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

Latest revision as of 23:41, 30 October 2024

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

1 Problem
2 Solution 1 (Area Addition)
3 Solution 2 (Area Addition)
4 Solution 3 (Similar Triangles)
5 Solution 4 (Similar Triangles)
6 Solution 5 (Similar Triangles)
7 Solution 6 (Similar Triangles)
8 Solution 7 (Coordinate Geometry)
9 Solution 8 (Coordinate Geometry)
10 Solution 9 (Pythagorean Theorem)
11 Solution 10 (Proportions)
12 Solution 11 (Bash)
13 Solution 12
14 Video Solution by Pi Academy (Fast and Easy with Area Addition)
15 Video Solution (#21-#25)
16 Video Solution by Richard Rusczyk
17 See Also

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 (Similar Triangles)

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(180); real labelscalefactor = 1.5; /* changes label-to-point distance */ // pen dps = linewidth(0.5) + fontsize(10); // defaultpen(dps); /* default pen style */ // pen dotstyle = black; /* point style */ real xmin = -1.6030465381283199, xmax = 7.095084767820557, ymin = -1.3624649422453508, ymax = 4.065350676871526; /* image dimensions */ /* draw figures */ draw((0,0)--(0,3)); draw((0,0)--(4,0)); draw((4,0)--(0,3)); draw((0,0.2857142857142857)--(0.2857142857142857,0.2857142857142857)); draw((0.2857142857142857,0.2857142857142857)--(0.2857142857142857,0)); draw((0.07142857142857142,0)--(1.4857142857142858,1.885714285714286)); label("$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 $AD=x$ . Note that $\triangle DEF$ 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 $GF:BF=3:5$ . We now have $3\left(4-\frac{1}{4}x\right)=5\left(2+\frac{5}{4}x\right)$ . Solving this equation, we get that $x=\frac{2}{7}$ so the area of $S$ is $\frac{4}{49}$ . The area of the triangle is $\frac{3\cdot 4}{2}=6$ so the fraction of field that is unplanted is $\frac{\frac{4}{49}}{6}=\frac{2}{147}$ . Thus, the fraction of the field that is planted is $1-\frac{2}{147}=\boxed{\textbf{(D) } \frac{145}{147}}$ .

~Heavytoothpaste

Solution 7 (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 8 (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 9 (Pythagorean Theorem)

Let the side length of the square be $a$ , and the lengths that the line from $S$ hits the hypotenuse be $x$ and $5-x$ . Also, connect the outermost vertex of $S$ to the vertices that $S$ isn't connected to. Note that the line that hits the hypotenuse must create a right angle, since it is the shortest possible distance. This creates two pairs of right triangles that share the same hypotenuse. This means that we can set up a system of equations using the Pythagorean Theorem: $\begin{align*} a^2+(4-a)^2&=(5-x)^2+2^2, \\ a^2+(3-a)^2&=x^2+2^2. \end{align*}$ After *some* algebra, we obtain $a=\frac{2}{7}$ , which gives the answer $\boxed{\textbf{(D) } \frac{145}{147}}$ .

Solution 10 (Proportions)

We name small triangle the triangle similar to given in which unplanted square $S$ is inscribed. The height of given triangle is 2.4 units so similarity coefficient is $\frac {2.4 - 2}{2.4} = \frac {1}{6}$ , the area is $\frac {1}{36}$ of total area.

The ratio of planted area in small triangle to the area of the square is $\frac {3}{8} + \frac {2}{3} = \frac {25}{24}.$

The fraction of planted area in small triangle is $\frac {25}{25+24} = \frac {25}{49}.$

Therefore, the fraction of the planted field is $\frac {25}{49} \cdot \frac {1}{36} + \frac {35}{36} = \boxed{\textbf{(D) } \frac{145}{147}}.$

vladimir.shelomovskii@gmail.com, vvsss

Solution 11 (Bash)

$[asy] size(240); pair A, B, C, D, F, X, Y, P, Q, M, N; A = origin; label(A, "$A$", SW); B = (4,0); label(B, "$B$", S); C = (0,3); label(C, "$C$", W); D = (2/7,2/7); label(D, "$D$", NE); F = foot(D,B,C); label(F, "$F$", NE); X = (2/7,39/14); label(X, "$X$", NE, red); Y = (76/21,2/7); label(Y, "$Y$", NE, red); P = foot(X,A,C); label(P, "$P$", W, red); Q = foot(Y,A,B); label(Q, "$Q$", S, red); M = (2/7,0); label(M, "$M$", S); N = (0,2/7); label(N, "$N$", W); fill(A--(2/7,0)--D--(0,2/7)--cycle, lightgray); draw(A--B--C--cycle); draw((2/7,0)--D--(0,2/7)); label("$x$", midpoint(A--M), S); label("$x$", midpoint(A--N), W); label("$2$", midpoint(D--F), SE); draw(D--F); draw(D--X, red); draw(D--Y, red); draw(X--P, red); draw(Y--Q, red); [/asy]$

Denote $A,B,C$ to be the three vertices of the triangular field. Also denote $A,M,D,N$ to be the vertices of the square $S$ . Let $X$ be on $BC$ such that $AC\parallel DX$ and $Y$ be on $BC$ such that $AB\parallel DY$ . Let $P$ and $Q$ be the foot of the altitudes from $X$ to $AC$ and from $Y$ to $AB$ respectively.

Note that $\triangle ABC \sim \triangle DYX \sim \triangle PXC \sim \triangle QBY$ . Thus, $PC = x \cdot \frac34$ and $QB = x \cdot \frac43$ , making $\begin{align*} DX &= 3-x-\dfrac{3}{4}x = 3-\dfrac{7}{4}x, \\ MQ &= 4-x-\dfrac{4}{3}x = 4-\dfrac{7}{3}x. \end{align*}$ Also from the similarity ratio is the fact that $CX = \frac54 x$ and $BY = \frac53 x$ , making $XY = 5 - \dfrac{5}{4}x - \dfrac{5}{3}x = 5 - \dfrac{35}{12}x.$ Computing the area of $\triangle XDY$ in two ways gives an equation for $x$ : $\begin{align*} \left(3-\dfrac{7}{4}x\right)\left(4-\dfrac{7}{3}x\right) &= 2\cdot \left(5 - \dfrac{35}{12}x\right) \\ 10-\dfrac{35}{6}x &= \dfrac{49}{12}x^2 - 14x + 12 \\ \dfrac{49}{12}x^2 - \dfrac{49}{6}x + 2 &= 0 \\ 49x^2 - 98x + 24 &= 0 \\ x&=\dfrac{2}{7} \text{ or } \dfrac{12}{7}. \end{align*}$ But $x=\dfrac{12}{7}$ is extraneous. Thus, the area of square $S = x^2 = \dfrac{4}{49}$ , making the portion of the field that is planted being $1 - \dfrac{\tfrac{4}{49}}{6} = 1 - \dfrac{2}{147} = \boxed{\textbf{(D) } \frac{145}{147}}.$

-Solution by sml1809

Solution 12

Let the square have side length $x$ . Note that when $x = \dfrac{12}{7}$ , the square is inscribed and touching the hypotenuse. Denote $f(x)$ to be the minimum distance from the square to the hypotenuse. Notably, $f$ is linear with respect to $x$ . $f(0) = \dfrac{12}{5}$ , because it is simply the length of the hypotenuse's altitude. Similarly, $f(\dfrac{12}{7}) = 0$ . We can find that $f(x) = \dfrac{12-7x}{5}$ . Setting this equal to $2$ , we get that $f(\dfrac{2}{7}) = 2$ . Therefore, the side has side length $\dfrac27$ , and has area $\dfrac{4}{49}$ . So, the unshaded area is $1 - \dfrac{\frac{4}{49}}{6} = \boxed{\dfrac{145}{147}}$ , or $(D)$ . ~Puck_0

Video Solution by Pi Academy (Fast and Easy with Area Addition)

https://youtu.be/Jdx74PGIgpw?si=kO6EGrKQQUkp22X1

~ Pi Academy

Video Solution (#21-#25)

https://youtube.com/playlist?list=PLpxy89D2tvVow8EoCSsNY3Y-2SwJly_SZ&si=aEJ3Ttjck10aCIUH

Video Solution by Richard Rusczyk

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

~ dolphin7

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

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