Difference between revisions of "2019 AMC 10A Problems/Problem 7"
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==Solution 2== | ==Solution 2== | ||
− | Like in Solution 1, we determine the coordinates of the three vertices of the triangle. Now, using the [[Shoelace Theorem]], we can directly find that the area is <math>\boxed{\textbf{(C) }6}</math>. | + | Like in Solution 1, we determine the coordinates of the three vertices of the triangle. The coordinates that we get are: |
+ | <math>(2,2)</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== | ||
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<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== | + | ==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, | The area of a triangle formed by three lines, | ||
<cmath>a_1x + a_2y + a_3 = 0</cmath> | <cmath>a_1x + a_2y + a_3 = 0</cmath> | ||
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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. | ||
− | |||
− | + | ==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 |
Latest revision as of 22:24, 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
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