Difference between revisions of "2005 AMC 10A Problems/Problem 23"
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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> | 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 | + | ==Solution 5== |
+ | <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 (Video) == | == Solution 5 (Video) == |
Revision as of 05:51, 16 July 2022
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 5
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 (Video)
Video solution: https://youtu.be/i6eooSSJF64
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.