Difference between revisions of "2017 AMC 12B Problems/Problem 18"
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Pi is 3.14 (talk | contribs) (→Video Solution (Similar Triangles)) |
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<math>\textbf{(A)}\ \frac{120}{37}\qquad\textbf{(B)}\ \frac{140}{39}\qquad\textbf{(C)}\ \frac{145}{39}\qquad\textbf{(D)}\ \frac{140}{37}\qquad\textbf{(E)}\ \frac{120}{31}</math> | <math>\textbf{(A)}\ \frac{120}{37}\qquad\textbf{(B)}\ \frac{140}{39}\qquad\textbf{(C)}\ \frac{145}{39}\qquad\textbf{(D)}\ \frac{140}{37}\qquad\textbf{(E)}\ \frac{120}{31}</math> | ||
− | ==Solution== | + | |
− | + | ==Solution 1== | |
+ | |||
+ | <asy> | ||
+ | /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ | ||
+ | import graph; size(8.865514650638614cm); | ||
+ | real labelscalefactor = 0.5; /* changes label-to-point distance */ | ||
+ | pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ | ||
+ | pen dotstyle = black; /* point style */ | ||
+ | real xmin = -6.36927122464312, xmax = 11.361758076634109, ymin = -3.789601803155515, ymax = 7.420015026296013; /* image dimensions */ | ||
+ | |||
+ | |||
+ | draw((-2.,0.)--(0.6486486486486486,1.8918918918918919)--(2.,0.)--cycle); | ||
+ | /* draw figures */ | ||
+ | draw(circle((0.,0.), 2.)); | ||
+ | draw((-2.,0.)--(5.,5.)); | ||
+ | draw((5.,5.)--(5.,0.)); | ||
+ | draw((5.,0.)--(-2.,0.)); | ||
+ | draw((-2.,0.)--(0.6486486486486486,1.8918918918918919)); | ||
+ | draw((0.6486486486486486,1.8918918918918919)--(2.,0.)); | ||
+ | draw((2.,0.)--(-2.,0.)); | ||
+ | draw((2.,0.)--(5.,5.)); | ||
+ | draw((0.,0.)--(5.,5.)); | ||
+ | /* dots and labels */ | ||
+ | dot((0.,0.),dotstyle); | ||
+ | label("$O$", (-0.10330578512396349,-0.39365890308038826), NE * labelscalefactor); | ||
+ | dot((-2.,0.),dotstyle); | ||
+ | label("$A$", (-2.2370398196844437,-0.42371149511645134), NE * labelscalefactor); | ||
+ | dot((2.,0.),dotstyle); | ||
+ | label("$B$", (2.045454545454548,-0.36360631104432517), NE * labelscalefactor); | ||
+ | dot((5.,0.),dotstyle); | ||
+ | label("$D$", (4.900450788880542,-0.42371149511645134), NE * labelscalefactor); | ||
+ | dot((5.,5.),dotstyle); | ||
+ | label("$E$", (5.06574004507889,5.15104432757325), NE * labelscalefactor); | ||
+ | dot((0.6486486486486486,1.8918918918918919),linewidth(3.pt) + dotstyle); | ||
+ | label("$C$", (0.48271975957926694,2.100706235912847), NE * labelscalefactor); | ||
+ | clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); | ||
+ | /* end of picture */ | ||
+ | </asy> | ||
+ | |||
+ | Let <math>O</math> be the center of the circle. Note that <math>EC + CA = EA = \sqrt{AD^2 + DE^2} = \sqrt{(2+2+3)^2 + 5^2} = \sqrt{74}</math>. However, by Power of a Point, <math>(EC)(EC + CA) = EO^2 - R^2 = (2+3)^2 + 5^2 - 2^2 = 25 + 25 - 4 = 46 \implies EC = \frac{46}{\sqrt{74}}</math>, so <math>AC = \sqrt{74} - \frac{46}{\sqrt{74}} = \frac{28}{\sqrt{74}}</math>. Now <math>BC = \sqrt{AB^2 - AC^2} = \sqrt{4^2 - \frac{28^2}{74}} = \sqrt{\frac{16 \cdot 74 - 28^2}{74}} = \sqrt{\frac{1184 - 784}{74}} = \frac{20}{\sqrt{74}}</math>. Since <math>\angle ACB = 90^{\circ}, [ABC] = \frac{1}{2} \cdot BC \cdot AC = \frac{1}{2} \cdot \frac{20}{\sqrt{74}} \cdot \frac{28}{\sqrt{74}} = \boxed{\textbf{(D)}\ \frac{140}{37}}</math>. | ||
+ | |||
+ | |||
+ | ==Solution 2: Similar triangles with Pythagorean== | ||
+ | <math>AB</math> is the diameter of the circle, so <math>\angle ACB</math> is a right angle, and therefore by AA similarity, <math>\triangle ACB \sim \triangle ADE</math>. | ||
+ | |||
+ | Because of this, <math>\frac{AC}{AD} = \frac{AB}{AE} \Longrightarrow \frac{AC}{2+2+3} = \frac{2+2}{\sqrt{7^2 + 5^2}}</math>, so <math>AC = \frac{28}{\sqrt{74}}</math>. | ||
+ | |||
+ | Likewise, <math>\frac{BC}{ED} = \frac{AB}{AE} \Longrightarrow \frac{BC}{5} = \frac{4}{\sqrt{74}}</math>, so <math>BC = \frac{20}{\sqrt{74}}</math>. | ||
+ | |||
+ | Thus the area of <math>\triangle ABC = \frac{1}{2} \cdot \frac{28}{\sqrt{74}} \cdot \frac{20}{\sqrt{74}} = \boxed{\textbf{(D)}\ \frac{140}{37}}</math>. | ||
+ | |||
+ | == Solution 2b: Area shortcut == | ||
+ | |||
+ | Because <math>AE</math> is <math>\sqrt{74}</math> and <math>AB</math> is <math>4</math>, the ratio of the sides is <math>\frac{\sqrt{74}}{4}</math>, meaning the ratio of the areas is thus <math>{(\frac{\sqrt{74}}{4})}^2 \implies \frac{74}{16} \implies \frac{37}{8}</math>. We then have the proportion <math>\frac{\frac{5*7}{2}}{[ABC]}=\frac{37}{8} \implies 37*[ABC]=140 \implies \boxed{\textbf{(D)}\ \frac{140}{37}}</math> | ||
+ | |||
+ | ==Solution 3: Similar triangles without Pythagorean== | ||
+ | Or, use similar triangles all the way, dispense with Pythagorean, and go for minimal calculation: | ||
+ | |||
+ | Draw <math>BF \parallel ED</math> with <math>F</math> on <math>AE</math>. <math>BF=5\times\frac{4}{7}=\frac{20}{7}</math>. | ||
+ | |||
+ | <math>[\triangle ABF]=\frac{1}{2} \times 4 \times \frac{20}{7}=\frac{40}{7}</math>. | ||
+ | |||
+ | <math>AC:CB:CF=49:35:25</math>. (<math>7:5</math> ratio applied twice) | ||
+ | |||
+ | <math>[\triangle ABC]=\frac{49}{49+25}[\triangle ABF]=\boxed{\textbf{(D)}\ \frac{140}{37}}</math>. | ||
+ | |||
+ | ==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. | ||
+ | |||
+ | We can find the area of <math>\triangle ABC</math> by finding the the altitude from line <math>AB</math> to point <math>C</math>. Realize that this altitude is the <math>y</math> coordinate of point <math>C</math> on the coordinate plane, since the respective base of <math>\triangle ABC</math> is on the <math>x</math>-axis. | ||
+ | |||
+ | 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>, 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: | ||
+ | |||
+ | <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 line <math>AE</math> at the origin, so we will ignore this root. | ||
+ | |||
+ | Expanding the expression and factoring, we get: | ||
+ | |||
+ | <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>50y(37y-70) = 0</math> | ||
+ | |||
+ | Our non-zero root is thus <math>\frac{70}{37}</math>. Calculating the area of <math>\triangle ABC</math> with <math>4</math> as the length of <math>AB</math> and <math>\frac{70}{37}</math> as the altitude, we get: | ||
+ | |||
+ | <math>\frac{(4)(\frac{70}{37})}{2} = \boxed{\textbf{(D)}\ \frac{140}{37}}</math>. | ||
+ | |||
+ | -Solution by Joeya | ||
+ | |||
+ | == Video Solution by OmegaLearn (Similar Triangles) == | ||
+ | https://youtu.be/NsQbhYfGh1Q?t=512 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2017|ab=B|num-b=17|num-a=19}} | {{AMC12 box|year=2017|ab=B|num-b=17|num-a=19}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] |
Latest revision as of 04:01, 23 January 2023
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 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.