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

m (Fixed formatting)
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==Solution 4==
 
==Solution 4==
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Like in the other solutions, we find, either using algebra or simply by drawing the lines on squared paper, that the three points of intersection are <math>(2, 2)</math>, <math>(4, 6)</math> and <math>(6, 4)</math>. We can now draw the bounding square with vertices <math>(2, 2)</math>, <math>(2, 6)</math>, <math>(6, 6)</math> and <math>(6, 2)</math>, and deduce that the triangle's area is <math>16-4-2-4=\boxed{\textbf{(C) }6}</math>.
+
Like in the other solutions, we find, either using algebra or simply by drawing the lines on squared paper, that the three points of intersection are <math>(2, 2)</math>, <math>(4, 6)</math>, and <math>(6, 4)</math>. We can now draw the bounding square with vertices <math>(2, 2)</math>, <math>(2, 6)</math>, <math>(6, 6)</math> and <math>(6, 2)</math>, and deduce that the triangle's area is <math>16-4-2-4=\boxed{\textbf{(C) }6}</math>.
 +
 
 +
==Solution 5==
 +
Like in other solutions, we find that the three points of intersection are  <math>(2, 2)</math>, <math>(4, 6)</math>, and <math>(6, 4)</math>. Using graph paper, we can see that this triangle has <math>6</math> boundary lattice points and <math>4</math> interior lattice points. By Pick's Theorem, the area is <math>\frac62 + 4 - 1 = \boxed{\textbf{(C) }6}</math>.
 +
 
 +
==Solution 6==
 +
Like in other solutions, we find the three points of intersection. Label these <math>A (2, 2)</math>, <math>B (4, 6)</math>, and <math>C (6, 4)</math>. By the Pythagorean Theorem, <math>AB = AC = 2\sqrt5</math> and <math>BC = 2\sqrt2</math>. By the Law of Cosines,
 +
<cmath>\cos A = \frac{(2\sqrt5)^2 + (2\sqrt5)^2 - (2\sqrt2)^2}{2 \cdot 2\sqrt5 \cdot 2\sqrt5} = \frac{20 + 20 - 8}{40} = \frac{32}{40} = \frac45</cmath>
 +
Therefore, <math>\sin A = \sqrt{1 - \cos^2 A} = \frac35</math>, so the area is <math>\frac12 bc \sin A = \frac12 \cdot 2\sqrt5 \cdot 2\sqrt5 \cdot \frac35 = \boxed{\textbf{(C) }6}</math>.
 +
 
 +
==Solution 7==
 +
Like in other solutions, we find that the three points of intersection are  <math>(2, 2)</math>, <math>(4, 6)</math>, and <math>(6, 4)</math>. The area of the triangle is half the absolute value of the determinant of the matrix determined by these points.
 +
<cmath>
 +
\frac12\begin{Vmatrix}
 +
2&2&1\\
 +
4&6&1\\
 +
6&4&1\\
 +
\end{Vmatrix} = \frac12|-12| = \frac12 \cdot 12 = \boxed{\textbf{(C) }6}</cmath>
 +
 
 +
 
 +
==Solution 8==
 +
Like in other solutions, we find the three points of intersection. Label these <math>A (2, 2)</math>, <math>B (4, 6)</math>, and <math>C (6, 4)</math>. Then vectors <math>\overrightarrow{AB} = \langle 2, 4 \rangle</math> and <math>\overrightarrow{AC} = \langle 4, 2 \rangle</math>. The area of the triangle is half the magnitude of the cross product of these two vectors.
 +
<cmath>
 +
\frac12\begin{Vmatrix}
 +
i&j&k\\
 +
2&4&0\\
 +
4&2&0\\
 +
\end{Vmatrix} = \frac12|-12k| = \frac12 \cdot 12 = \boxed{\textbf{(C) }6}</cmath>
  
 
==See Also==
 
==See Also==

Revision as of 22:04, 29 March 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

Let's first work out the slope-intercept form of all three lines: $(x,y)=(2,2)$ and $y=\frac{x}{2} + b$ implies $2=\frac{2}{2} +b=1+b$ so $b=1$, while $y=2x + c$ implies $2= 2 \cdot 2+c=4+c$ so $c=-2$. Also, $x+y=10$ implies $y=-x+10$. Thus the lines are $y=\frac{x}{2} +1, y=2x-2,$ and $y=-x+10$. Now we find the intersection points between each of the lines with $y=-x+10$, which are $(6,4)$ and $(4,6)$. Using the distance formula and then the Pythagorean Theorem, we see that we have an isosceles triangle with base $2\sqrt{2}$ and height $3\sqrt{2}$, whose area is $\boxed{\textbf{(C) }6}$.

Solution 2

Like in Solution 1, we determine the coordinates of the three vertices of the triangle. Now, using the Shoelace Theorem, we can directly find that the area is $\boxed{\textbf{(C) }6}$.

Solution 3

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

Solution 4

Like in the other solutions, we find, either using algebra or simply by drawing the lines on squared paper, that the three points of intersection are $(2, 2)$, $(4, 6)$, and $(6, 4)$. We can now draw the bounding square with vertices $(2, 2)$, $(2, 6)$, $(6, 6)$ and $(6, 2)$, and deduce that the triangle's area is $16-4-2-4=\boxed{\textbf{(C) }6}$.

Solution 5

Like in other solutions, we find that the three points of intersection are $(2, 2)$, $(4, 6)$, and $(6, 4)$. Using graph paper, we can see that this triangle has $6$ boundary lattice points and $4$ interior lattice points. By Pick's Theorem, the area is $\frac62 + 4 - 1 = \boxed{\textbf{(C) }6}$.

Solution 6

Like in other solutions, we find the three points of intersection. Label these $A (2, 2)$, $B (4, 6)$, and $C (6, 4)$. By the Pythagorean Theorem, $AB = AC = 2\sqrt5$ and $BC = 2\sqrt2$. By the Law of Cosines, \[\cos A = \frac{(2\sqrt5)^2 + (2\sqrt5)^2 - (2\sqrt2)^2}{2 \cdot 2\sqrt5 \cdot 2\sqrt5} = \frac{20 + 20 - 8}{40} = \frac{32}{40} = \frac45\] Therefore, $\sin A = \sqrt{1 - \cos^2 A} = \frac35$, so the area is $\frac12 bc \sin A = \frac12 \cdot 2\sqrt5 \cdot 2\sqrt5 \cdot \frac35 = \boxed{\textbf{(C) }6}$.

Solution 7

Like in other solutions, we find that the three points of intersection are $(2, 2)$, $(4, 6)$, and $(6, 4)$. The area of the triangle is half the absolute value of the determinant of the matrix determined by these points. \[\frac12\begin{Vmatrix} 2&2&1\\ 4&6&1\\ 6&4&1\\ \end{Vmatrix} = \frac12|-12| = \frac12 \cdot 12 = \boxed{\textbf{(C) }6}\]


Solution 8

Like in other solutions, we find the three points of intersection. Label these $A (2, 2)$, $B (4, 6)$, and $C (6, 4)$. Then vectors $\overrightarrow{AB} = \langle 2, 4 \rangle$ and $\overrightarrow{AC} = \langle 4, 2 \rangle$. The area of the triangle is half the magnitude of the cross product of these two vectors. \[\frac12\begin{Vmatrix} i&j&k\\ 2&4&0\\ 4&2&0\\ \end{Vmatrix} = \frac12|-12k| = \frac12 \cdot 12 = \boxed{\textbf{(C) }6}\]

See Also

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

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