Difference between revisions of "2019 AMC 10A Problems/Problem 7"
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==Solution 1== | ==Solution 1== | ||
Let's first work out the slope-intercept form of all three lines: | 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=1+b</math> so <math>b=1</math>, while <math>y=2x + c</math> implies <math>2= 2 \cdot 2+c=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>. | + | <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 \boxed{\textbf{(C) }6}<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, we determine the coordinates of the three vertices of the triangle. Now, using the [[Shoelace Theorem]], we can directly find that the area is < | + | 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== | ||
− | Like in the other solutions, solve the systems of equations to see that the triangle's two other vertices are at < | + | 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 < | + | 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}} | ||
− | |||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 15:49, 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.
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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.