Difference between revisions of "2005 AMC 10A Problems/Problem 23"
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Let <math>AB</math> be a diameter of a circle and let <math>C</math> be a point on <math>AB</math> with <math>2\cdot AC=BC</math>. Let <math>D</math> and <math>E</math> be points on the circle such that <math>DC \perp AB</math> and <math>DE</math> is a second diameter. What is the ratio of the area of <math>\triangle DCE</math> to the area of <math>\triangle ABD</math>? | Let <math>AB</math> be a diameter of a circle and let <math>C</math> be a point on <math>AB</math> with <math>2\cdot AC=BC</math>. Let <math>D</math> and <math>E</math> be points on the circle such that <math>DC \perp AB</math> and <math>DE</math> is a second diameter. What is the ratio of the area of <math>\triangle DCE</math> to the area of <math>\triangle ABD</math>? | ||
− | <math> \ | + | <asy> |
+ | unitsize(2.5cm); | ||
+ | defaultpen(fontsize(10pt)+linewidth(.8pt)); | ||
+ | dotfactor=3; | ||
+ | pair O=(0,0), C=(-1/3.0), B=(1,0), A=(-1,0); | ||
+ | pair D=dir(aCos(C.x)), E=(-D.x,-D.y); | ||
+ | draw(A--B--D--cycle); | ||
+ | draw(D--E--C); | ||
+ | draw(unitcircle,white); | ||
+ | drawline(D,C); | ||
+ | dot(O); | ||
+ | clip(unitcircle); | ||
+ | draw(unitcircle); | ||
+ | label("$E$",E,SSE); | ||
+ | label("$B$",B,E); | ||
+ | label("$A$",A,W); | ||
+ | label("$D$",D,NNW); | ||
+ | label("$C$",C,SW); | ||
+ | draw(rightanglemark(D,C,B,2));</asy> | ||
+ | |||
+ | <math> \textbf{(A) } \frac{1}{6}\qquad \textbf{(B) } \frac{1}{4}\qquad \textbf{(C) } \frac{1}{3}\qquad \textbf{(D) } \frac{1}{2}\qquad \textbf{(E) } \frac{2}{3} </math> | ||
==Solution 1== | ==Solution 1== | ||
[[File:Circlenc1.png]] | [[File:Circlenc1.png]] | ||
− | + | WLOG, Let us assume that the diameter is of length <math>1</math>. | |
− | <math> | + | The length of <math>AC</math> is <math>\frac{1}{3}</math> and <math>CO</math> is <math>\frac{1}{2}-\frac{1}{3} = \frac{1}{6}</math>. |
− | + | <math>OD</math> is the radius of the circle, which is <math>\frac{1}{2}</math>, so using the [[Pythagorean Theorem]] the height <math>CD</math> of <math>\triangle DCO</math> is <math>\sqrt{\left(\frac{1}{2}\right)^2-\left(\frac{1}{6}\right)^2} = \frac{\sqrt{2}}{3}</math>. This is also the height of the <math>\triangle ABD</math>. | |
− | The | + | The area of <math>\triangle DCO</math> is <math>\frac{1}{2}\cdot\frac{1}{6}\cdot\frac{\sqrt{2}}{3}</math> = <math>\frac{\sqrt{2}}{36}</math>. |
− | + | The height of <math>\triangle DCE</math> can be found using the area of <math>\triangle DCO</math> and <math>DO</math> as base. | |
− | The diameter is the base for both the triangles <math>\triangle DCE</math> and <math>\triangle ABD</math> | + | Hence, the height of <math>\triangle DCE</math> is <math>\dfrac{\dfrac{\sqrt{2}}{36}}{\dfrac{1}{2}\cdot\dfrac{1}{2}}</math> = <math>\dfrac{\sqrt{2}}{9}</math>. |
+ | |||
+ | The diameter is the base for both the triangles <math>\triangle DCE</math> and <math>\triangle ABD</math>, | ||
Hence, the ratio of the area of <math>\triangle DCE</math> to the area of <math>\triangle ABD</math> is | Hence, the ratio of the area of <math>\triangle DCE</math> to the area of <math>\triangle ABD</math> is | ||
− | <math>\ | + | <math>\dfrac{\dfrac{\sqrt{2}}{9}}{\dfrac{\sqrt{2}}{3}}</math> = <math>\boxed{\textbf{(C) }\frac{1}{3}}</math> |
==Solution 2== | ==Solution 2== | ||
Line 39: | Line 61: | ||
<math>OD=r, OC=\frac{1}{3}r</math>. | <math>OD=r, OC=\frac{1}{3}r</math>. | ||
− | Since <math>m\angle DCO=m\angle DFC=90^\circ</math>, then <math>\triangle DCO\ | + | Since <math>m\angle DCO=m\angle DFC=90^\circ</math>, then <math>\triangle DCO\sim \triangle DFC</math>. So the ratio of the two altitudes is <math>\frac{CF}{DC}=\frac{OC}{DO}=\boxed{\textbf{(C) }\frac{1}{3}}</math> |
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | Say the center of the circle is point <math>O</math>; | ||
+ | Without loss of generality, assume <math>AC=2</math>, so <math>CB=4</math> and the diameter and radius are <math>6</math> and <math>3</math>, respectively. Therefore, <math>CO=1</math>, and <math>DO=3</math>. | ||
+ | The area of <math>\triangle DCE</math> can be expressed as <math>\frac{1}{2}(CD)(6)\text{sin }(CDE).</math> <math>\frac{1}{2}(CD)(6)</math> happens to be the area of <math>\triangle ABD</math>. Furthermore, <math>\text{sin } CDE = \frac{CO}{DO},</math> or <math>\frac{1}{3}.</math> Therefore, the ratio is <math>\boxed{\textbf{(C) }\frac{1}{3}}.</math> | ||
+ | |||
+ | ==Solution 4== | ||
+ | <asy> | ||
+ | unitsize(2.5cm); | ||
+ | defaultpen(fontsize(10pt)+linewidth(.8pt)); | ||
+ | dotfactor=3; | ||
+ | pair O=(0,0), C=(-1/3.0), B=(1,0), A=(-1,0); | ||
+ | pair D=dir(aCos(C.x)), E=(-D.x,-D.y); | ||
+ | draw(A--B--D--cycle); | ||
+ | draw(D--E--C); | ||
+ | draw(unitcircle,white); | ||
+ | drawline(D,C); | ||
+ | dot(O); | ||
+ | clip(unitcircle); | ||
+ | draw(unitcircle); | ||
+ | label("$E$",E,SSE); | ||
+ | label("$B$",B,E); | ||
+ | label("$A$",A,W); | ||
+ | label("$D$",D,NNW); | ||
+ | label("$C$",C,SW); | ||
+ | draw(rightanglemark(D,C,B,2)); | ||
+ | </asy> | ||
+ | |||
+ | |||
+ | |||
+ | Let the point G be the reflection of point <math>D</math> across <math>\overline{AB}</math>. (Point G is on the circle). | ||
+ | |||
+ | |||
+ | Let <math>AC=x</math>, then <math>BC=2x</math>. The diameter is <math>3x</math>. To find <math>DC</math>, there are two ways (presented here): | ||
+ | |||
+ | 1. Since <math>\overline{AB}</math> is the diameter, <math>CD=CG</math>. Using power of points, | ||
+ | <cmath>AC\cdot BC=x\cdot2x=2x^{2}=CD^{2} \longrightarrow CD=x\sqrt{2}</cmath> | ||
+ | 2. Use the geometric mean theorem, | ||
+ | <cmath>AC\cdot BC=x\cdot2x=2x^{2}=CD^{2} \longrightarrow CD=x\sqrt{2}</cmath> | ||
+ | (These are the same equations but obtained through different formulae) | ||
+ | |||
+ | |||
+ | Therefore <math>DG=2x\sqrt{2}</math>. Since <math>\overline{DE}</math> is a diameter, <math>\triangle DGE</math> is right. By the Pythagorean theorem, | ||
+ | <cmath>DE^{2}=GD^{2}+GE^{2} \longrightarrow \left(3x\right)^{2}=\left(2x\sqrt{2}\right)^{2}+GE^{2}</cmath> | ||
+ | <cmath>9x^{2}=8x^{2}+GE^{2} \longrightarrow GE^{2}=x^{2} \longrightarrow GE=x</cmath> | ||
+ | |||
+ | |||
+ | As established before, <math>\angle DGE</math> is right (if you are unsure, look up "inscribed angle theorem", this is a special case of the theorem where the central angle measures <math>180^{\circ}</math>) so <math>GE=x</math> is the altitude of <math>\triangle DCE</math>, and <math>DC=x\sqrt{2}</math> is the base. Therefore | ||
+ | <cmath>\left[DCE\right]=\frac{1}{2}\cdot DC\cdot GE=\frac{1}{2}\cdot x\sqrt{2}\cdot x=\frac{x^{2}\sqrt{2}}{2}</cmath> | ||
+ | |||
+ | |||
+ | <math>AB=3x</math> is the base of <math>\triangle ABD</math> and <math>CD=x\sqrt{2}</math> is the height. | ||
+ | <cmath>\left[ABD\right]=\frac{1}{2}\cdot3x\cdot x\sqrt{2}=\frac{3x^{2}\sqrt{2}}{2}</cmath> | ||
+ | |||
+ | |||
+ | The required ratio is | ||
+ | <cmath>\frac{\left[DCE\right]}{\left[ABD\right]}=\frac{\frac{x^{2}\sqrt{2}}{2}}{\frac{3x^{2}\sqrt{2}}{2}}=\frac{x^{2}\sqrt{2}}{2}\cdot\frac{2}{3x^{2}\sqrt{2}}=\frac{x^{2}\sqrt{2}}{3x^{2}\sqrt{2}}=\frac{1}{3}</cmath> | ||
+ | The answer is <math>\boxed{\textbf{(C) } \frac{1}{3}}</math>. | ||
+ | |||
+ | |||
+ | |||
+ | ~JH. L | ||
+ | |||
+ | == Solution 5 == | ||
+ | |||
+ | Assume the diameter is <math>3</math>. | ||
+ | |||
+ | <math>AC = 1</math> | ||
+ | |||
+ | Get the height <math>CD = \sqrt{(AC)(BC)} = \sqrt2</math> via power of a point. | ||
+ | |||
+ | <math>CO = AO - AC = 1/2</math>. | ||
+ | |||
+ | By altitude of right triangle <math>\triangle CDO</math>: Altitude from <math>C</math> to <math>DE</math> is same as altitude from <math>C</math> to <math>DO</math> is <math>\frac{(CO)(CD)}{DO} = \frac{(1/2)(\sqrt2)}{\frac 3 2}</math>. | ||
+ | |||
+ | <math>\triangle DCE</math> and <math>\triangle ABD</math> have the same (diameter) hypotenuse length, so their area ratio is their altitude ratio is | ||
+ | <math>\frac {\frac{ (1/2)(\sqrt2) } {\frac 3 2}} {\sqrt2} = \boxed{1/3}</math>. | ||
+ | |||
+ | ~oinava | ||
+ | |||
+ | == Video solution == | ||
+ | https://youtu.be/i6eooSSJF64 | ||
− | ==See | + | ==See Also== |
{{AMC10 box|year=2005|ab=A|num-b=22|num-a=24}} | {{AMC10 box|year=2005|ab=A|num-b=22|num-a=24}} | ||
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
+ | [[Category:Area Ratio Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 20:57, 9 July 2024
Contents
Problem
Let be a diameter of a circle and let
be a point on
with
. Let
and
be points on the circle such that
and
is a second diameter. What is the ratio of the area of
to the area of
?
Solution 1
WLOG, Let us assume that the diameter is of length .
The length of is
and
is
.
is the radius of the circle, which is
, so using the Pythagorean Theorem the height
of
is
. This is also the height of the
.
The area of is
=
.
The height of can be found using the area of
and
as base.
Hence, the height of is
=
.
The diameter is the base for both the triangles and
,
Hence, the ratio of the area of to the area of
is
=
Solution 2
Since and
share a base, the ratio of their areas is the ratio of their altitudes. Draw the altitude from
to
.
.
Since , then
. So the ratio of the two altitudes is
Solution 3
Say the center of the circle is point ;
Without loss of generality, assume
, so
and the diameter and radius are
and
, respectively. Therefore,
, and
.
The area of
can be expressed as
happens to be the area of
. Furthermore,
or
Therefore, the ratio is
Solution 4
Let the point G be the reflection of point across
. (Point G is on the circle).
Let , then
. The diameter is
. To find
, there are two ways (presented here):
1. Since is the diameter,
. Using power of points,
2. Use the geometric mean theorem,
(These are the same equations but obtained through different formulae)
Therefore . Since
is a diameter,
is right. By the Pythagorean theorem,
As established before, is right (if you are unsure, look up "inscribed angle theorem", this is a special case of the theorem where the central angle measures
) so
is the altitude of
, and
is the base. Therefore
is the base of
and
is the height.
The required ratio is
The answer is
.
~JH. L
Solution 5
Assume the diameter is .
Get the height via power of a point.
.
By altitude of right triangle : Altitude from
to
is same as altitude from
to
is
.
and
have the same (diameter) hypotenuse length, so their area ratio is their altitude ratio is
.
~oinava
Video solution
See Also
2005 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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 |
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