Difference between revisions of "2018 AMC 10A Problems/Problem 23"
(\square ◽ and □, 534) |
MRENTHUSIASM (talk | contribs) (→Solution 9 (Pythagorean Theorem)) |
||
(96 intermediate revisions by 30 users not shown) | |||
Line 1: | Line 1: | ||
+ | {{duplicate|[[2018 AMC 10A Problems/Problem 23|2018 AMC 10A #23]] and [[2018 AMC 12A Problems/Problem 17|2018 AMC 12A #17]]}} | ||
+ | |||
== Problem == | == 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 <math>S</math> so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortest distance from <math>S</math> to the hypotenuse is 2 units. What fraction of the field is planted? | + | Farmer Pythagoras has a field in the shape of a right triangle. The right triangle's legs have lengths <math>3</math> and <math>4</math> units. In the corner where those sides meet at a right angle, he leaves a small unplanted square <math>S</math> so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortest distance from <math>S</math> to the hypotenuse is <math>2</math> units. What fraction of the field is planted? |
<asy> | <asy> | ||
− | + | /* Edited by MRENTHUSIASM */ | |
− | + | size(160); | |
− | + | pair A, B, C, D, F; | |
− | label("$4$", ( | + | A = origin; |
− | label("$3$", ( | + | B = (4,0); |
− | label("$2$", ( | + | C = (0,3); |
− | label("$S$", ( | + | D = (2/7,2/7); |
− | draw( | + | 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)== |
+ | 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> | ||
+ | /* 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 <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 | ||
+ | |||
+ | ==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>. | 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>. | ||
− | + | 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>. | ||
− | + | 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 | + | ==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 <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>\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>\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)=\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 | + | ==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>. | ||
+ | |||
+ | ~Heavytoothpaste | ||
+ | |||
+ | ==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 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>. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | ==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>. | ||
+ | |||
+ | ==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 | + | ==Solution 11 (Bash)== |
<asy> | <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); | |
− | label("$ | + | 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); | |
− | label("$ | + | P = foot(X,A,C); label(P, "$P$", W, red); |
− | label("$ | + | Q = foot(Y,A,B); label(Q, "$Q$", S, red); |
− | label("$ | + | M = (2/7,0); label(M, "$M$", S); |
− | label("$ | + | N = (0,2/7); label(N, "$N$", W); |
− | label("$ | + | |
− | + | fill(A--(2/7,0)--D--(0,2/7)--cycle, lightgray); | |
− | draw(( | + | draw(A--B--C--cycle); |
− | label("$ | + | draw((2/7,0)--D--(0,2/7)); |
− | label("$ | + | label("$x$", midpoint(A--M), S); |
− | label("$ | + | 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> | </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> | ||
+ | |||
+ | -Solution by sml1809 | ||
− | + | == Video Solution by Richard Rusczyk == | |
− | + | https://www.youtube.com/watch?v=p9npzq4FY_Y | |
− | + | ~ dolphin7 | |
==See Also== | ==See Also== | ||
Line 76: | Line 267: | ||
{{AMC12 box|year=2018|ab=A|num-b=16|num-a=18}} | {{AMC12 box|year=2018|ab=A|num-b=16|num-a=18}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] |
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.
Contents
- 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 Video Solution by Richard Rusczyk
- 14 See Also
Problem
Farmer Pythagoras has a field in the shape of a right triangle. The right triangle's legs have lengths and
units. In the corner where those sides meet at a right angle, he leaves a small unplanted square
so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortest distance from
to the hypotenuse is
units. What fraction of the field is planted?
Solution 1 (Area Addition)
Note that the hypotenuse of the field is and the area of the field is
Let
be the side-length of square
We partition the field into a red triangle, a yellow triangle, and a green triangle, as shown below:
Let the brackets denote areas. By area addition, we set up an equation for
from which
Therefore, the answer is
~MRENTHUSIASM
Solution 2 (Area Addition)
Let the square have side length . Connect the upper-right vertex of square
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
.
Square
has area
, and the two thin triangle regions have area
and
. The final triangular region with the hypotenuse as its base and height
has area
. Thus, we have
Solving gives
. The area of
is
and the desired ratio is
.
Alternatively, once you get , you can avoid computation by noticing that there is a denominator of
, so the answer must have a factor of
in the denominator, which only
does.
Solution 3 (Similar Triangles)
Let the square have side length . 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
. Using the side-to-side ratios of these triangles, we can find that the length of the larger similar triangle is
. Now, let's extend this larger similar right triangle to the left until it hits the side of length
. Now, the length is
, and using the ratios of the side lengths, the height is
. Looking at the diagram, if we add the height of this triangle to the side length of the square, we'd get
, so
So, the area of the square is
.
Now comes the easy part--finding the ratio of the areas: .
Solution 4 (Similar Triangles)
On the diagram above, find two smaller triangles similar to the large one with side lengths
,
, and
; consequently, the segments with length
and
.
With being the side length of the square, we need to find an expression for
. Using the hypotenuse, we can see that
. Simplifying,
, or
.
A different calculation would yield , so
. In other words,
, while to check,
. As such,
, and
.
Finally, we get , to finish. As a proportion of the triangle with area
, the answer would be
, so
is correct.
Solution 5 (Similar Triangles)
Let the side length of the square be . First off, let us make a similar triangle with the segment of length
and the top-right corner of
. Therefore, the longest side of the smaller triangle must be
. We then do operations with that side in terms of
. We subtract
from the bottom, and
from the top. That gives us the equation of
. Solving,
Thus,
, so the fraction of the triangle (area
) covered by the square is
. The answer is then
.
Solution 6 (Similar Triangles)
Let
. Note that
is a
triangle, so
and
.
. We know that
is
from the problem so
.
is also a
triangle with
. We now have
. Solving this equation, we get that
so the area of
is
. The area of the triangle is
so the fraction of field that is unplanted is
. Thus, the fraction of the field that is planted is
.
~Heavytoothpaste
Solution 7 (Coordinate Geometry)
We use coordinate geometry. Let the right angle be at and the hypotenuse be the line
for
. Denote the position of
as
, and by the point to line distance formula, we know that
Solving this, we get
. Obviously
, so
, and from here, the rest of the solution follows to get
.
Solution 8 (Coordinate Geometry)
Let the right angle be at , the point
be the far edge of the unplanted square and the hypotenuse be the line
. Since the line from
to the hypotenuse is the shortest possible distance, we know this line, call it line
, is perpendicular to the hypotenuse and therefore has a slope of
.
Since we know , we can see that the line rises by
and moves to the right by
to meet the hypotenuse. (Let
and the rise be
and the run be
and then solve.) Therefore, line
intersects the hypotenuse at the point
. Plugging into the equation for the hypotenuse we have
, and after a bit of computation we get
.
Solution 9 (Pythagorean Theorem)
Let the side length of the square be , and the lengths that the line from
hits the hypotenuse be
and
. Also, connect the outermost vertex of
to the vertices that
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:
After *some* algebra, we obtain
, which gives the answer
.
Solution 10 (Proportions)
We name small triangle the triangle similar to given in which unplanted square is inscribed.
The height of given triangle is 2.4 units so similarity coefficient is
, the area is
of total area.
The ratio of planted area in small triangle to the area of the square is
The fraction of planted area in small triangle is
Therefore, the fraction of the planted field is
vladimir.shelomovskii@gmail.com, vvsss
Solution 11 (Bash)
Denote to be the three vertices of the triangular field. Also denote
to be the vertices of the square
. Let
be on
such that
and
be on
such that
. Let
and
be the foot of the altitudes from
to
and from
to
respectively.
Note that . Thus,
and
, making
Also from the similarity ratio is the fact that
and
, making
Computing the area of
in two ways gives an equation for
:
But
is extraneous. Thus, the area of square
, making the portion of the field that is planted being
-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.