Difference between revisions of "2019 AMC 10A Problems/Problem 7"
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<math>(4,6)</math>. Now, using the [[Shoelace Theorem]], we can directly find that the area is <math>\boxed{\textbf{(C) }6}</math>. | <math>(4,6)</math>. Now, using the [[Shoelace Theorem]], we can directly find that the area is <math>\boxed{\textbf{(C) }6}</math>. | ||
+ | ==Super fast solution= | ||
+ | Simply transform <math>(2,2)</math> to the origin, which would change <math>x+y=10</math> to <math>x+y=6</math>. Now the three points are <math>(0,0)</math>, <math>(2,4)</math>, <math>(4,2)</math>. Now, because it is at the origin, we can easily take the half the discriminant: <math>\frac{16-4}{2}=6\rightarrow \boxed{\textbf{(C) }6} | ||
+ | ~N828335 | ||
==Solution 3== | ==Solution 3== | ||
− | Like in the other solutions, solve the systems of equations to see that the triangle's two other vertices are at <math>(4, 6)< | + | 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== | ==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)< | + | 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== | ==Solution 5== | ||
− | Like in other solutions, we find that the three points of intersection are <math>(2, 2)< | + | 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== | ==Solution 6== | ||
− | Like in other solutions, we find the three points of intersection. Label these <math>A (2, 2)< | + | 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> | <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< | + | 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== | ==Solution 7== | ||
− | Like in other solutions, we find that the three points of intersection are <math>(2, 2)< | + | 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> | <cmath> | ||
\frac12\begin{Vmatrix} | \frac12\begin{Vmatrix} | ||
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==Solution 8== | ==Solution 8== | ||
− | Like in other solutions, we find the three points of intersection. Label these <math>A (2, 2)< | + | 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> | <cmath> | ||
\frac12\begin{Vmatrix} | \frac12\begin{Vmatrix} | ||
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==Solution 9== | ==Solution 9== | ||
− | Like in other solutions, we find that the three points of intersection are <math>(2, 2)< | + | 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> | <cmath>\frac{2\sqrt2 \cdot \sqrt{80-8}}{4} = \frac{\sqrt{576}}{4} = \frac{24}{4} = \boxed{\textbf{(C) }6}</cmath> | ||
==Solution 10 (Trig) == | ==Solution 10 (Trig) == | ||
− | Like in other solutions, we find the three points of intersection. Label these <math>A (2, 2)< | + | 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> | <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< | + | 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>2R = \frac{a}{\sin A} = \frac{2\sqrt2}{\frac35} = \frac{10\sqrt2}{3}</cmath> | ||
<cmath>R = \frac{5\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} | + | 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}$. |
==Solution 11== | ==Solution 11== |
Revision as of 17:28, 8 February 2020
- The following problem is from both the 2019 AMC 10A #7 and 2019 AMC 12A #5, so both problems redirect to this page.
Problem
Two lines with slopes and intersect at . What is the area of the triangle enclosed by these two lines and the line
Solution 1
Let's first work out the slope-intercept form of all three lines: and implies so , while implies so . Also, implies . Thus the lines are and . Now we find the intersection points between each of the lines with , which are and . Using the distance formula and then the Pythagorean Theorem, we see that we have an isosceles triangle with base and height , whose area is .
Solution 2
Like in Solution 1, we determine the coordinates of the three vertices of the triangle. The coordinates that we get are: . Now, using the Shoelace Theorem, we can directly find that the area is .
=Super fast solution
Simply transform to the origin, which would change to . Now the three points are , , . Now, because it is at the origin, we can easily take the half the discriminant: (4, 6)(6, 4)s = \sqrt{2} + \sqrt{20}\boxed{\textbf{(C) }6}$.
==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)(6, 4)(2, 2)(2, 6)(6, 6)(6, 2)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)(6, 4)64\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)C (6, 4)AB = AC = 2\sqrt5BC = 2\sqrt2\sin A = \sqrt{1 - \cos^2 A} = \frac35\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)(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)C (6, 4)\overrightarrow{AB} = \langle 2, 4 \rangle\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)(6, 4)2\sqrt22\sqrt5bl\frac{b\sqrt{4l^2-b^2}}{4}b = 2\sqrt2l = 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)C (6, 4)AB = AC = 2\sqrt5BC = 2\sqrt2\sin A = \sqrt{1 - \cos^2 A} = \frac35\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, is the absolute value of Plugging in the three lines, the area is the absolute value of 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.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.