2019 AMC 10A Problems/Problem 7

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. The coordinates that we get are: $(2,2)$ $(6,4)$ $(4,6)$ . Now, using the Shoelace Theorem, we can directly find that the area is $\boxed{\textbf{(C) }6}$ .

Super fast solution

Simply transform $(2,2)$ to the origin, which would change $x+y=10$ to $x+y=6$ . Now the three points are $(0,0)$ , $(2,4)$ , $(4,2)$ . Now, because it is at the origin, we can easily take the half the discriminant: $\frac{16-4}{2}=6\rightarrow \boxed{\textbf{(C) }6}$ $~N828335 ==Solution 3== Like in the other solutions, solve the systems of equations to see that the triangle's two other vertices are at$ (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$ (Error compiling LaTeX. Unknown error_msg)(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$ (Error compiling LaTeX. Unknown error_msg)(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$ (Error compiling LaTeX. Unknown error_msg)A (2, 2) $,$ B (4, 6) $, and$ C (6, 4) $. By the Pythagorean Theorem,$ AB = AC = 2\sqrt5 $and$ BC = 2\sqrt2 $. By the Law of Cosines, <cmath>\cos A = \frac{(2\sqrt5)^2 + (2\sqrt5)^2 - (2\sqrt2)^2}{2 \cdot 2\sqrt5 \cdot 2\sqrt5} = \frac{20 + 20 - 8}{40} = \frac{32}{40} = \frac45</cmath> Therefore,$ \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$ (Error compiling LaTeX. Unknown error_msg)(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. <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$ (Error compiling LaTeX. Unknown error_msg)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. <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>

==Solution 9== Like in other solutions, we find that the three points of intersection are$ (Error compiling LaTeX. Unknown error_msg)(2, 2) $,$ (4, 6) $, and$ (6, 4) $. By the Pythagorean theorem, this is an isosceles triangle with base$ 2\sqrt2 $and equal length$ 2\sqrt5 $. The area of an isosceles triangle with base$ b $and equal length$ l $is$ \frac{b\sqrt{4l^2-b^2}}{4} $. Plugging in$ b = 2\sqrt2 $and$ l = 2\sqrt5$, <cmath>\frac{2\sqrt2 \cdot \sqrt{80-8}}{4} = \frac{\sqrt{576}}{4} = \frac{24}{4} = \boxed{\textbf{(C) }6}</cmath>

==Solution 10 (Trig) == Like in other solutions, we find the three points of intersection. Label these$ (Error compiling LaTeX. Unknown error_msg)A (2, 2) $,$ B (4, 6) $, and$ C (6, 4) $. By the Pythagorean Theorem,$ AB = AC = 2\sqrt5 $and$ BC = 2\sqrt2 $. By the Law of Cosines, <cmath>\cos A = \frac{(2\sqrt5)^2 + (2\sqrt5)^2 - (2\sqrt2)^2}{2 \cdot 2\sqrt5 \cdot 2\sqrt5} = \frac{20 + 20 - 8}{40} = \frac{32}{40} = \frac45</cmath> Therefore,$ \sin A = \sqrt{1 - \cos^2 A} = \frac35 $. By the extended Law of Sines, <cmath>2R = \frac{a}{\sin A} = \frac{2\sqrt2}{\frac35} = \frac{10\sqrt2}{3}</cmath> <cmath>R = \frac{5\sqrt2}{3}</cmath> Then the area is$ \frac{abc}{4R} = \frac{2\sqrt2 \cdot 2\sqrt5^2}{4 \cdot \frac{5\sqrt2}{3}} = \boxed{\textbf{(C) }6}$.

Solution 11

The area of a triangle formed by three lines, $a_1x + a_2y + a_3 = 0$ $b_1x + b_2y + b_3 = 0$ $c_1x + c_2y + c_3 = 0$ is the absolute value of $\frac12 \cdot \frac{1}{(b_1c_2-b_2c_1)(a_1c_2-a_2c_1)(a_1b_2-a_2b_1)} \cdot \begin{vmatrix} a_1&a_2&a_3\\ b_1&b_2&b_3\\ c_1&c_2&c_3\\ \end{vmatrix}^2$ Plugging in the three lines, $-x + 2y - 2 = 0$ $-2x + y + 2 = 0$ $x + y - 10 = 0$ the area is the absolute value of $\frac12 \cdot \frac{1}{(-2-1)(-1-2)(-1+4)} \cdot \begin{vmatrix} -1&2&-2\\ -2&1&2\\ 1&1&-10\\ \end{vmatrix}^2 = \frac12 \cdot \frac{1}{27} \cdot 18^2 = \boxed{\textbf{(C) }6}$ Source: Orrick, Michael L. “THE AREA OF A TRIANGLE FORMED BY THREE LINES.” Pi Mu Epsilon Journal, vol. 7, no. 5, 1981, pp. 294–298. JSTOR, www.jstor.org/stable/24336991.

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

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

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