Difference between revisions of "2013 AMC 10A Problems/Problem 18"
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==Problem== | ==Problem== | ||
− | Let points <math>A = (0, 0)</math>, <math>B = (1, 2)</math>, <math>C=(3, 3)</math>, and <math>D = (4, 0)</math>. Quadrilateral <math>ABCD</math> is cut into equal area pieces by a line passing through <math>A</math>. This line intersects <math>\overline{CD}</math> at point <math>(\frac{p}{q}, \frac{r}{s})</math>, where these fractions are in lowest terms. What is <math>p+q+r+s</math>? | + | Let points <math>A = (0, 0)</math>, <math>B = (1, 2)</math>, <math>C=(3, 3)</math>, and <math>D = (4, 0)</math>. Quadrilateral <math>ABCD</math> is cut into equal area pieces by a line passing through <math>A</math>. This line intersects <math>\overline{CD}</math> at point <math>\left(\frac{p}{q}, \frac{r}{s}\right)</math>, where these fractions are in lowest terms. What is <math>p+q+r+s</math>? |
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[[Category: Introductory Geometry Problems]] | [[Category: Introductory Geometry Problems]] | ||
− | ==Solution== | + | ==Solution 1== |
− | + | <center><asy> | |
+ | size(8cm); | ||
+ | pair A, B, C, D, E, EE; | ||
+ | A = (0,0); | ||
+ | B = (1,2); | ||
+ | C = (3,3); | ||
+ | D = (4,0); | ||
+ | E = (27/8,15/8); | ||
+ | EE = (27/8,0); | ||
+ | draw(A--B--C--D--A--E); | ||
+ | draw(E--EE,linetype("8 8")); | ||
+ | dot(A); | ||
+ | dot(B); | ||
+ | dot(C); | ||
+ | dot(D); | ||
+ | dot(E); | ||
+ | draw(rightanglemark(E,EE,D,4)); | ||
+ | label("A",A,SW); | ||
+ | label("B",B,NW); | ||
+ | label("C",C,NE); | ||
+ | label("D",D,SE); | ||
+ | label("E",E,NE); | ||
+ | label("$4$",(A+D)/2,S); | ||
+ | label("$\frac{27}{8}$",(A+EE)/2,S); | ||
+ | label("$\frac{15}{8}$",(E+EE)/2,W); | ||
+ | </asy></center> | ||
+ | |||
+ | First, we shall find the area of quadrilateral <math>ABCD</math>. This can be done in any of three ways: | ||
+ | |||
+ | Pick's Theorem: <math>[ABCD] = I + \dfrac{B}{2} - 1 = 5 + \dfrac{7}{2} - 1 = \dfrac{15}{2}.</math> | ||
+ | |||
+ | Splitting: Drop perpendiculars from <math>B</math> and <math>C</math> to the x-axis to divide the quadrilateral into triangles and trapezoids, and so the area is <math>1 + 5 + \dfrac{3}{2} = \dfrac{15}{2}.</math> | ||
+ | |||
+ | Shoelace Method: The area is half of <math>|1 \cdot 3 - 2 \cdot 3 - 3 \cdot 4| = 15</math>, or <math>\dfrac{15}{2}</math>. | ||
+ | |||
+ | <math>[ABCD] = \frac{15}{2}</math>. Therefore, each equal piece that the line separates <math>ABCD</math> into must have an area of <math>\frac{15}{4}</math>. | ||
Call the point where the line through <math>A</math> intersects <math>\overline{CD}</math> <math>E</math>. We know that <math>[ADE] = \frac{15}{4} = \frac{bh}{2}</math>. Furthermore, we know that <math>b = 4</math>, as <math>AD = 4</math>. Thus, solving for <math>h</math>, we find that <math>2h = \frac{15}{4}</math>, so <math>h = \frac{15}{8}</math>. This gives that the y coordinate of E is <math>\frac{15}{8}</math>. | Call the point where the line through <math>A</math> intersects <math>\overline{CD}</math> <math>E</math>. We know that <math>[ADE] = \frac{15}{4} = \frac{bh}{2}</math>. Furthermore, we know that <math>b = 4</math>, as <math>AD = 4</math>. Thus, solving for <math>h</math>, we find that <math>2h = \frac{15}{4}</math>, so <math>h = \frac{15}{8}</math>. This gives that the y coordinate of E is <math>\frac{15}{8}</math>. | ||
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Line CD can be expressed as <math>y = -3x+12</math>, so the <math>x</math> coordinate of E satisfies <math>\frac{15}{8} = -3x + 12</math>. Solving for <math>x</math>, we find that <math>x = \frac{27}{8}</math>. | Line CD can be expressed as <math>y = -3x+12</math>, so the <math>x</math> coordinate of E satisfies <math>\frac{15}{8} = -3x + 12</math>. Solving for <math>x</math>, we find that <math>x = \frac{27}{8}</math>. | ||
− | From this, we know that <math>E = (\frac{27}{8}, \frac{15}{8})</math>. <math>27 + 15 + 8 + 8 = \boxed{\textbf{(B) }58}</math> | + | From this, we know that <math>E = \left(\frac{27}{8}, \frac{15}{8}\right)</math>. <math>27 + 15 + 8 + 8 = \boxed{\textbf{(B) }58}</math> |
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | <center><asy> | ||
+ | size(8cm); | ||
+ | pair A, B, C, D, E, F; | ||
+ | A = (0,0); | ||
+ | B = (1,2); | ||
+ | C = (3,3); | ||
+ | D = (4,0); | ||
+ | E = (27/8,15/8); | ||
+ | F = (27/8,0); | ||
+ | draw(A--B--C--D--A--E); | ||
+ | draw(E--F,linetype("8 8")); | ||
+ | dot(A); | ||
+ | dot(B); | ||
+ | dot(C); | ||
+ | dot(D); | ||
+ | dot(E); | ||
+ | draw(rightanglemark(E,F,D,4)); | ||
+ | label("A",A,SW); | ||
+ | label("B",B,NW); | ||
+ | label("C",C,NE); | ||
+ | label("D",D,SE); | ||
+ | label("E",E,NE); | ||
+ | label("F",F,S); | ||
+ | label("$4$",(A+D)/2,S); | ||
+ | label("$x$",(A+F)/2,S); | ||
+ | label("$\frac{15}{8}$",(E+F)/2,W); | ||
+ | </asy></center> | ||
+ | Let the point where the altitude from <math>E</math> to <math>\overline{AD}</math> be labeled <math>F</math>. | ||
+ | Following the steps above, you can find that the height of <math>\triangle ADE</math> is <math>\frac{15}{8}</math>, and from there split the base into two parts, <math>x</math>, and <math>4-x</math>, such that <math>x</math> is the segment from the origin to the point <math>F</math>, and <math>4-x</math> is the segment from point <math>F</math> to point <math>D</math>. Then, by the Pythagorean Theorem, <math>x=\frac{27}{8}</math>, and the answer is <math>\boxed{\textbf{(B) }58}</math> | ||
==See Also== | ==See Also== |
Latest revision as of 11:18, 18 April 2020
Contents
Problem
Let points , , , and . Quadrilateral is cut into equal area pieces by a line passing through . This line intersects at point , where these fractions are in lowest terms. What is ?
Solution 1
First, we shall find the area of quadrilateral . This can be done in any of three ways:
Pick's Theorem:
Splitting: Drop perpendiculars from and to the x-axis to divide the quadrilateral into triangles and trapezoids, and so the area is
Shoelace Method: The area is half of , or .
. Therefore, each equal piece that the line separates into must have an area of .
Call the point where the line through intersects . We know that . Furthermore, we know that , as . Thus, solving for , we find that , so . This gives that the y coordinate of E is .
Line CD can be expressed as , so the coordinate of E satisfies . Solving for , we find that .
From this, we know that .
Solution 2
Let the point where the altitude from to be labeled . Following the steps above, you can find that the height of is , and from there split the base into two parts, , and , such that is the segment from the origin to the point , and is the segment from point to point . Then, by the Pythagorean Theorem, , and the answer is
See Also
2013 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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 |
2013 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 12 |
Followed by Problem 14 |
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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