Difference between revisions of "2019 AMC 10A Problems/Problem 7"

Revision as of 18:37, 10 February 2019

The following problem is from both the 2019 AMC 10A #7 and 2019 AMC 12A #5, so both problems redirect to this page.

Problem

Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10 ?$

$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$

Solution 1

The two lines are $y=2x-2$ and $y = \frac{x}{2}+1$ , which intersect the third line at $(4,6)$ and $(6,4)$ . So we have an isosceles triangle with base $2\sqrt{2}$ and height $3\sqrt{2} \implies \boxed{\textbf{(C) }6}$ .

Solution 2

Like in Solution 1, let's first calculate the slope-intercept form of all three lines: $(x,y)=(2,2)$ and $y=x/2 + b$ implies $2=2/2 +b=1+b$ so b=1, while $y=2x + c$ implies $2= 2*2+c=4+c$ so c=-2. Also, $x+y=10$ implies $y=-x+10$ . Thus the lines are $y=x/2 +1, y=2x-2,$ and $y=-x+10$ . Now we find the intersections between each of the lines with $y=-x+10$ , which are $(6,4)$ and $(4,6)$ . Applying the Shoelace Theorem, we can find that the solution is $6 \implies \boxed{\textbf{(C) }6}.$

Solution 3

Like the other solutions, solve the systems to see that the triangles two other points are at $(4, 6)$ and $(6, 4)$ . The apply Heron's Formula. The semi-perimeter will be $s = \sqrt{2} + \sqrt{20}$ . The area then reduces nicely to a difference of squares, making it $6 \implies \boxed{\textbf{(C) }6}.$

2019 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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

2019 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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.

@@ Line 12: / Line 12: @@
 ==Solution 2==
 Like in Solution 1, let's first calculate the slope-intercept form of all three lines:
-<math>(x,y)=(2,2)</math>
+<math>(x,y)=(2,2)</math> and <math>y=x/2 + b</math> implies <math>2=2/2 +b=1+b</math> so b=1, while <math>y=2x + c</math> implies <math>2= 2*2+c=4+c</math> so c=-2. Also, <math>x+y=10</math> implies <math>y=-x+10</math>. Thus the lines are <math>y=x/2 +1, y=2x-2,</math> and <math>y=-x+10</math>.
-<math>y=x/2 + b</math> becomes <math>2=2/2 +b=1+b</math> so b=1,
-<math>y=2x + c</math> becomes <math>2= 2*2+c=4+c</math> so c=-2, and
-<math>x+y=10</math> becomes <math>y=-x+10</math>.
-(<math>y=x/2 +1, y=2x-2,</math> and <math>y=-x+10</math>.)
 Now we find the intersections between each of the lines with <math>y=-x+10</math>, which are <math>(6,4)</math> and <math>(4,6)</math>. Applying the Shoelace Theorem, we can find that the solution is <math>6 \implies \boxed{\textbf{(C) }6}.</math>

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