Difference between revisions of "2017 AMC 12B Problems/Problem 18"
Pi is 3.14 (talk | contribs) (→Video Solution (Similar Triangles)) |
Mathophobia (talk | contribs) (→Solution 4 (Coordinate Geometry)) |
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-Solution by Joeya | -Solution by Joeya | ||
+ | |||
+ | ==Solution 5 (No sqrts)== | ||
+ | Slope of AC is 5/7 | ||
+ | As stated in other solutions AB is the diameter, ABC is right. | ||
+ | |||
+ | Let CF be an altitude of ABC. | ||
+ | |||
+ | AF:CF = CF:BF = 7:5 | ||
+ | |||
+ | We can set AF = 49, CF = 35, BF = 25 and scale back later | ||
+ | |||
+ | Then the radius is AB/2 = (AF+BF)/2 = 74/2 = 37. | ||
+ | |||
+ | So the radius is 37 and the height of ABC is 35. | ||
+ | |||
+ | If we scale it back so that our radius is 2, our height is 70/37. | ||
+ | |||
+ | Area of ABC is bh/2 = <math>\frac{(4)(\frac{70}{37})}{2}</math> = <math>\boxed{\textbf{(D)}\ \frac{140}{37}}</math>. | ||
+ | |||
+ | -mathophobia | ||
== Video Solution by OmegaLearn (Similar Triangles) == | == Video Solution by OmegaLearn (Similar Triangles) == |
Revision as of 20:40, 30 September 2024
Contents
[hide]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
Solution 5 (No sqrts)
Slope of AC is 5/7 As stated in other solutions AB is the diameter, ABC is right.
Let CF be an altitude of ABC.
AF:CF = CF:BF = 7:5
We can set AF = 49, CF = 35, BF = 25 and scale back later
Then the radius is AB/2 = (AF+BF)/2 = 74/2 = 37.
So the radius is 37 and the height of ABC is 35.
If we scale it back so that our radius is 2, our height is 70/37.
Area of ABC is bh/2 = =
.
-mathophobia
Video Solution by OmegaLearn (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.