Difference between revisions of "2019 AMC 10A Problems/Problem 7"
(→BEST solution) |
Coltsfan10 (talk | contribs) m (→BEST solution) |
||
Line 19: | Line 19: | ||
==BEST solution== | ==BEST solution== | ||
− | Simply | + | Simply translate <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: <cmath>\frac12\begin{Vmatrix} |
4&2\\ | 4&2\\ | ||
2&4\\ | 2&4\\ |
Revision as of 13:41, 21 March 2020
- 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
.
BEST solution
Simply translate 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:
~N828335
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.
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.