Difference between revisions of "2014 AMC 10A Problems/Problem 16"
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<math> \textbf{(A)}\ \dfrac1{12}\qquad\textbf{(B)}\ \dfrac{\sqrt3}{18}\qquad\textbf{(C)}\ \dfrac{\sqrt2}{12}\qquad\textbf{(D)}\ \dfrac{\sqrt3}{12}\qquad\textbf{(E)}\ \dfrac16 </math> | <math> \textbf{(A)}\ \dfrac1{12}\qquad\textbf{(B)}\ \dfrac{\sqrt3}{18}\qquad\textbf{(C)}\ \dfrac{\sqrt2}{12}\qquad\textbf{(D)}\ \dfrac{\sqrt3}{12}\qquad\textbf{(E)}\ \dfrac16 </math> | ||
+ | [[Category: Introductory Geometry Problems]] | ||
− | ==Solution== | + | ==Solution 1== |
− | === | + | Denote <math>D=(0,0)</math>. Then <math>A= (0,2), F = \left(\frac12,0\right), H = \left(\frac12,1\right)</math>. Let the intersection of <math>AF</math> and <math>DH</math> be <math>X</math>, and the intersection of <math>BF</math> and <math>CH</math> be <math>Y</math>. Then we want to find the coordinates of <math>X</math> so we can find <math>XY</math>. From our points, the slope of <math>AF</math> is <math>\bigg(\dfrac{-2}{\tfrac12}\bigg) = -4</math>, and its <math>y</math>-intercept is just <math>2</math>. Thus the equation for <math>AF</math> is <math>y = -4x + 2</math>. We can also quickly find that the equation of <math>DH</math> is <math>y = 2x</math>. Setting the equations equal, we have <math>2x = -4x +2 \implies x = \frac13</math>. Because of symmetry, we can see that the distance from <math>Y</math> to <math>BC</math> is also <math>\frac13</math>, so <math>XY = 1 - 2 \cdot \frac13 = \frac13</math>. Now the area of the kite is simply the product of the two diagonals over <math>2</math>. Since the length <math>HF = 1</math>, our answer is <math>\dfrac{\dfrac{1}{3} \cdot 1}{2} = \boxed{\textbf{(E)} \: \dfrac16}</math>. |
− | |||
− | + | <asy> | |
+ | import graph; | ||
+ | size(9cm); | ||
+ | pen dps = fontsize(10); defaultpen(dps); | ||
+ | pair D = (0,0); | ||
+ | pair F = (1/2,0); | ||
+ | pair C = (1,0); | ||
+ | pair G = (0,1); | ||
+ | pair E = (1,1); | ||
+ | pair A = (0,2); | ||
+ | pair B = (1,2); | ||
+ | pair H = (1/2,1); | ||
− | == | + | // do not look |
+ | pair X = (1/3,2/3); | ||
+ | pair Y = (2/3,2/3); | ||
− | Let the area of the shaded region be <math>x</math>. Let the other two vertices of the kite be <math>I</math> and <math>J</math> with <math>I</math> closer to <math>AD</math> than <math>J</math>. Note that <math> [ABCD] = [ABF] + [DCH] - x + [ADI] + [BCJ]</math>. The area of <math>ABF</math> is <math>1</math> and the area of <math>DCH</math> is <math>\dfrac{1}{2}</math>. We will solve for the areas of <math>ADI</math> and <math>BCJ</math> in terms of x by noting that the area of each triangle is the length of the perpendicular from <math>I</math> to <math>AD</math> and <math>J</math> to <math>BC</math> respectively. Because the area of <math>x</math> = <math>\dfrac{1}{2} | + | draw(A--B--C--D--cycle); |
+ | draw(G--E); | ||
+ | draw(A--F--B); | ||
+ | draw(D--H--C); | ||
+ | filldraw(H--X--F--Y--cycle,grey); | ||
+ | draw(X--Y,dashed); | ||
+ | |||
+ | |||
+ | |||
+ | label("$A\: (0,2)$",A,NW); | ||
+ | label("$B$",B,NE); | ||
+ | label("$C$",C,SE); | ||
+ | label("$D \: (0,0)$",D,SW); | ||
+ | label("$E$",E,E); | ||
+ | label("$F\: (\frac12,0)$",F,S); | ||
+ | label("$G$",G,W); | ||
+ | label("$H \: (\frac12,1)$",H,N); | ||
+ | label("$Y$",Y,E); | ||
+ | label("$X$",X,W); | ||
+ | |||
+ | |||
+ | label("$\frac12$",(0.25,0),S); | ||
+ | label("$\frac12$",(0.75,0),S); | ||
+ | label("$1$",(1,0.5),E); | ||
+ | label("$1$",(1,1.5),E); | ||
+ | </asy> | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | Let the area of the shaded region be <math>x</math>. Let the other two vertices of the kite be <math>I</math> and <math>J</math> with <math>I</math> closer to <math>AD</math> than <math>J</math>. Note that <math> [ABCD] = [ABF] + [DCH] - x + [ADI] + [BCJ]</math>. The area of <math>ABF</math> is <math>1</math> and the area of <math>DCH</math> is <math>\dfrac{1}{2}</math>. We will solve for the areas of <math>ADI</math> and <math>BCJ</math> in terms of x by noting that the area of each triangle is the length of the perpendicular from <math>I</math> to <math>AD</math> and <math>J</math> to <math>BC</math> respectively. Because the area of <math>x</math> = <math>\dfrac{1}{2} \cdot IJ</math> based on the area of a kite formula, <math>\dfrac{ab}{2}</math> for diagonals of length <math>a</math> and <math>b</math>, <math>IJ = 2x</math>. So each perpendicular is length <math>\dfrac{1-2x}{2}</math>. So taking our numbers and plugging them into <math> [ABCD] =[ABF] + [DCH] - x + [ADI] + [BCJ]</math> gives us <math>2 = \dfrac{5}{2} - 3x</math> Solving this equation for <math>x</math> gives us <math> x = \boxed{\textbf{(E)} \: \frac{1}{6}}</math> | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | From the diagram in Solution 1, let <math>e</math> be the height of <math>XHY</math> and <math>f</math> be the height of <math>XFY</math>. It is clear that their sum is <math>1</math> as they are parallel to <math>GD</math>. Let <math>k</math> be the ratio of the sides of the similar triangles <math>XFY</math> and <math>AFB</math>, which are similar because <math>XY</math> is parallel to <math>AB</math> and the triangles share angle <math>F</math>. Then <math>k = f/2</math>, as 2 is the height of <math>AFB</math>. Since <math>XHY</math> and <math>DHC</math> are similar for the same reasons as <math>XFY</math> and <math>AFB</math>, the height of <math>XHY</math> will be equal to the base, like in <math>DHC</math>, making <math>XY = e</math>. However, <math>XY</math> is also the base of <math>XFY</math>, so <math>k = e / AB</math> where <math>AB = 1</math> so <math>k = e</math>. Subbing into <math>k = f/2</math> gives a system of linear equations, <math>e + f = 1</math> and <math>e = f/2</math>. Solving yields <math>e = XY = 1/3</math> and <math>f = \frac{2}{3}</math>, and since the area of the kite is simply the product of the two diagonals over <math>2</math> and <math>HF = 1</math>, our answer is <math>\frac{\frac{1}{3} \cdot 1}{2} = \boxed{\textbf{(E)} \: \dfrac16}</math>. | ||
+ | |||
+ | ==Solution 4== | ||
+ | Let the unmarked vertices of the shaded area be labeled <math>I</math> and <math>J</math>, with <math>I</math> being closer to <math>GD</math> than <math>J</math>. Noting that kite <math>HJFI</math> can be split into triangles <math>HJI</math> and <math>JIF</math>. | ||
+ | |||
+ | Lemma: The distance from line segment <math>JI</math> to <math>H</math> is half the distance from <math>JI</math> to <math>F</math> | ||
+ | |||
+ | Proof: Drop perpendiculars of triangles <math>HJI</math> and <math>JIF</math> to line <math>JI</math>, and let the point of intersection be <math>Q</math>. Note that <math>HJI</math> and <math>JIF</math> are similar to <math>HDC</math> and <math>ABF</math>, respectively. Now, the ratio of <math>DC</math> to <math>HF</math> is <math>1:1</math>, which shows that the ratio of <math>JI</math> to <math>HQ</math> is <math>1:1</math>, because of similar triangles as described above. Similarly, the ratio of <math>JI</math> to <math>FQ</math> is <math>1:2</math>. Since these two triangles contain the same base, <math>JI</math>, the ratio of <math>HQ:FQ = 1:2</math>. | ||
+ | |||
+ | Because kite <math>HJFI</math> is orthodiagonal, we multiply <math>\frac{1\cdot\tfrac{1}{3}}{2} = \boxed{\textbf{(E)} \: \frac{1}{6}}</math> | ||
+ | |||
+ | ~Lemma proof by sakshamsethi | ||
==See Also== | ==See Also== | ||
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{{AMC10 box|year=2014|ab=A|num-b=15|num-a=17}} | {{AMC10 box|year=2014|ab=A|num-b=15|num-a=17}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
+ | |||
+ | [[Category: Introductory Geometry Problems]] |
Latest revision as of 12:05, 7 September 2021
Problem
In rectangle , , , and points , , and are midpoints of , , and , respectively. Point is the midpoint of . What is the area of the shaded region?
Solution 1
Denote . Then . Let the intersection of and be , and the intersection of and be . Then we want to find the coordinates of so we can find . From our points, the slope of is , and its -intercept is just . Thus the equation for is . We can also quickly find that the equation of is . Setting the equations equal, we have . Because of symmetry, we can see that the distance from to is also , so . Now the area of the kite is simply the product of the two diagonals over . Since the length , our answer is .
Solution 2
Let the area of the shaded region be . Let the other two vertices of the kite be and with closer to than . Note that . The area of is and the area of is . We will solve for the areas of and in terms of x by noting that the area of each triangle is the length of the perpendicular from to and to respectively. Because the area of = based on the area of a kite formula, for diagonals of length and , . So each perpendicular is length . So taking our numbers and plugging them into gives us Solving this equation for gives us
Solution 3
From the diagram in Solution 1, let be the height of and be the height of . It is clear that their sum is as they are parallel to . Let be the ratio of the sides of the similar triangles and , which are similar because is parallel to and the triangles share angle . Then , as 2 is the height of . Since and are similar for the same reasons as and , the height of will be equal to the base, like in , making . However, is also the base of , so where so . Subbing into gives a system of linear equations, and . Solving yields and , and since the area of the kite is simply the product of the two diagonals over and , our answer is .
Solution 4
Let the unmarked vertices of the shaded area be labeled and , with being closer to than . Noting that kite can be split into triangles and .
Lemma: The distance from line segment to is half the distance from to
Proof: Drop perpendiculars of triangles and to line , and let the point of intersection be . Note that and are similar to and , respectively. Now, the ratio of to is , which shows that the ratio of to is , because of similar triangles as described above. Similarly, the ratio of to is . Since these two triangles contain the same base, , the ratio of .
Because kite is orthodiagonal, we multiply
~Lemma proof by sakshamsethi
See Also
2014 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.