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

m (→‎Solution 1 (Area Addition))
(→‎Solution 9 (Pythagorean Theorem))
(46 intermediate revisions by 7 users not shown)
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−
{{duplicate|[[2018 AMC 12A Problems|2018 AMC 12A #17]] and [[2018 AMC 10A Problems|2018 AMC 10A #23]]}}
+
{{duplicate|[[2018 AMC 10A Problems/Problem 23|2018 AMC 10A #23]] and [[2018 AMC 12A Problems/Problem 17|2018 AMC 12A #17]]}}
  
 
== Problem ==
 
== Problem ==
Line 6: Line 6:
  
 
<asy>
 
<asy>
−
draw((0,0)--(4,0)--(0,3)--(0,0));
+
/* Edited by MRENTHUSIASM */
−
draw((0,0)--(0.3,0)--(0.3,0.3)--(0,0.3)--(0,0));
+
size(160);
−
fill(origin--(0.3,0)--(0.3,0.3)--(0,0.3)--cycle, gray);
+
pair A, B, C, D, F;
−
label("$4$", (2,0), N);
+
A = origin;
−
label("$3$", (0,1.5), E);
+
B = (4,0);
−
label("$2$", (.8,1), E);
+
C = (0,3);
−
label("$S$", (0.15,0.15));
+
D = (2/7,2/7);
−
draw((0.3,0.3)--(1.4,1.9), dashed);
+
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>
 
</asy>
  
Line 23: Line 31:
 
We partition the field into a red triangle, a yellow triangle, and a green triangle, as shown below:
 
We partition the field into a red triangle, a yellow triangle, and a green triangle, as shown below:
 
<asy>
 
<asy>
−
fill((0,0)--(0,3)--(0.3,0.3)--cycle,red);
+
/* Edited by MRENTHUSIASM */
−
fill((0,0)--(4,0)--(0.3,0.3)--cycle,yellow);
+
size(180);
−
fill((0,3)--(4,0)--(0.3,0.3)--cycle,green);
+
pair A, B, C, D, F;
−
draw((0,0)--(4,0)--(0,3)--(0,0));
+
A = origin;
−
draw((0,0)--(0.3,0)--(0.3,0.3)--(0,0.3)--(0,0));
+
B = (4,0);
−
label("$5$", (2,1.5), NE);
+
C = (0,3);
−
label("$4$", (2,0), S);
+
D = (2/7,2/7);
−
label("$3$", (0,1.5), W);
+
F = foot(D,B,C);
−
label("$2$", (.8,1), E);
+
fill(A--D--C--cycle, red);
−
label("$S$", (0.15,0.15));
+
fill(A--D--B--cycle, yellow);
−
draw((0.3,0.3)--(1.4,1.9), dashed);
+
fill(B--D--C--cycle, green);
−
draw((0.3,0.3)--(4,0), dashed);
+
draw(A--B--C--cycle);
−
draw((0.3,0.3)--(0,3), dashed);
+
label("$5$", midpoint(B--C), NE);
−
draw((0,0)--(0.3,0.3), dashed);
+
label("$4$", midpoint(A--B), S);
−
label("$\small{x}$", (0.15,0.3), N);
+
label("$3$", midpoint(A--C), W);
−
label("$\small{x}$", (0.3,0.15), E);
+
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>
 
</asy>
 
Let the brackets denote areas. By area addition, we set up an equation for <math>x:</math>
 
Let the brackets denote areas. By area addition, we set up an equation for <math>x:</math>
Line 51: Line 65:
 
Let the square have side length <math>x</math>. Connect the upper-right vertex of square <math>S</math> 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 <math>6</math>.
 
Let the square have side length <math>x</math>. Connect the upper-right vertex of square <math>S</math> 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 <math>6</math>.
 
<asy>
 
<asy>
−
draw((0,0)--(4,0)--(0,3)--(0,0));
+
/* Edited by MRENTHUSIASM */
−
draw((0,0)--(0.3,0)--(0.3,0.3)--(0,0.3)--(0,0));
+
size(180);
−
fill(origin--(0.3,0)--(0.3,0.3)--(0,0.3)--cycle, gray);
+
pair A, B, C, D, F;
−
label("$4$", (2,0), S);
+
A = origin;
−
label("$3$", (0,1.5), W);
+
B = (4,0);
−
label("$2$", (.8,1), E);
+
C = (0,3);
−
label("$S$", (0.15,0.15));
+
D = (2/7,2/7);
−
draw((0.3,0.3)--(1.4,1.9), dashed);
+
F = foot(D,B,C);
−
draw((0.3,0.3)--(4,0), dashed);
+
fill(A--(2/7,0)--D--(0,2/7)--cycle, lightgray);
−
draw((0.3,0.3)--(0,3), dashed);
+
draw(A--B--C--cycle);
−
label("$\small{x}$", (0.15,0.3), N);
+
label("$5$", midpoint(B--C), NE);
−
label("$\small{x}$", (0.3,0.15), E);
+
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>
 
</asy>
 
Square <math>S</math> has area <math>x^2</math>, and the two thin triangle regions have area <math>\dfrac{x(3-x)}{2}</math> and <math>\dfrac{x(4-x)}{2}</math>. The final triangular region with the hypotenuse as its base and height <math>2</math> has area <math>5</math>. Thus, we have <cmath>x^2+\dfrac{x(3-x)}{2}+\dfrac{x(4-x)}{2}+5=6.</cmath>
 
Square <math>S</math> has area <math>x^2</math>, and the two thin triangle regions have area <math>\dfrac{x(3-x)}{2}</math> and <math>\dfrac{x(4-x)}{2}</math>. The final triangular region with the hypotenuse as its base and height <math>2</math> has area <math>5</math>. Thus, we have <cmath>x^2+\dfrac{x(3-x)}{2}+\dfrac{x(4-x)}{2}+5=6.</cmath>
−
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{\textbf{(D) } \frac{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-\tfrac{4}{49}}{6}=\boxed{\textbf{(D) } \frac{145}{147}}</math>.
  
 
Alternatively, once you get <math>x=\frac{2}{7}</math>, you can avoid computation by noticing that there is a denominator of <math>7</math>, so the answer must have a factor of <math>7</math> in the denominator, which only <math>\frac{145}{147}</math> does.
 
Alternatively, once you get <math>x=\frac{2}{7}</math>, you can avoid computation by noticing that there is a denominator of <math>7</math>, so the answer must have a factor of <math>7</math> in the denominator, which only <math>\frac{145}{147}</math> does.
  
 
==Solution 3 (Similar Triangles)==
 
==Solution 3 (Similar Triangles)==
−
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 <math>2</math> smaller similar triangles that share a side of length <math>2</math>. Using the side-to-side ratios of these triangles, we can find that the length of the larger similar triangle is <math>\frac{5}{3}(2)=\frac{10}{3}</math>. Now, let's extend this larger similar right triangle to the left until it hits the side of length <math>3</math>. 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}\left(\frac{10}{3}+s\right)=\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 <math>3</math>, 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 } \left(\frac{2}{7}\right)^2=\frac{4}{49}.</cmath>
+
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 two smaller similar triangles that share a side of length <math>2</math>. Using the side-to-side ratios of these triangles, we can find that the length of the larger similar triangle is <math>\frac{5}{3}(2)=\frac{10}{3}</math>. Now, let's extend this larger similar right triangle to the left until it hits the side of length <math>3</math>. 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}\left(\frac{10}{3}+s\right)=\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 <math>3</math>, so
−
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{\textbf{(D) } \frac{145}{147}}</math>.
+
<cmath>\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*}</cmath>
 +
So, the area of the square is <math>\left(\frac{2}{7}\right)^2=\frac{4}{49}</math>.
 +
 
 +
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{\textbf{(D) } \frac{145}{147}}</math>.
  
 
==Solution 4 (Similar Triangles)==
 
==Solution 4 (Similar Triangles)==
 
<asy>
 
<asy>
−
draw((0,0)--(4,0)--(0,3)--(0,0));
+
/* Edited by MRENTHUSIASM */
−
draw((0,0)--(0.3,0)--(0.3,0.3)--(0,0.3)--(0,0));
+
size(180);
−
fill(origin--(0.3,0)--(0.3,0.3)--(0,0.3)--cycle, gray);
+
pair A, B, C, D, F;
−
draw((0.3,0.3)--(3.6,0.3), dashed);
+
A = origin;
−
draw((0.3,2.7)--(0.3,0.3), dashed);
+
B = (4,0);
−
label("$S$", (-0.05,-0.05), NE);
+
C = (0,3);
−
draw((0.3,0.3)--(1.41,1.91));
+
D = (2/7,2/7);
−
draw((1.63,1.78)--(1.48,1.56));
+
F = foot(D,B,C);
−
draw((1.28,1.70)--(1.48,1.56));
+
fill(A--(2/7,0)--D--(0,2/7)--cycle, lightgray);
−
label("$4$", (2,0), S);
+
draw(A--B--C--cycle);
−
label("$3$", (0,1.5), W);
+
label("$4$", midpoint(A--B), S);
−
label("$\frac{10}{3}$", (2,0.3), N);
+
label("$3$", midpoint(A--C), W);
−
label("$\frac{5}{2}$", (0.3,1.5), E);
+
label("$2$", midpoint(D--F), SE);
−
label("$2$", (1,1.2), E);
+
label("$S$", midpoint(A--D));
−
draw((3.6,0)--(3.6,0.3), dashed);
+
label("$\ell$", midpoint((0,2/7)--D), N);
−
draw((0,2.7)--(0.3,2.7), dashed);
+
label("$\ell$", midpoint((2/7,0)--D), E);
−
label("$\small{l}$", (3.6,0.15), W);
+
label("$\ell$", midpoint((2/7,2/7+5/2)--(0,2/7+5/2)), S);
−
label("$\small{l}$", (0.15,2.7), S);
+
label("$\ell$", midpoint((2/7+10/3,2/7)--(2/7+10/3,0)), W);
−
label("$\small{l}$", (0.3,0.15), E);
+
label("$\frac{5}{2}$", midpoint((2/7,2/7+5/2)--D), E);
−
label("$\small{l}$", (0.15,0.3), N);
+
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>
 
</asy>
 
On the diagram above, find two smaller triangles similar to the large one with side lengths <math>3</math>, <math>4</math>, and <math>5</math>; consequently, the segments with length <math>\frac{5}{2}</math> and <math>\frac{10}{3}</math>.
 
On the diagram above, find two smaller triangles similar to the large one with side lengths <math>3</math>, <math>4</math>, and <math>5</math>; consequently, the segments with length <math>\frac{5}{2}</math> and <math>\frac{10}{3}</math>.
  
−
With <math>l</math> being the side length of the square, we need to find an expression for <math>l</math>. Using the hypotenuse, we can see that <math>\frac{3}{2}+\frac{8}{3}+\frac{5}{4}l+\frac{5}{3}l=5</math>. Simplifying, <math>\frac{35}{12}l=\frac{5}{6}</math>, or <math>l=2/7</math>.
+
With <math>\ell</math> being the side length of the square, we need to find an expression for <math>\ell</math>. Using the hypotenuse, we can see that <math>\frac{3}{2}+\frac{8}{3}+\frac{5}{4}\ell+\frac{5}{3}\ell=5</math>. Simplifying, <math>\frac{35}{12}\ell=\frac{5}{6}</math>, or <math>\ell=\frac27</math>.
  
−
A different calculation would yield <math>l+\frac{3}{4}l+\frac{5}{2}=3</math>, so <math>\frac{7}{4}l=\frac{1}{2}</math>. In other words, <math>l=\frac{2}{7}</math>, while to check, <math>l+\frac{4}{3}l+\frac{10}{3}=4</math>. As such, <math>\frac{7}{3}l=\frac{2}{3}</math>, and <math>l=\frac{2}{7}</math>.
+
A different calculation would yield <math>\ell+\frac{3}{4}\ell+\frac{5}{2}=3</math>, so <math>\frac{7}{4}\ell=\frac{1}{2}</math>. In other words, <math>\ell=\frac{2}{7}</math>, while to check, <math>\ell+\frac{4}{3}\ell+\frac{10}{3}=4</math>. As such, <math>\frac{7}{3}\ell=\frac{2}{3}</math>, and <math>\ell=\frac{2}{7}</math>.
  
−
Finally, we get <math>A(\Square S)=l^2=\frac{4}{49}</math>, to finish. As a proportion of the triangle with area <math>6</math>, the answer would be <math>1-\frac{4}{49\cdot6}=1-\frac{2}{147}=\frac{145}{147}</math>, so <math>\boxed{\textbf{(D) } \frac{145}{147}}</math> is correct.
+
Finally, we get <math>A(\Square S)=\ell^2=\frac{4}{49}</math>, to finish. As a proportion of the triangle with area <math>6</math>, the answer would be <math>1-\frac{4}{49\cdot6}=1-\frac{2}{147}=\frac{145}{147}</math>, so <math>\boxed{\textbf{(D) } \frac{145}{147}}</math> is correct.
  
 
==Solution 5 (Similar Triangles)==
 
==Solution 5 (Similar Triangles)==
 
Let the side length of the square be <math>x</math>. First off, let us make a similar triangle with the segment of length <math>2</math> and the top-right corner of <math>S</math>. Therefore, the longest side of the smaller triangle must be <math>2 \cdot \frac54 = \frac52</math>. We then do operations with that side in terms of <math>x</math>. We subtract <math>x</math> from the bottom, and <math>\frac{3x}{4}</math> from the top. That gives us the equation of <math>3-\frac{7x}{4} = \frac{5}{2}</math>. Solving, <cmath>12-7x = 10 \implies x = \frac{2}{7}.</cmath>
 
Let the side length of the square be <math>x</math>. First off, let us make a similar triangle with the segment of length <math>2</math> and the top-right corner of <math>S</math>. Therefore, the longest side of the smaller triangle must be <math>2 \cdot \frac54 = \frac52</math>. We then do operations with that side in terms of <math>x</math>. We subtract <math>x</math> from the bottom, and <math>\frac{3x}{4}</math> from the top. That gives us the equation of <math>3-\frac{7x}{4} = \frac{5}{2}</math>. Solving, <cmath>12-7x = 10 \implies x = \frac{2}{7}.</cmath>
 +
Thus, <math>x^2 = \frac{4}{49}</math>, so the fraction of the triangle (area <math>6</math>) covered by the square is <math>\frac{2}{147}</math>. The answer is then <math>\boxed{\textbf{(D) } \frac{145}{147}}</math>.
 +
 +
==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 <math>AD=x</math>. Note that <math>\triangle DEF</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=3:5</math>. We now have <math>3\left(4-\frac{1}{4}x\right)=5\left(2+\frac{5}{4}x\right)</math>. Solving this equation, we get that <math>x=\frac{2}{7}</math> so the area of <math>S</math> is <math>\frac{4}{49}</math>. The area of the triangle is <math>\frac{3\cdot 4}{2}=6</math> so the fraction of field that is unplanted is <math>\frac{\frac{4}{49}}{6}=\frac{2}{147}</math>. Thus, the fraction of the field that is planted is <math>1-\frac{2}{147}=\boxed{\textbf{(D) } \frac{145}{147}}</math>.
  
−
Thus, <math>x^2 = \frac{4}{49}</math>, so the fraction of the triangle (area <math>6</math>) covered by the square is <math>\frac{2}{147}</math>. The answer is then <math>\boxed{\textbf{(D) } \frac{145}{147}}</math>.
+
~Heavytoothpaste
  
−
==Solution 6 (Coordinate Geometry)==
+
==Solution 7 (Coordinate Geometry)==
 
We use coordinate geometry. Let the right angle be at <math>(0,0)</math> and the hypotenuse be the line <math>3x+4y = 12</math> for <math>0\le x\le 3</math>. Denote the position of <math>S</math> as <math>(s,s)</math>, and by the point to line distance formula, we know that
 
We use coordinate geometry. Let the right angle be at <math>(0,0)</math> and the hypotenuse be the line <math>3x+4y = 12</math> for <math>0\le x\le 3</math>. Denote the position of <math>S</math> as <math>(s,s)</math>, and by the point to line distance formula, we know that
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
Line 117: Line 182:
 
Solving this, we get <math>s=\frac{22}{7}, \frac{2}{7}</math>. Obviously <math>s<\frac{22}{7}</math>, so <math>s = \frac{2}{7}</math>, and from here, the rest of the solution follows to get <math>\boxed{\textbf{(D) } \frac{145}{147}}</math>.
 
Solving this, we get <math>s=\frac{22}{7}, \frac{2}{7}</math>. Obviously <math>s<\frac{22}{7}</math>, so <math>s = \frac{2}{7}</math>, and from here, the rest of the solution follows to get <math>\boxed{\textbf{(D) } \frac{145}{147}}</math>.
  
−
==Solution 7 (Coordinate Geometry)==
+
==Solution 8 (Coordinate Geometry)==
 
Let the right angle be at <math>(0,0)</math>, the point <math>(x,x)</math> be the far edge of the unplanted square and the hypotenuse be the line <math>y=-\frac{3}{4}x+3</math>. Since the line from <math>(x,x)</math> to the hypotenuse is the shortest possible distance, we know this line, call it line <math>\l</math>, is perpendicular to the hypotenuse and therefore has a slope of <math>\frac{4}{3}</math>.  
 
Let the right angle be at <math>(0,0)</math>, the point <math>(x,x)</math> be the far edge of the unplanted square and the hypotenuse be the line <math>y=-\frac{3}{4}x+3</math>. Since the line from <math>(x,x)</math> to the hypotenuse is the shortest possible distance, we know this line, call it line <math>\l</math>, is perpendicular to the hypotenuse and therefore has a slope of <math>\frac{4}{3}</math>.  
  
 
Since we know <math>m=\frac{4}{3}</math> , we can see that the line rises by <math>\frac{8}{5}</math> and moves to the right by <math>\frac{6}{5}</math> to meet the hypotenuse. (Let <math>2 = 5x</math> and the rise be <math>4x</math> and the run be <math>3x</math> and then solve.) Therefore, line <math>\l</math> intersects the hypotenuse at the point <math>\left(x+\frac{6}{5}, x+\frac{8}{5}\right)</math>. Plugging into the equation for the hypotenuse we have <math>x=\frac{2}{7}</math> , and after a bit of computation we get <math>\boxed{\textbf{(D) } \frac{145}{147}}</math>.
 
Since we know <math>m=\frac{4}{3}</math> , we can see that the line rises by <math>\frac{8}{5}</math> and moves to the right by <math>\frac{6}{5}</math> to meet the hypotenuse. (Let <math>2 = 5x</math> and the rise be <math>4x</math> and the run be <math>3x</math> and then solve.) Therefore, line <math>\l</math> intersects the hypotenuse at the point <math>\left(x+\frac{6}{5}, x+\frac{8}{5}\right)</math>. Plugging into the equation for the hypotenuse we have <math>x=\frac{2}{7}</math> , and after a bit of computation we get <math>\boxed{\textbf{(D) } \frac{145}{147}}</math>.
  
−
==Hardness of Problem==
+
==Solution 9 (Pythagorean Theorem)==
 +
Let the side length of the square be <math>a</math>, and the lengths that the line from <math>S</math> hits the hypotenuse be <math>x</math> and <math>5-x</math>. Also, connect the outermost vertex of <math>S</math> to the vertices that <math>S</math> 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:
 +
<cmath>\begin{align*}
 +
a^2+(4-a)^2&=(5-x)^2+2^2, \\
 +
a^2+(3-a)^2&=x^2+2^2.
 +
\end{align*}</cmath>
 +
After *some* algebra, we obtain <math>a=\frac{2}{7}</math>, which gives the answer <math>\boxed{\textbf{(D) } \frac{145}{147}}</math>.
  
−
The problem itself requires the drawing of a few obvious lines and algebra, although the image deceives the solver.
+
==Solution 10 (Proportions)==
 +
We name <i><b>small triangle</b></i> the  triangle similar to given in which unplanted square <math>S</math> is inscribed.
 +
The height of given triangle is 2.4 units so similarity coefficient is <math>\frac {2.4 - 2}{2.4} = \frac {1}{6}</math> , the  area is <math>\frac {1}{36}</math> of total area.
 +
 
 +
The ratio of planted area in <i><b>small triangle</b></i> to the area of the square is <math>\frac {3}{8} + \frac {2}{3} = \frac {25}{24}.</math>
 +
 
 +
The fraction of planted area in <i><b>small triangle</b></i> is <math>\frac {25}{25+24} = \frac {25}{49}.</math>
 +
 
 +
Therefore, the fraction of the planted  field is <math>\frac {25}{49} \cdot  \frac {1}{36} + \frac {35}{36} = \boxed{\textbf{(D) } \frac{145}{147}}.</math>
 +
 
 +
'''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 <math>A,B,C</math> to be the three vertices of the triangular field. Also denote <math>A,M,D,N</math> to be the vertices of the square <math>S</math>. Let <math>X</math> be on <math>BC</math> such that <math>AC\parallel DX</math> and <math>Y</math> be on <math>BC</math> such that <math>AB\parallel DY</math>. Let <math>P</math> and <math>Q</math> be the foot of the altitudes from <math>X</math> to <math>AC</math> and from <math>Y</math> to <math>AB</math> respectively.
 +
 
 +
Note that <math>\triangle ABC \sim \triangle DYX \sim \triangle PXC \sim \triangle QBY</math>. Thus, <math>PC = x \cdot \frac34</math> and <math>QB = x \cdot \frac43</math>, making
 +
<cmath>\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*}</cmath>
 +
Also from the similarity ratio is the fact that <math>CX = \frac54 x</math> and <math>BY = \frac53 x</math>, making
 +
<cmath>XY = 5 - \dfrac{5}{4}x - \dfrac{5}{3}x = 5 - \dfrac{35}{12}x.</cmath>
 +
Computing the area of <math>\triangle XDY</math> in two ways gives an equation for <math>x</math>:
 +
<cmath>\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*}</cmath>
 +
But <math>x=\dfrac{12}{7}</math> is extraneous. Thus, the area of square <math>S = x^2 = \dfrac{4}{49}</math>, making the portion of the field that is planted being <cmath>1 - \dfrac{\tfrac{4}{49}}{6} = 1 - \dfrac{2}{147} = \boxed{\textbf{(D) } \frac{145}{147}}.</cmath>
  
−
Comment: Hardness and difficulty are relative/subjective to the problem solver. It can be useful in personal situations to label questions with a hardness scale, but it may not suffice for other problem solvers. Please don't feel pressured to spontaneously know concepts that are difficult for you. Additionally, nothing is 'obvious' because it's all relative to the problem solver and individual minds. Happy mathing!
+
-Solution by sml1809
  
 
== Video Solution by Richard Rusczyk ==
 
== Video Solution by Richard Rusczyk ==

Revision as of 12:30, 23 September 2023

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 (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

Video Solution by Richard Rusczyk

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

~ dolphin7

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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