = foot(D,B,C); fill(A--(2/7,0)--D--(0,2/7)--cycle, lightgray); draw(A--B--C--cycle); draw((2/7,0)--D--(0,2/7)); label("$4$", midpoint(A--B), N); label("$3$", midpoint(A--C), E); label("$2$", midpoint(D--F), SE); label("$S$", midpoint(A--D)); draw(D--F, 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 (Area Addition)

Note that the hypotenuse of the field is $5,$ and the area of the field is $6.$ Let $x$ be the side-length of square $S.$

We partition the field into a red triangle, a yellow triangle, and a green triangle, as shown below: $[asy] /* Edited by MRENTHUSIASM */ size(180); pair A, B, C, D, F; A = origin; B = (4,0); C = (0,3); D = (2/7,2/7); F = foot(D,B,C); fill(A--D--C--cycle, red); fill(A--D--B--cycle, yellow); fill(B--D--C--cycle, green); draw(A--B--C--cycle); label("$5$", midpoint(B--C), NE); label("$4$", midpoint(A--B), S); label("$3$", midpoint(A--C), W); label("$2$", midpoint(D--F), SE); label("$S$", midpoint(A--D)); label("$x$", midpoint((0,2/7)--D), N); label("$x$", midpoint((2/7,0)--D), E); draw((2/7,0)--D--(0,2/7)); draw(A--D^^B--D^^C--D, dashed); draw(D--F, dashed); [/asy]$ Let the brackets denote areas. By area addition, we set up an equation for $x:$ $\begin{align*} [\text{Red Triangle}]+[\text{Yellow Triangle}]+[\text{Green Triangle}]&=[\text{Field}] \\ \frac{3x}{2}+\frac{4x}{2}+\frac{5\cdot2}{2}&=6, \end{align*}$ from which $x=\frac27.$ Therefore, the answer is $\frac{[\text{Field}]-[S]}{[\text{Field}]}=\frac{6-x^2}{6}=\boxed{\textbf{(D) } \frac{145}{147}}.$ ~MRENTHUSIASM

Solution 2 (Area Addition)

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] /* Edited by MRENTHUSIASM */ size(180); pair A, B, C, D, F; A = origin; B = (4,0); C = (0,3); D = (2/7,2/7); F = foot(D,B,C); fill(A--(2/7,0)--D--(0,2/7)--cycle, lightgray); draw(A--B--C--cycle); label("$5$", midpoint(B--C), NE); label("$4$", midpoint(A--B), S); label("$3$", midpoint(A--C), W); label("$2$", midpoint(D--F), SE); label("$S$", midpoint(A--D)); label("$x$", midpoint((0,2/7)--D), N); label("$x$", midpoint((2/7,0)--D), E); draw((2/7,0)--D--(0,2/7)); draw(B--D^^C--D, dashed); draw(D--F, dashed); [/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 $\dfrac{6-\tfrac{4}{49}}{6}=\boxed{\textbf{(D) } \frac{145}{147}}$ .

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 $\frac{145}{147}$ does.

Solution 3 (Similar Triangles)

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 two 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 $\frac{3}{4}\left(\frac{10}{3}+s\right)=\frac{5}{2}+\frac{3s}{4}$ . Looking at the diagram, if we add the height of this triangle to the side length of the square, we'd get $3$ , so $\begin{align*} \frac{5}{2}+\frac{3s}{4}+s&=3 \\ \frac{5}{2}+\frac{7s}{4}&=3 \\ \frac{7s}{4}&=\frac{1}{2} \\ s&=\frac{2}{7}. \end{align*}$ So, the area of the square is $\left(\frac{2}{7}\right)^2=\frac{4}{49}$ .

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

Solution 4 (Similar Triangles)

$[asy] /* Edited by MRENTHUSIASM */ size(180); pair A, B, C, D, F; A = origin; B = (4,0); C = (0,3); D = (2/7,2/7); F = foot(D,B,C); fill(A--(2/7,0)--D--(0,2/7)--cycle, lightgray); draw(A--B--C--cycle); label("$4$", midpoint(A--B), S); label("$3$", midpoint(A--C), W); label("$2$", midpoint(D--F), SE); label("$S$", midpoint(A--D)); label("$\ell$", midpoint((0,2/7)--D), N); label("$\ell$", midpoint((2/7,0)--D), E); label("$\ell$", midpoint((2/7,2/7+5/2)--(0,2/7+5/2)), S); label("$\ell$", midpoint((2/7+10/3,2/7)--(2/7+10/3,0)), W); label("$\frac{5}{2}$", midpoint((2/7,2/7+5/2)--D), E); label("$\frac{10}{3}$", midpoint((2/7+10/3,2/7)--D), N); draw((2/7,0)--D--(0,2/7)); draw((2/7,2/7+5/2)--D^^(2/7+10/3,2/7)--D, dashed); draw((2/7,2/7+5/2)--(0,2/7+5/2)^^(2/7+10/3,2/7)--(2/7+10/3,0), dashed); draw(D--F, dashed); [/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 $\ell$ being the side length of the square, we need to find an expression for $\ell$ . Using the hypotenuse, we can see that $\frac{3}{2}+\frac{8}{3}+\frac{5}{4}\ell+\frac{5}{3}\ell=5$ . Simplifying, $\frac{35}{12}\ell=\frac{5}{6}$ , or $\ell=\frac27$ .

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

Finally, we get $A(\Square S)=\ell^2=\frac{4}{49}$ , 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 $\boxed{\textbf{(D) } \frac{145}{147}}$ is correct.

Solution 5 (Similar Triangles)

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 $\boxed{\textbf{(D) } \frac{145}{147}}$ .

Solution 6 (Coordinate Geometry)

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 $\begin{align*} \frac{|3s+4s-12|}{5} &= 2 \\ |7s-12| &= 10 \end{align*}$ Solving this, we get $s=\frac{22}{7}, \frac{2}{7}$ . Obviously $s<\frac{22}{7}$ , so $s = \frac{2}{7}$ , and from here, the rest of the solution follows to get $\boxed{\textbf{(D) } \frac{145}{147}}$ .

Solution 7 (Coordinate Geometry)

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 $\left(x+\frac{6}{5}, x+\frac{8}{5}\right)$ . Plugging into the equation for the hypotenuse we have $x=\frac{2}{7}$ , and after a bit of computation we get $\boxed{\textbf{(D) } \frac{145}{147}}$ .

Solution 8

$[asy]/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 1.5; /* changes label-to-point distance */ pen dps = linewidth(0.5) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -1.6030465381283199, xmax = 7.095084767820557, ymin = -1.3624649422453508, ymax = 4.065350676871526; /* image dimensions */ /* draw figures */ draw((0,0)--(0,3), linewidth(1.5)); draw((0,0)--(4,0), linewidth(1.5)); draw((4,0)--(0,3), linewidth(1.5)); draw((0,0.2857142857142857)--(0.2857142857142857,0.2857142857142857), linewidth(1.5)); draw((0.2857142857142857,0.2857142857142857)--(0.2857142857142857,0), linewidth(1.5)); draw((0.07142857142857142,0)--(1.4857142857142858,1.885714285714286), linewidth(1.5)); label("$A$",(0, 0),SW*labelscalefactor); label("$B$",(4,0),SE*labelscalefactor); label("$C$",(0, 3),N*labelscalefactor); label("$D$",(0.2857142857142857,0),S*labelscalefactor); label("$E$",(0.2857142857142857, 0.2857142857142857),E*labelscalefactor); label("$F$",(0.0714285714, 0),S*labelscalefactor); label("$G$", (1.49, 1.89), NE*labelscalefactor); /* dots and labels */ clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy]$

Let $AD=x$ . $\triangle FDE$ is a $3-4-5$ triangle so $EF=\frac{5}{4}x$ and $FD=\frac{3}{4}x$ . $BF=BD+FD=4-x+\frac{3}{4}x=4-\frac{1}{4}x$ . We know that $GE$ is $2$ from the problem so $GF=2+\frac{5}{4}x$ . $\triangle FGB$ is also a $3-4-5$ triangle with $GF:BF=4:5$ . We now have $4(x-\frac{1}{4})$

Video Solution by Richard Rusczyk

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

~ dolphin7

@@ Line 181: / Line 181: @@
   /* end of picture */</asy>
-Let <math>AD=x</math>. <math>\triangle FDE</math> is a <math>3-4-5</math> triangle so <math>EF=\frac{5}{4}x</math> and <math>FD=\frac{3}{4}x</math>. <math>BF=BD+FD=4-x+\frac{3}{4}x=4-\frac{1}{4}x</math>. We know that <math>GE</math> is 2 from the problem so <math>GF=2+\frac{5}{4}x</math>
+Let <math>AD=x</math>. <math>\triangle FDE</math> is a <math>3-4-5</math> triangle so <math>EF=\frac{5}{4}x</math> and <math>FD=\frac{3}{4}x</math>. <math>BF=BD+FD=4-x+\frac{3}{4}x=4-\frac{1}{4}x</math>. We know that <math>GE</math> is <math>2</math> from the problem so <math>GF=2+\frac{5}{4}x</math>. <math>\triangle FGB</math> is also a <math>3-4-5</math> triangle with <math>GF:BF=4:5</math>. We now have <math>4(x-\frac{1}{4})</math>
 == Video Solution by Richard Rusczyk ==

2018 AMC 10A (Problems • Answer Key • Resources)
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All AMC 10 Problems and Solutions

2018 AMC 12A (Problems • Answer Key • Resources)
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