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

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\end{vmatrix}^2 = \frac12 \cdot \frac{1}{27} \cdot 18^2 = \boxed{\textbf{(C) }6}</cmath>
 
\end{vmatrix}^2 = \frac12 \cdot \frac{1}{27} \cdot 18^2 = \boxed{\textbf{(C) }6}</cmath>
 
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.
 
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.
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==Solution 12 (Heron's Formula) ==
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Like in other solutions, we find that our triangle is isosceles with legs of <math>2\sqrt5</math> and base <math>2\sqrt2</math>. Then, the semi - perimeter of our triangle is, <cmath>\frac{4\sqrt5+2\sqrt2}{2} = 2\sqrt5 + \sqrt2.</cmath> Applying Heron's formula, we find that the area of this triangle is equivalent to <cmath>\sqrt{{(2\sqrt5+\sqrt2)}{(2\sqrt5-\sqrt2)}{(2)}} = \sqrt{{(20-2)}{(2)}} = \boxed{\textbf{(C) }6}.</cmath>
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~rbcubed13
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==Video Solution 1==
 
==Video Solution 1==
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~savannahsolver
 
~savannahsolver
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==Solution 12 (Heron's Formula) ==
 
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Like in other solutions, we find that our triangle is isosceles with legs of <math>2\sqrt5</math> and base <math>2\sqrt2</math>. Then, the semi - perimeter of our triangle is, <cmath>\frac{4\sqrt5+2\sqrt2}{2} = 2\sqrt5 + \sqrt2.</cmath> Applying Heron's formula, we find that the area of this triangle is equivalent to <cmath>\sqrt{{(2\sqrt5+\sqrt2)}{(2\sqrt5-\sqrt2)}{(2)}} = \sqrt{{(20-2)}{(2)}} = \boxed{\textbf{(C) }6}.</cmath>
 
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~rbcubed13
 

Revision as of 21:24, 10 November 2023

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 $\boxed{\textbf{(C) }6}$.

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. \[\frac12\begin{Vmatrix} i&j&k\\ 2&4&0\\ 4&2&0\\ \end{Vmatrix} = \frac12|-12k| = \frac12 \cdot 12 = \boxed{\textbf{(C) }6}\]

Solution 9

Like in other solutions, we find that the three points of intersection are $(2, 2)$, $(4, 6)$, 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 $l = 2\sqrt5$, \[\frac{2\sqrt2 \cdot \sqrt{80-8}}{4} = \frac{\sqrt{576}}{4} = \frac{24}{4} = \boxed{\textbf{(C) }6}\]

Solution 10 (Trig)

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, $\sin A = \sqrt{1 - \cos^2 A} = \frac35$. By the extended Law of Sines, \[2R = \frac{a}{\sin A} = \frac{2\sqrt2}{\frac35} = \frac{10\sqrt2}{3}\] \[R = \frac{5\sqrt2}{3}\] 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.


Solution 12 (Heron's Formula)

Like in other solutions, we find that our triangle is isosceles with legs of $2\sqrt5$ and base $2\sqrt2$. Then, the semi - perimeter of our triangle is, \[\frac{4\sqrt5+2\sqrt2}{2} = 2\sqrt5 + \sqrt2.\] Applying Heron's formula, we find that the area of this triangle is equivalent to \[\sqrt{{(2\sqrt5+\sqrt2)}{(2\sqrt5-\sqrt2)}{(2)}} = \sqrt{{(20-2)}{(2)}} = \boxed{\textbf{(C) }6}.\]

~rbcubed13


Video Solution 1

https://youtu.be/KWs9FpLSi5A

Education, the Study of Everything




Video Solution 2

https://youtu.be/D5FEuT5ExmU

~savannahsolver

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