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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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.
Contents
[hide]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 .
Solution 3
Like in the other solutions, solve the systems of equations to see that the triangle's two other vertices are at and . Then apply Heron's Formula: the semi-perimeter will be , so the area reduces nicely to a difference of squares, making it .
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 , , and . We can now draw the bounding square with vertices , , and , and deduce that the triangle's area is .
Solution 5
Like in other solutions, we find that the three points of intersection are , , and . Using graph paper, we can see that this triangle has boundary lattice points and interior lattice points. By Pick's Theorem, the area is .
Solution 6
Like in other solutions, we find the three points of intersection. Label these , , and . By the Pythagorean Theorem, and . By the Law of Cosines, Therefore, , so the area is .
Solution 7
Like in other solutions, we find that the three points of intersection are , , and . The area of the triangle is half the absolute value of the determinant of the matrix determined by these points.
Solution 8
Like in other solutions, we find the three points of intersection. Label these , , and . Then vectors and . The area of the triangle is half the magnitude of the cross product of these two vectors.
Solution 9
Like in other solutions, we find that the three points of intersection are , , and . By the Pythagorean theorem, this is an isosceles triangle with base and equal length . The area of an isosceles triangle with base and equal length is . Plugging in and ,
Solution 10 (Trig)
Like in other solutions, we find the three points of intersection. Label these , , and . By the Pythagorean Theorem, and . By the Law of Cosines, Therefore, . By the extended Law of Sines, Then the area is .
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.
Solution 12 (Heron's Formula)
Like in other solutions, we find that our triangle is isosceles with legs of and base . Then, the semi - perimeter of our triangle is, Applying Heron's formula, we find that the area of this triangle is equivalent to
~rbcubed13
Video Solution 1
Education, the Study of Everything
Video Solution 2
~savannahsolver