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

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==Solution 1==
 
==Solution 1==
−
The two lines are <math>y=2x-2</math> and <math>y = \frac{x}{2}+1</math>, which intersect the third line at <math>(4,6)</math> and <math>(6,4)</math>. So we have an isosceles triangle with base <math>2\sqrt{2}</math> and height <math>3\sqrt{2} \implies \boxed{\textbf{(C) }6}</math>.
+
Let's first work out the slope-intercept form of all three lines:
 +
<math>(x,y)=(2,2)</math> and <math>y=\frac{x}{2} + b</math> implies <math>2=\frac{2}{2} +b=> 2=1+b</math> so <math>b=1</math>, while <math>y=2x + c</math> implies <math>2= 2 \cdot 2+c=> 2=4+c</math> so <math>c=-2</math>. Also, <math>x+y=10</math> implies <math>y=-x+10</math>. Thus the lines are <math>y=\frac{x}{2} +1, y=2x-2,</math> and <math>y=-x+10</math>.
 +
Now we find the intersection points between each of the lines with <math>y=-x+10</math>, which are <math>(6,4)</math> and <math>(4,6)</math>. Using the distance formula and then the Pythagorean Theorem, we see that we have an isosceles triangle with base <math>2\sqrt{2}</math> and height <math>3\sqrt{2}</math>, whose area is <math>\boxed{\textbf{(C) }6}</math>.
  
 
==Solution 2==
 
==Solution 2==
−
Like in Solution 1, let's first calculate the slope-intercept form of all three lines:
+
Like in Solution 1, we determine the coordinates of the three vertices of the triangle. The coordinates that we get are:
−
<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>(2,2)</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>
+
<math>(6,4)</math>
 +
<math>(4,6)</math>. Now, using the [[Shoelace Theorem]], we can directly find that the area is <math>\boxed{\textbf{(C) }6}</math>.
  
 
==Solution 3==
 
==Solution 3==
−
Like the other solutions, solve the systems to see that the triangles two other points are at <math>(4, 6)</math> and <math>(6, 4)</math>. The apply Heron's Formula. The semi-perimeter will be <math>s = \sqrt{2} + \sqrt{20}</math>. The area then reduces nicely to a difference of squares, making it <math>6 \implies \boxed{\textbf{(C) }6}.</math>
+
Like in the other solutions, solve the systems of equations to see that the triangle's two other vertices are at <math>(4, 6)</math> and <math>(6, 4)</math>. Then apply Heron's Formula: the semi-perimeter will be <math>s = \sqrt{2} + \sqrt{20}</math>, so the area reduces nicely to a difference of squares, making it <math>\boxed{\textbf{(C) }6}</math>.
 +
 
 +
==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 <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>
 +
 
 +
==Solution 9==
 +
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>. By the Pythagorean theorem, this is an isosceles triangle with base <math>2\sqrt2</math> and equal length <math>2\sqrt5</math>. The area of an isosceles triangle with base <math>b</math> and equal length <math>l</math> is <math>\frac{b\sqrt{4l^2-b^2}}{4}</math>. Plugging  in <math>b = 2\sqrt2</math> and <math>l = 2\sqrt5</math>,
 +
<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 <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>. 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 <math>\frac{abc}{4R} = \frac{2\sqrt2 \cdot 2\sqrt5^2}{4 \cdot \frac{5\sqrt2}{3}} = \boxed{\textbf{(C) }6}</math>.
 +
 
 +
==Solution 11==
 +
The area of a triangle formed by three lines,
 +
<cmath>a_1x + a_2y + a_3 = 0</cmath>
 +
<cmath>b_1x + b_2y + b_3 = 0</cmath>
 +
<cmath>c_1x + c_2y + c_3 = 0</cmath>
 +
is the absolute value of
 +
<cmath>\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</cmath>
 +
Plugging in the three lines,
 +
<cmath>-x + 2y - 2 = 0</cmath>
 +
<cmath>-2x + y + 2 = 0</cmath>
 +
<cmath>x + y - 10 = 0</cmath>
 +
the area is the absolute value of
 +
<cmath>\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}</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.
 +
 
 +
 
 +
==Solution 12 (Heron's Formula) ==
 +
 
 +
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>
 +
 
 +
~rbcubed13
 +
 
 +
 
 +
 
 +
==Video Solution 1==
 +
 
 +
https://youtu.be/KWs9FpLSi5A
 +
 
 +
Education, the Study of Everything
 +
 
 +
 
 +
 
 +
 
 +
 
 +
 
 +
==Video Solution 2==
 +
https://youtu.be/D5FEuT5ExmU
 +
 
 +
~savannahsolver
  
 
==See Also==
 
==See Also==
  
 
{{AMC10 box|year=2019|ab=A|num-b=6|num-a=8}}
 
{{AMC10 box|year=2019|ab=A|num-b=6|num-a=8}}
−
{{AMC12 box|year=2019|ab=A|num-b=4|num-a=6}}
 
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 14:49, 1 August 2024

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 $2=\frac{2}{2} +b=> 2=1+b$ so $b=1$, while $y=2x + c$ implies $2= 2 \cdot 2+c=> 2=4+c$ 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

See Also

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

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