Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Problem: LaTeX'ed the problem.)
(→‎Solution 9 (Pythagorean Theorem))
 
(52 intermediate revisions by 7 users not shown)
Line 1: Line 1:
−
{{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("$S$", (0,0), NE);
+
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>
  
 
<math>\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} </math>
 
<math>\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} </math>
  
−
==Solution 1==
+
==Solution 1 (Area Addition)==
−
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>.
+
Note that the hypotenuse of the field is <math>5,</math> and the area of the field is <math>6.</math> Let <math>x</math> be the side-length of square <math>S.</math>
  
 +
We partition the field into a red triangle, a yellow triangle, and a green triangle, as shown below:
 
<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("$S$", (0,0), NE);
+
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--D--C--cycle, red);
−
draw((0.3,0.3)--(0,3), dashed);
+
fill(A--D--B--cycle, yellow);
−
label("$\small{x}$", (0.15,0.3), N);
+
fill(B--D--C--cycle, green);
−
label("$\small{x}$", (0.3,0.15), E);
+
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>
 
</asy>
 +
Let the brackets denote areas. By area addition, we set up an equation for <math>x:</math>
 +
<cmath>\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*}</cmath>
 +
from which <math>x=\frac27.</math> Therefore, the answer is <cmath>\frac{[\text{Field}]-[S]}{[\text{Field}]}=\frac{6-x^2}{6}=\boxed{\textbf{(D) } \frac{145}{147}}.</cmath>
 +
~MRENTHUSIASM
  
−
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>
+
==Solution 2 (Area Addition)==
 +
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>
 +
/* 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 <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-\tfrac{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-\dfrac{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)==
 +
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
 +
<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>.
  
−
==Solution 2 (Similar to Solution 1)==
+
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>.
−
Same drawing as before:
 
  
 +
==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;
−
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("$S$", (0,0), NE);
+
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("$\small{a}$", (0.15,0.3), N);
+
label("$4$", midpoint(A--B), S);
−
label("$\small{a}$", (0.3,0.15), E);
+
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>
 
</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>.
  
−
Let's assign <math>a</math> as the side length of box S. We then get each of the smaller triangle areas.
+
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>.
−
The sum of all the triangular areas(not including the box) is equal to
 
−
<math>\frac{(3-a) \cdot a}{2} + \frac{(4-a) \cdot a}{2} + \frac{5 \cdot 2}{2} = \frac{3 \cdot 4}{2} - a^2</math>
 
  
−
You can solve for <math>a=\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>.
  
−
Then, the ratio would be <math>1-\dfrac{{\frac{2}{7}}^2}{6}</math> which is equal to <math>\boxed{\textbf{(D) } \frac{145}{147}}</math>.
+
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.
  
−
~Starshooter11
+
==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>
 +
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 3 (Similar to Solution 1)==
+
==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 */
  
−
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:
+
/* draw figures */
−
<cmath>\dfrac{3x}{2}+\dfrac{4x}{2}+\dfrac{5\cdot2}{2}=6,</cmath> which solves to <math>x=\dfrac{2}{7},</math> the same as the first solution (but easier to calculate). The final answer is <math>\boxed{\textbf{(D) } \frac{145}{147}}</math>.
+
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>.
  
−
==Solution 4==
+
~Heavytoothpaste
−
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 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 <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 3. 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 3, 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>
 
  
−
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 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
 +
<cmath>\begin{align*}
 +
\frac{|3s+4s-12|}{5} &= 2 \\
 +
|7s-12| &= 10
 +
\end{align*}</cmath>
 +
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 5==
+
==Solution 8 (Coordinate Geometry)==
−
<asy>
+
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>.
−
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 <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>.
+
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>.
  
−
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>.
+
==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>.
  
−
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>.
+
==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.
  
−
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.
+
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>
  
−
==Solution 6==
+
The fraction of planted area in <i><b>small triangle</b></i> is <math>\frac {25}{25+24} = \frac {25}{49}.</math>
−
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>.
+
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>
  
−
==Solution 7 (Coordinate Geometry)==
+
'''vladimir.shelomovskii@gmail.com, vvsss'''
−
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>\frac{|3s+4s-12|}{5} = 2</cmath> <cmath>\Rightarrow |7s-12| = 10</cmath> 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 8 (Coordinate Geometry)==
+
==Solution 11 (Bash)==
−
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>.
+
<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);
  
−
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>(x+\frac{6}{5}, x+\frac{8}{5})</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>.
+
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>
  
−
==Hardness of Problem==
+
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.
  
−
The problem itself requires the drawing of a few obvious lines and algebra, although the image deceives the solver.
+
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 ==

Latest revision as of 13:30, 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_23&oldid=198529"