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Difference between revisions of "2009 AMC 10B Problems/Problem 16"
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\text{(E) } \frac {\sqrt3}{2}
 
\text{(E) } \frac {\sqrt3}{2}
 
</math>
 
</math>
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[[Category: Introductory Geometry Problems]]
  
 
== Solution ==
 
== Solution ==
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</asy>
 
</asy>
  
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As <math>\triangle ABC</math> is equilateral, we have <math>\angle BAC = \angle BCA = 60^\circ</math>, hence <math>\angle OAC = \angle OCA = 30^\circ</math>. Then <math>\angla AOC = 120^\circ</math>, and from symmetry we have <math>\angle AOB = \angle COB = 60^\circ</math>. Finally this gives us <math>\angle ABO = \angle CBO = 30^\circ</math>.
+
As <math>\triangle ABC</math> is equilateral, we have <math>\angle BAC = \angle BCA = 60^\circ</math>, hence <math>\angle OAC = \angle OCA = 30^\circ</math>. Then <math>\angle AOC = 120^\circ</math>, and from symmetry we have <math>\angle AOB = \angle COB = 60^\circ</math>. Thus, this gives us <math>\angle ABO = \angle CBO = 30^\circ</math>.
  
−
We know that <math>DO = AO</math>, as <math>D</math> lies on the circle. From <math>\triangle ABO</math> we also have <math>AO = BO \sin 30^\circ = \frac{BO}2</math>, Hence <math>DO = \frac{BO}2</math>, therefore <math>BD = BO - DO = \frac{BO}2</math>, and <math>\frac{BD}{BO} = \boxed{\frac 12}</math>.
+
We know that <math>DO = AO</math>, as <math>D</math> lies on the circle. From <math>\triangle ABO</math> we also have <math>AO = BO \sin 30^\circ = \frac{BO}2</math>, Hence <math>DO = \frac{BO}2</math>, therefore <math>BD = BO - DO = \frac{BO}2</math>, and <math>\frac{BD}{BO} = \boxed{\frac 12 \Longrightarrow B}</math>.
  
 
=== Solution 2 ===
 
=== Solution 2 ===
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We then have <math>\angle SCD = \angle SAD = 30^\circ</math>, hence <math>D</math> is the incenter of <math>\triangle ABC</math>, and as <math>\triangle ABC</math> is equilateral, <math>D</math> is also its centroid. Hence <math>2 \cdot SD = BD</math>, and as <math>SD = SO</math>, we have <math>2\cdot SD = SD + SO = OD</math>, therefore <math>BD=OD</math>, and as before we conclude that <math>\frac{BD}{BO} = \boxed{\frac 12}</math>.
 
We then have <math>\angle SCD = \angle SAD = 30^\circ</math>, hence <math>D</math> is the incenter of <math>\triangle ABC</math>, and as <math>\triangle ABC</math> is equilateral, <math>D</math> is also its centroid. Hence <math>2 \cdot SD = BD</math>, and as <math>SD = SO</math>, we have <math>2\cdot SD = SD + SO = OD</math>, therefore <math>BD=OD</math>, and as before we conclude that <math>\frac{BD}{BO} = \boxed{\frac 12}</math>.
 +
 +
== Solution 3==
 +
<math>\triangle BAO\cong\triangle BCO</math> by SSS congruence, so <math>\angle ABO = \angle CBO = \frac{60}{2} = 30 ^\circ</math>. Since <math>BA</math> is tangent to the circle, it is perpendicular to <math>AO</math>. This means that <math>\triangle BAO</math> is a 30-60-90 triangle. The ratio of the side-lengths of a 30-60-90 triangle is <math>1:\sqrt3:2</math>, so <math>BO=2</math>. <math>BD = BO-DO</math>. Since <math>DO</math> is the radius of the circle, <math>DO=1</math> and <math>BD = 2-1=1</math>. Hence, <math>\frac{BD}{BO} = \frac{1}{2}</math>, and the answer is <math>\boxed{\textbf{(B) } \frac{1}{2}}</math> ~azc1027
  
 
== See Also ==
 
== See Also ==

Latest revision as of 18:59, 25 November 2023

Problem

Points $A$ and $C$ lie on a circle centered at $O$, each of $\overline{BA}$ and $\overline{BC}$ are tangent to the circle, and $\triangle ABC$ is equilateral. The circle intersects $\overline{BO}$ at $D$. What is $\frac{BD}{BO}$?

$\text{(A) } \frac {\sqrt2}{3} \qquad \text{(B) } \frac {1}{2} \qquad \text{(C) } \frac {\sqrt3}{3} \qquad \text{(D) } \frac {\sqrt2}{2} \qquad \text{(E) } \frac {\sqrt3}{2}$

Solution

Solution 1

[asy] unitsize(1.5cm); defaultpen(0.8); pair B=(0,0), A=(3,0), C=3*dir(60), O=intersectionpoint( C -- (C+3*dir(-30)), A -- (A+3*dir(90)) ); pair D=intersectionpoint(B--O, circle(O,length(A-O))); draw(circle(O,length(A-O))); draw(A--B--C--O--A); draw(B--O); draw(rightanglemark(B,A,O)); draw(rightanglemark(B,C,O)); draw(A--C, dashed ); label("$B$",B,SW); label("$A$",A,S); label("$C$",C,NW); label("$O$",O,NE); label("$D$",D,(S+SSW)); [/asy]

As $\triangle ABC$ is equilateral, we have $\angle BAC = \angle BCA = 60^\circ$, hence $\angle OAC = \angle OCA = 30^\circ$. Then $\angle AOC = 120^\circ$, and from symmetry we have $\angle AOB = \angle COB = 60^\circ$. Thus, this gives us $\angle ABO = \angle CBO = 30^\circ$.

We know that $DO = AO$, as $D$ lies on the circle. From $\triangle ABO$ we also have $AO = BO \sin 30^\circ = \frac{BO}2$, Hence $DO = \frac{BO}2$, therefore $BD = BO - DO = \frac{BO}2$, and $\frac{BD}{BO} = \boxed{\frac 12 \Longrightarrow B}$.

Solution 2

[asy] unitsize(1.5cm); defaultpen(0.8); pair B=(0,0), A=(3,0), C=3*dir(60), O=intersectionpoint( C -- (C+3*dir(-30)), A -- (A+3*dir(90)) ); pair D=intersectionpoint(B--O, circle(O,length(A-O))); draw(circle(O,length(A-O))); draw(A--B--C--O--A); draw(B--O); draw(rightanglemark(B,A,O)); draw(rightanglemark(B,C,O)); draw(A--C--D--A, dashed ); pair Sp = intersectionpoint(D--O,A--C); label("$B$",B,SW); label("$A$",A,S); label("$C$",C,NW); label("$O$",O,NE); label("$D$",D,(S+SSW)); label("$S$",Sp,(S+SSW)); [/asy]

As in the previous solution, we find out that $\angle AOB = \angle COB = 60^\circ$. Hence $\triangle AOD$ and $\triangle COD$ are both equilateral.

We then have $\angle SCD = \angle SAD = 30^\circ$, hence $D$ is the incenter of $\triangle ABC$, and as $\triangle ABC$ is equilateral, $D$ is also its centroid. Hence $2 \cdot SD = BD$, and as $SD = SO$, we have $2\cdot SD = SD + SO = OD$, therefore $BD=OD$, and as before we conclude that $\frac{BD}{BO} = \boxed{\frac 12}$.

Solution 3

$\triangle BAO\cong\triangle BCO$ by SSS congruence, so $\angle ABO = \angle CBO = \frac{60}{2} = 30 ^\circ$. Since $BA$ is tangent to the circle, it is perpendicular to $AO$. This means that $\triangle BAO$ is a 30-60-90 triangle. The ratio of the side-lengths of a 30-60-90 triangle is $1:\sqrt3:2$, so $BO=2$. $BD = BO-DO$. Since $DO$ is the radius of the circle, $DO=1$ and $BD = 2-1=1$. Hence, $\frac{BD}{BO} = \frac{1}{2}$, and the answer is $\boxed{\textbf{(B) } \frac{1}{2}}$ ~azc1027

See Also

2009 AMC 10B (Problems • Answer Key • Resources)
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Problem 15 		Followed by
Problem 17
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