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 "2013 AMC 10A Problems/Problem 18"

(→‎Problem)
m (→‎Solution 2)
 
(19 intermediate revisions by 9 users not shown)
Line 1: Line 1:
 
==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>?
  
  
Line 7: Line 7:
 
[[Category: Introductory Geometry Problems]]
 
[[Category: Introductory Geometry Problems]]
  
−
==Solution==
+
==Solution 1==
  
−
First, various area formulas (shoelace, splitting, etc) allow us to find that <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>.
+
<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 Theorem]]: 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>.   
Line 15: Line 50:
 
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 21:19, 2 November 2023

Problem

Let points $A = (0, 0)$, $B = (1, 2)$, $C=(3, 3)$, and $D = (4, 0)$. Quadrilateral $ABCD$ is cut into equal area pieces by a line passing through $A$. This line intersects $\overline{CD}$ at point $\left(\frac{p}{q}, \frac{r}{s}\right)$, where these fractions are in lowest terms. What is $p+q+r+s$?


$\textbf{(A)}\ 54\qquad\textbf{(B)}\ 58\qquad\textbf{(C)}\ 62\qquad\textbf{(D)}\ 70\qquad\textbf{(E)}\ 75$

Solution 1

[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]

First, we shall find the area of quadrilateral $ABCD$. This can be done in any of three ways:

Pick's Theorem: $[ABCD] = I + \dfrac{B}{2} - 1 = 5 + \dfrac{7}{2} - 1 = \dfrac{15}{2}.$

Splitting: Drop perpendiculars from $B$ and $C$ to the x-axis to divide the quadrilateral into triangles and trapezoids, and so the area is $1 + 5 + \dfrac{3}{2} = \dfrac{15}{2}.$

Shoelace Theorem: The area is half of $|1 \cdot 3 - 2 \cdot 3 - 3 \cdot 4| = 15$, or $\dfrac{15}{2}$.

$[ABCD] = \frac{15}{2}$. Therefore, each equal piece that the line separates $ABCD$ into must have an area of $\frac{15}{4}$.

Call the point where the line through $A$ intersects $\overline{CD}$ $E$. We know that $[ADE] = \frac{15}{4} = \frac{bh}{2}$. Furthermore, we know that $b = 4$, as $AD = 4$. Thus, solving for $h$, we find that $2h = \frac{15}{4}$, so $h = \frac{15}{8}$. This gives that the y coordinate of E is $\frac{15}{8}$.

Line CD can be expressed as $y = -3x+12$, so the $x$ coordinate of E satisfies $\frac{15}{8} = -3x + 12$. Solving for $x$, we find that $x = \frac{27}{8}$.

From this, we know that $E = \left(\frac{27}{8}, \frac{15}{8}\right)$. $27 + 15 + 8 + 8 = \boxed{\textbf{(B) }58}$

Solution 2

[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]

Let the point where the altitude from $E$ to $\overline{AD}$ be labeled $F$. Following the steps above, you can find that the height of $\triangle ADE$ is $\frac{15}{8}$, and from there split the base into two parts, $x$, and $4-x$, such that $x$ is the segment from the origin to the point $F$, and $4-x$ is the segment from point $F$ to point $D$. Then, by the Pythagorean Theorem, $x=\frac{27}{8}$, and the answer is $\boxed{\textbf{(B) }58}$

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

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=2013_AMC_10A_Problems/Problem_18&oldid=200663"