Difference between revisions of "2005 AMC 12A Problems/Problem 15"
(→Solution 1) |
(Minor math mistake in the last solution. Said that the ratio of the area of triangle DEC to area of triangle DCX is 1/3, when it's the ratio of the area of triangle DEC to the area of the triangle ABD that's 1/3.) |
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(57 intermediate revisions by 11 users not shown) | |||
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===Solution 1=== | ===Solution 1=== | ||
− | Notice that the bases of both triangles are diameters of the circle. Hence the ratio of the areas is just the ratio of the heights of the triangles, or <math>\frac{ | + | Notice that the bases of both triangles are diameters of the circle. Hence the ratio of the areas is just the ratio of the heights of the triangles, or <math>\frac{CF}{CD}</math> (<math>F</math> is the foot of the [[perpendicular]] from <math>C</math> to <math>DE</math>). |
− | Call the radius <math>r</math>. Then <math>AC = \frac 13(2r) = \frac 23r</math>, <math>CO = \frac 13r</math>. Using the [[Pythagorean Theorem]] in <math>\triangle OCD</math>, we get <math>\frac{1}{ | + | Call the radius <math>r</math>. Then <math>AC = \frac 13(2r) = \frac 23r</math>, <math>CO = \frac 13r</math>. Using the [[Pythagorean Theorem]] in <math>\triangle OCD</math>, we get <math>\frac{1}{9}r^2 + CD^2 = r^2 \Longrightarrow CD = \frac{2\sqrt{2}}3r</math>. |
Now we have to find <math>CF</math>. Notice <math>\triangle OCD \sim \triangle OFC</math>, so we can write the [[proportion]]: | Now we have to find <math>CF</math>. Notice <math>\triangle OCD \sim \triangle OFC</math>, so we can write the [[proportion]]: | ||
Line 38: | Line 38: | ||
By the [[Pythagorean Theorem]] in <math>\triangle OFC</math>, we have <math>\left(\frac{1}{9}r\right)^2 + CF^2 = \left(\frac{1}{3}r\right)^2 \Longrightarrow CF = \sqrt{\frac{8}{81}r^2} = \frac{2\sqrt{2}}{9}r</math>. | By the [[Pythagorean Theorem]] in <math>\triangle OFC</math>, we have <math>\left(\frac{1}{9}r\right)^2 + CF^2 = \left(\frac{1}{3}r\right)^2 \Longrightarrow CF = \sqrt{\frac{8}{81}r^2} = \frac{2\sqrt{2}}{9}r</math>. | ||
− | Our answer is <math>\frac{ | + | Our answer is <math>\frac{CF}{CD} = \frac{\frac{2\sqrt{2}}{9}r}{\frac{2\sqrt{2}}{3}r} = \frac 13 \Longrightarrow \mathrm{(C)}</math>. |
===Solution 2=== | ===Solution 2=== | ||
− | Let the | + | Let the center of the circle be <math>O</math>. |
Note that <math>2 \cdot AC = BC \Rightarrow 3 \cdot AC = AB</math>. | Note that <math>2 \cdot AC = BC \Rightarrow 3 \cdot AC = AB</math>. | ||
Line 57: | Line 57: | ||
Then the area of <math>\triangle ABD</math> is <math>\frac{1}{2} AB \cdot CD = \frac{2\sqrt{2}}{3}r^2</math>. Similarly, the area of <math>\triangle DCE</math> is <math>\frac{1}{2}(r-AC) \cdot 2 \cdot CD = \frac{2\sqrt{2}}{9}r^2</math>, so the desired ratio is <math>\frac{\frac{2\sqrt{2}}{9}r^2}{\frac{2\sqrt{2}}{3}r^2} = \frac{1}{3} \Longrightarrow \mathrm{(C)}</math> | Then the area of <math>\triangle ABD</math> is <math>\frac{1}{2} AB \cdot CD = \frac{2\sqrt{2}}{3}r^2</math>. Similarly, the area of <math>\triangle DCE</math> is <math>\frac{1}{2}(r-AC) \cdot 2 \cdot CD = \frac{2\sqrt{2}}{9}r^2</math>, so the desired ratio is <math>\frac{\frac{2\sqrt{2}}{9}r^2}{\frac{2\sqrt{2}}{3}r^2} = \frac{1}{3} \Longrightarrow \mathrm{(C)}</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); | ||
+ | pair H=(e.x,0); | ||
+ | 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,NE); | ||
+ | label("$A$",A,W); | ||
+ | label("$D$",D,NNW); | ||
+ | label("$C$",C,SW); | ||
+ | label("$H$",H,SE); | ||
+ | draw(e--H,dashed); | ||
+ | label("O",(0,0),NE); | ||
+ | label("1",(C--O),N); | ||
+ | label("1",(H--O),N); | ||
+ | label("2",(A--C),N); | ||
+ | label("2",(H--B),N); | ||
+ | label("3",(O--D),NE); | ||
+ | label("3",(O--e),NE); | ||
+ | label("$2\sqrt{2}$",(D--C),W); | ||
+ | label("$2\sqrt{2}$",(H--e),E); | ||
+ | draw(rightanglemark(e,(e.x,0),A,2)); | ||
+ | draw(rightanglemark(D,C,B,2));</asy> | ||
+ | |||
+ | Let the center of the circle be <math>O</math>. | ||
+ | Without loss of generality, let the radius of the circle be equal to <math>3</math>. Thus, <math>AO=3</math> and <math>OB=3</math>. As a consequence of <math>2(AC)=BC</math>, <math>AC=2</math> and <math>CO=1</math>. Also, we know that <math>DO</math> and <math>OE</math> are both equal to <math>3</math> due to the fact that they are both radii. Thus from the Pythagorean Theorem, we have DC being equal to <math>\sqrt{3^2-1^2}</math> or <math>2\sqrt{2}</math>. Now we know that the area of <math>[ABD]</math> is equal to <math>\frac{(3+2+1)(2\sqrt{2})}{2}</math> or <math>6\sqrt{2}</math>. Know we need to find the area of <math>[DCE]</math>. By simple inspection <math>[COD]</math> <math>\cong</math> <math>[HOE]</math> due to angles being equal and CPCTC. Thus <math>HE=2\sqrt{2}</math> and <math>OH=1</math>. Know we know the area of <math>[CHE]=\frac{(1+1)(2\sqrt{2})}{2}</math> or <math>2\sqrt{2}</math>. We also know that the area of <math>[OHE]=\frac{(1)(2\sqrt{2})}{2}</math> or <math>\sqrt{2}</math>. Thus the area of <math>[COE]=2\sqrt{2}-\sqrt{2}</math> or <math>\sqrt{2}</math>. We also can calculate the area of <math>[DOC]</math> to be <math>\frac{(1)(2\sqrt{2})}{2}</math> or <math>\sqrt{2}</math>. Thus <math>[DCE]</math> is equal to <math>[COE]</math> + <math>[DOC] </math> or <math>\sqrt{2}+\sqrt{2}</math> or <math>2\sqrt{2}</math>. The ratio between <math>[DCE]</math> and <math>[ABD]</math> is equal to <math>\frac{2\sqrt{2}}{6\sqrt{2}}</math> or <math>\frac{1}{3}</math> <math>\Longrightarrow \mathrm{(C)}</math>. | ||
+ | ===Solution 5=== | ||
+ | We will use the shoelace formula. Our origin is the center of the circle. Denote the ordered pair for <math>D (x,y)</math>, and notice how <math>E</math> is a 180 degree rotation of <math>D</math>, using the rotation matrix formula we get <math>E = (-x,-y)</math>. WLOG say that this circle has radius <math>3</math>. We can now find points <math>A</math>, <math>B</math>, and <math>C</math> which are <math>(-3,0)</math>, <math>(3,0)</math>, and <math>(-1,0)</math> respectively. | ||
+ | By shoelace the area of <math>CED</math> is <math>Y</math>, and the area of <math>ADB</math> is <math>3Y</math>. Using division we get that the answer is <math> \mathrm{(C)}</math>. | ||
+ | |||
+ | ===Solution 6 (Mass Points)=== | ||
+ | <asy> | ||
+ | /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ | ||
+ | import graph; size(7cm); | ||
+ | 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 = -4.24313994860289, xmax = 6.360350402147026, ymin = -8.17642986522568, ymax = 4.1323989018072735; /* image dimensions */ | ||
+ | pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); | ||
+ | /* draw figures */ | ||
+ | draw(circle((0.9223776980185863,-1.084225871478444), 3.171161249925393), linewidth(2) + wrwrwr); | ||
+ | draw((-0.5623104956355617,1.717909856970905)--(-0.39884850438732933,-3.9670413347199647), linewidth(2) + wrwrwr); | ||
+ | draw((-2.2487343499798245,-1.1018908670262892)--(4.0935388680341624,-1.0849377800575102), linewidth(2) + wrwrwr); | ||
+ | draw((-0.4813673187299407,-1.0971666418223938)--(2.1729847859273126,-3.904641555130568), linewidth(2) + wrwrwr); | ||
+ | draw((-0.5623104956355617,1.717909856970905)--(2.1561002471522333,-3.887757016355488), linewidth(2) + wrwrwr); | ||
+ | draw((-2.2487343499798245,-1.1018908670262892)--(-0.5623104956355617,1.717909856970905), linewidth(2) + wrwrwr); | ||
+ | draw((-0.5623104956355617,1.717909856970905)--(4.0935388680341624,-1.0849377800575102), linewidth(2) + wrwrwr); | ||
+ | draw(circle((-2.858607769046372,-1.1524617347926092), 0.45463011998128017), linewidth(2) + wrwrwr); | ||
+ | draw(circle((4.790088296064635,-1.144019465405069), 0.45463011998128006), linewidth(2) + wrwrwr); | ||
+ | draw(circle((-0.9168858099122313,-1.6336710898823752), 0.45463011998127983), linewidth(2) + wrwrwr); | ||
+ | draw(circle((1.4976032349241348,-1.3128648531558642), 0.4546301199812797), linewidth(2) + wrwrwr); | ||
+ | draw(circle((-0.815578577261755,2.334195522261307), 0.4546301199812809), linewidth(2) + wrwrwr); | ||
+ | draw(circle((2.6119827940793803,-4.5546962979711285), 0.45463011998128033), linewidth(2) + wrwrwr); | ||
+ | /* dots and labels */ | ||
+ | dot((0.9223776980185863,-1.084225871478444),dotstyle); | ||
+ | dot((-0.5623104956355617,1.717909856970905),dotstyle); | ||
+ | label("$D$", (-0.49477234053524466,1.8867552447217006), NE * labelscalefactor); | ||
+ | dot((-0.39884850438732933,-3.9670413347199647),dotstyle); | ||
+ | dot((-2.2487343499798245,-1.1018908670262892),dotstyle); | ||
+ | label("A", (-2.183226218043193,-0.9329627307165757), NE * labelscalefactor); | ||
+ | dot((4.0935388680341624,-1.0849377800575102),dotstyle); | ||
+ | label("B", (4.165360361386693,-0.9160781919414961), NE * labelscalefactor); | ||
+ | dot((-0.4813673187299407,-1.0971666418223938),linewidth(4pt) + dotstyle); | ||
+ | label("C", (-0.41034964665984724,-0.9667318082667347), NE * labelscalefactor); | ||
+ | dot((2.1729847859273126,-3.904641555130568),dotstyle); | ||
+ | dot((2.1561002471522333,-3.887757016355488),dotstyle); | ||
+ | label("E", (2.2236384022525524,-3.7189116286046926), NE * labelscalefactor); | ||
+ | label("2", (-2.9261459241466903,-1.2031153511178476), NE * labelscalefactor,wrwrwr); | ||
+ | label("1", (4.722550140964316,-1.186230812342768), NE * labelscalefactor,wrwrwr); | ||
+ | label("3", (-0.9844239650125497,-1.6758824368200735), NE * labelscalefactor,wrwrwr); | ||
+ | label("4", (1.4300650798238166,-1.355076200093563), NE * labelscalefactor,wrwrwr); | ||
+ | label("2", (-0.8831167323620728,2.2919841753236083), NE * labelscalefactor,wrwrwr); | ||
+ | label("2", (2.5444446389790625,-4.5969076449088275), NE * labelscalefactor,wrwrwr); | ||
+ | clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); | ||
+ | /* end of picture */ | ||
+ | </asy> | ||
+ | |||
+ | We set point <math>A</math> as a mass of 2. This means that point <math>B</math> has a mass of <math>1</math> since <math>2\times{AC} = 1\times{BC}</math>. This implies that point <math>C</math> has a mass of <math>2+1 = 3</math> and the center of the circle has a mass of <math>3+1 = 4</math>. After this, we notice that points <math>D</math> and <math>E</math> both must have a mass of <math>2</math> since <math>2+2 = 4</math> and they are both radii of the circle. | ||
+ | |||
+ | To find the ratio of the areas, we do the reciprocal of the multiplication of the mass points of DCE and the multiplication of ABD divided by each other. Which is simply <math>\frac{3\times2\times2}{2\times2\times1}</math> which is <math>\boxed{\frac{1}{3}}</math> (the reciprocal of 3) | ||
+ | |||
+ | -Brudder | ||
+ | |||
+ | ===Solution 7 (Slight Trigonometry)=== | ||
+ | <asy> | ||
+ | unitsize(3.5cm); | ||
+ | defaultpen(fontsize(12pt)+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,NE); | ||
+ | label("$A$",A,W); | ||
+ | label("$D$",D,NNW); | ||
+ | label("$C$",C,SW); | ||
+ | label("O",(0,0),NE); | ||
+ | label("x",(C--O),N); | ||
+ | label("3x",(B--O),N); | ||
+ | label("2x",(A--C),N); | ||
+ | label("3x",(O--D),NE); | ||
+ | label("3x",(O--E),NE); | ||
+ | label("$2x\sqrt{2}$",(D--C),W); | ||
+ | draw(rightanglemark(D,C,B,2));</asy> | ||
+ | |||
+ | Let the center of the circle be <math>O</math>. | ||
+ | The area of <math>ADB</math> is <math>\frac{1}{2} \cdot 6x \cdot 2x\sqrt{2}</math> = <math>6x^2\sqrt{2}</math>. The area of <math>DCE</math> is <math>\frac{1}{2} \cdot DC \cdot DE \cdot</math> sin <math>CDE</math>. We find sin <math>CDE</math> is <math>\frac{OC}{OD}</math> = <math>\frac{1}{3}</math>. Substituting <math>DC = 2x\sqrt{2}</math> and <math>DE = 6x</math>, we get <math>[DCE]</math> = <math>\frac{1}{2} \cdot 2x\sqrt{2} \cdot 6x \cdot \frac{1}{3}</math> = <math>2x^2\sqrt{2}</math>. Hence, the ratio between the areas of <math>DCE</math> and <math>ADB</math> is equal to <math>\frac{2x^2\sqrt{2}}{6x^2\sqrt{2}}</math> or <math>\frac{1}{3}</math> = <math>\boxed{(C)}</math>. | ||
+ | |||
+ | ~Math_Genius_164 | ||
+ | |||
+ | ===Solution 8=== | ||
+ | |||
+ | In my opinion, the solution below is the easiest and quickest. | ||
+ | |||
+ | Since both <math>DE</math> and <math>AB</math> are diameters, they intersect at the center of the circle. Call this center <math>O</math>. WLOG, let <math>AC=2, CO=1,OB=3</math>. Call the point where the extension of <math>DC</math> hits the circle <math>F</math>. Notice that <math>\angle DCB = 90^\circ</math>. This implies that <math>DC=CF</math>. WOLG, let <math>DC=CF=1</math>. Then, <math>[DCE]=\frac 12 bh = \frac 12 * DF*CO = \frac 12 *2*1=1</math> and <math>[ABD]=\frac 12 bh = \frac 12 *AB*CD = \frac 12 *6*1 = 3</math>. Thus, the answer is <math>\frac{1}{3}</math> = <math>\boxed{(C)}</math>. | ||
+ | |||
+ | Solution by franzliszt | ||
+ | ===Solution 9 (Slick construction)=== | ||
+ | <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), X = (1/3.0); | ||
+ | pair D=dir(aCos(C.x)), E=(-D.x,-D.y); | ||
+ | draw(A--B--D--cycle); | ||
+ | draw(D--E--C); | ||
+ | draw(D--X); | ||
+ | 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); | ||
+ | label("$X$",X,NE); | ||
+ | draw(rightanglemark(D,C,B,2));</asy> | ||
+ | Let <math>X</math> be the reflection of <math>C</math> over the center <math>O.</math> Since <math>\triangle DXO\cong\triangle ECO</math> by SAS, it follows that the area of <math>\triangle DEC</math> is equal to the area of <math>\triangle DCX.</math> However, we know that | ||
+ | <cmath>BC = 2AC\implies CO=\frac{1}{2}AC\implies CX=AC,</cmath> | ||
+ | so the ratio of the area of <math>\triangle DEC</math> to the area of <math>\triangle ABD</math> is <math>\frac{\frac{AC\cdot DC}{2}}{\frac{3AC\cdot DC}{2}}=\frac{1}{3}\implies \boxed{(C)}</math> | ||
== See also == | == See also == |
Latest revision as of 17:02, 18 August 2023
Contents
Problem
Let be a diameter of a circle and
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
Solution 1
Notice that the bases of both triangles are diameters of the circle. Hence the ratio of the areas is just the ratio of the heights of the triangles, or (
is the foot of the perpendicular from
to
).
Call the radius . Then
,
. Using the Pythagorean Theorem in
, we get
.
Now we have to find . Notice
, so we can write the proportion:
![$\frac{OF}{OC} = \frac{OC}{OD}$](http://latex.artofproblemsolving.com/0/2/2/02291691270e57eea057ea561416318b8de9fc01.png)
![$\frac{OF}{\frac{1}{3}r} = \frac{\frac{1}{3}r}{r}$](http://latex.artofproblemsolving.com/6/e/1/6e1dcd67ba34e71f9f880510df1fd482941f6e37.png)
![$OF = \frac 19r$](http://latex.artofproblemsolving.com/5/a/4/5a4dcf36a069d28f1265cf8c1b825c9ae46c0cbd.png)
By the Pythagorean Theorem in , we have
.
Our answer is .
Solution 2
Let the center of the circle be .
Note that .
is midpoint of
.
is midpoint of
Area of
Area of
Area of
Area of
.
Solution 3
Let be the radius of the circle. Note that
so
.
By Power of a Point Theorem, , and thus
Then the area of is
. Similarly, the area of
is
, so the desired ratio is
Solution 4
Let the center of the circle be .
Without loss of generality, let the radius of the circle be equal to
. Thus,
and
. As a consequence of
,
and
. Also, we know that
and
are both equal to
due to the fact that they are both radii. Thus from the Pythagorean Theorem, we have DC being equal to
or
. Now we know that the area of
is equal to
or
. Know we need to find the area of
. By simple inspection
due to angles being equal and CPCTC. Thus
and
. Know we know the area of
or
. We also know that the area of
or
. Thus the area of
or
. We also can calculate the area of
to be
or
. Thus
is equal to
+
or
or
. The ratio between
and
is equal to
or
.
Solution 5
We will use the shoelace formula. Our origin is the center of the circle. Denote the ordered pair for , and notice how
is a 180 degree rotation of
, using the rotation matrix formula we get
. WLOG say that this circle has radius
. We can now find points
,
, and
which are
,
, and
respectively.
By shoelace the area of
is
, and the area of
is
. Using division we get that the answer is
.
Solution 6 (Mass Points)
We set point as a mass of 2. This means that point
has a mass of
since
. This implies that point
has a mass of
and the center of the circle has a mass of
. After this, we notice that points
and
both must have a mass of
since
and they are both radii of the circle.
To find the ratio of the areas, we do the reciprocal of the multiplication of the mass points of DCE and the multiplication of ABD divided by each other. Which is simply which is
(the reciprocal of 3)
-Brudder
Solution 7 (Slight Trigonometry)
Let the center of the circle be .
The area of
is
=
. The area of
is
sin
. We find sin
is
=
. Substituting
and
, we get
=
=
. Hence, the ratio between the areas of
and
is equal to
or
=
.
~Math_Genius_164
Solution 8
In my opinion, the solution below is the easiest and quickest.
Since both and
are diameters, they intersect at the center of the circle. Call this center
. WLOG, let
. Call the point where the extension of
hits the circle
. Notice that
. This implies that
. WOLG, let
. Then,
and
. Thus, the answer is
=
.
Solution by franzliszt
Solution 9 (Slick construction)
Let
be the reflection of
over the center
Since
by SAS, it follows that the area of
is equal to the area of
However, we know that
so the ratio of the area of
to the area of
is
See also
2005 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 14 |
Followed by Problem 16 |
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.