Difference between revisions of "2017 AMC 12B Problems/Problem 18"
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<math>[\triangle ABC]=\frac{49}{49+25}[\triangle ABF]=\boxed{\textbf{(D)}\ \frac{140}{37}}</math>. | <math>[\triangle ABC]=\frac{49}{49+25}[\triangle ABF]=\boxed{\textbf{(D)}\ \frac{140}{37}}</math>. | ||
− | ==Solution 4 | + | ==Solution 4 (Coordinate Geometry)== |
Let <math>A</math> be at the origin <math>(0, 0)</math> of a coordinate plane, with <math>B</math> being located at <math>(4, 0)</math>, etc. | Let <math>A</math> be at the origin <math>(0, 0)</math> of a coordinate plane, with <math>B</math> being located at <math>(4, 0)</math>, etc. | ||
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Using the diagram in solution one, the equation for circle <math>O</math> is <math>(x-2)^2+y^2 = 4</math>. | Using the diagram in solution one, the equation for circle <math>O</math> is <math>(x-2)^2+y^2 = 4</math>. | ||
− | The equation for line <math>AE</math> is then <math>y = \frac{5}{7}x</math>, | + | The equation for line <math>AE</math> is then <math>y = \frac{5}{7}x</math>, therefore <math>x = \frac{7}{5}y</math>. |
Substituting <math>\frac{7}{5}y</math> for <math>x</math> in the equation for circle <math>O</math>, we get: | Substituting <math>\frac{7}{5}y</math> for <math>x</math> in the equation for circle <math>O</math>, we get: | ||
− | <math>(\frac{7}{5}y-2)^2+y^2 = 4</math> | + | <math>\left(\frac{7}{5}y-2\right)^2+y^2 = 4</math> |
− | We can solve for <math>y</math> to yield the <math>y</math> coordinate of point <math>C</math> in the coordinate plane, since this is the point of intersection of the circle and line <math>AE</math>. Note that one root will yield the intersection of the circle and <math>AE</math> at the origin, so we will ignore this root. | + | We can solve for <math>y</math> to yield the <math>y</math> coordinate of point <math>C</math> in the coordinate plane, since this is the point of intersection of the circle and line <math>AE</math>. Note that one root will yield the intersection of the circle and line <math>AE</math> at the origin, so we will ignore this root. |
Expanding the expression and factoring, we get: | Expanding the expression and factoring, we get: | ||
− | <math>(\frac{49}{25}y^2-\frac{28}{5}y+4)+y^2 = 4</math> | + | <math>\left(\frac{49}{25}y^2-\frac{28}{5}y+4\right)+y^2 = 4</math> |
<math>\frac{74}{25}y^2-\frac{28}{5}y = 0</math> | <math>\frac{74}{25}y^2-\frac{28}{5}y = 0</math> | ||
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-Solution by Joeya | -Solution by Joeya | ||
+ | |||
+ | == Video Solution (Similar Triangles) == | ||
+ | https://youtu.be/NsQbhYfGh1Q?t=512 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
==See Also== | ==See Also== |
Latest revision as of 21:27, 13 February 2021
Contents
Problem
The diameter of a circle of radius is extended to a point outside the circle so that . Point is chosen so that and line is perpendicular to line . Segment intersects the circle at a point between and . What is the area of ?
Solution 1
Let be the center of the circle. Note that . However, by Power of a Point, , so . Now . Since .
Solution 2: Similar triangles with Pythagorean
is the diameter of the circle, so is a right angle, and therefore by AA similarity, .
Because of this, , so .
Likewise, , so .
Thus the area of .
Solution 2b: Area shortcut
Because is and is , the ratio of the sides is , meaning the ratio of the areas is thus . We then have the proportion
Solution 3: Similar triangles without Pythagorean
Or, use similar triangles all the way, dispense with Pythagorean, and go for minimal calculation:
Draw with on . .
.
. ( ratio applied twice)
.
Solution 4 (Coordinate Geometry)
Let be at the origin of a coordinate plane, with being located at , etc.
We can find the area of by finding the the altitude from line to point . Realize that this altitude is the coordinate of point on the coordinate plane, since the respective base of is on the -axis.
Using the diagram in solution one, the equation for circle is .
The equation for line is then , therefore .
Substituting for in the equation for circle , we get:
We can solve for to yield the coordinate of point in the coordinate plane, since this is the point of intersection of the circle and line . Note that one root will yield the intersection of the circle and line at the origin, so we will ignore this root.
Expanding the expression and factoring, we get:
Our non-zero root is thus . Calculating the area of with as the length of and as the altitude, we get:
.
-Solution by Joeya
Video Solution (Similar Triangles)
https://youtu.be/NsQbhYfGh1Q?t=512
~ pi_is_3.14
See Also
2017 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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.