Difference between revisions of "2007 iTest Problems/Problem 15"

(Problem)
(Solution to Problem 15 — drawing took a while)
 
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== Problem ==
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==Problem==
  
Form a pentagon by taking a square of side length <math>1</math> and an equilateral triangle of side length <math>1</math> and placing the triangle so that one of its sides coincides with a side of the square. Then "circumscribe" a circle around the pentagon, passing through three of its vertices, so that the circle passes through exactly one vertex of the equilateral triangle, and exactly two vertices of the square. What is the radius of the circle?
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Form a pentagon by taking a square of side length 1 and an equilateral triangle of side length 1, and
 +
placing the triangle so that one of its sides coincides with a side of the square. Then "circumscribe" a circle around the
 +
pentagon, passing through three of its vertices, so that the circle passes through exactly one of the vertices of the
 +
equilateral triangle, and through exactly two vertices of the square. What is the radius of the circle?
  
<math>\textbf{(A) }\dfrac23\qquad
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<math>\text{(A) }\frac{2}{3}\qquad
\textbf{(B) }\dfrac34\qquad
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\text{(B) }\frac{3}{4}\qquad
\textbf{(C) }1\qquad
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\text{(C) }1\qquad
\textbf{(D) }\dfrac54\qquad
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\text{(D) }\frac{5}{4}\qquad
\textbf{(E) }\dfrac43\qquad </math>
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\text{(E) }\frac{4}{3}\qquad
<math>\textbf{(F) }\dfrac{\sqrt2}2\qquad
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\text{(F) }\frac{\sqrt{2}}{2}\qquad
\textbf{(G) }\dfrac{\sqrt3}2\qquad
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\text{(G) }\frac{\sqrt{3}}{2}\qquad
\textbf{(H) }\sqrt2\qquad
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\text{(H) }\sqrt{2}\qquad</math>
\textbf{(I) }\sqrt3\qquad
 
\textbf{(J) }\dfrac{1+\sqrt3}2\qquad </math>
 
<math>\textbf{(K) }\dfrac{2+\sqrt6}2\qquad
 
\textbf{(L) }\dfrac76\qquad
 
\textbf{(M) }\dfrac{2+\sqrt6}4\qquad
 
\textbf{(N) }\dfrac45\qquad
 
\textbf{(O) }2007\qquad </math>
 
  
== Solution ==
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<math>\text{(I) }\sqrt{3}\qquad
 +
\text{(J) }\frac{1+\sqrt{3}}{2}\qquad
 +
\text{(K) }\frac{2+\sqrt{6}}{2}\qquad
 +
\text{(L) }\frac{7}{6}\qquad
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\text{(M) }\frac{2+\sqrt{6}}{4}\qquad
 +
\text{(N) }\frac{4}{5}\qquad
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\text{(O) }2007\qquad</math>
 +
 
 +
==Solutions==
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 +
===Solution 1===
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 +
<asy>
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draw(circle((0,0),100));
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pair O=(0,0), A=(0,100), B=(-50,13.397), C=(50,13.397), D=(50,-86.603), E=(-50,-86.603);
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draw(B--E--D--C--A--B);
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dot(A);
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label("$A$",A,N);
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dot(B);
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label("$B$",B,NW);
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dot(C);
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label("$C$",C,NE);
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dot(E);
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label("$E$",E,SW);
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dot(D);
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label("$D$",D,SE);
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dot((0,0));
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label("$O$",(0,-5),S);
 +
 
 +
draw(O--A);
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draw(O--B);
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draw(O--C);
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draw(O--D);
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draw(O--E);
 +
 
 +
</asy>
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Let <math>A,B,C</math> be points of the equilateral triangle, <math>B,C,D,E</math> be points of the square, and <math>O</math> be the center of the circle.
 +
 
 +
<br>
 +
By the Base Angle Theorem, <math>\angle OED = \angle ODE</math>, so <math>\angle BEO = \angle CDO</math>.  Since <math>OD = OE</math>, by SAS Congruency, <math>\triangle BEO \cong \triangle CDO</math>, so <math>OB = OC</math>.
 +
 
 +
<br>
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By SSS Congruency, <math>\triangle ABO \cong \triangle ACO</math>, so <math>\angle BAO = \angle CAO = 30^\circ</math>.
 +
 
 +
<br>
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Finally, by SSS Congruency, <math>\triangle ABO \cong \triangle EBO</math>, so <math>\angle BEO = 30^\circ</math>.  Thus, <math>\angle OED = 60^\circ</math>, making <math>\triangle OED</math> equilateral, so the radius of the circle is <math>\boxed{\textbf{(C) }1}</math>.
 +
 
 +
===Solution 2===
 +
 
 +
<asy>
 +
draw(circle((0,0),100));
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draw((-50,-86.603)--(50,-86.603)--(50,13.397)--(50,13.397)--(0,100)--(-50,13.397)--(-50,-86.603));
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draw((-50,-86.603)--(0,100)--(50,-86.603));
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</asy>
 +
 
 +
Draw lines connecting the top vertex of the equilateral triangle to the bottom two vertices of the square (as seen in the diagram).  This means that the circle is the [[circumcircle]] of the triangle shown above.
 +
 
 +
<br>
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From the [[Law of Cosines]], the distance of each of the remaining sides is <math>\sqrt{1 + 1 - 2\cdot \cos(150^\circ)} = \sqrt{2 + \sqrt{3}}</math>.
 +
 
 +
<br>
 +
Now we need to find the length of the circumradius.  It can be done in a number of ways, but we’re going to be using the area formula <math>A = \tfrac{abc}{4R}</math>.
 +
 
 +
<br>
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The area of the triangle is <math>(1 + \tfrac{\sqrt{3}}{4}) - (2 \cdot \tfrac 12 \cdot \sin{150^\circ}) = \tfrac{2 + \sqrt{3}}{4}</math>.  The product of the sides of the triangle is <math>1 \cdot (\sqrt{2 + \sqrt{3}})^2 = 2 + \sqrt{3}</math>.  Plugging the values in the area formula yields
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<cmath>\frac{2 + \sqrt{3}}{4} = \frac{2 + \sqrt{3}}{4R}</cmath>
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<cmath>R = 1</cmath>
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The length of the radius of the circle is <math>\boxed{\textbf{(C) }1}</math>.
 +
 
 +
==See Also==
 +
{{iTest box|year=2007|num-b=14|num-a=16}}
 +
 
 +
[[Category:Intermediate Geometry Problems]]

Latest revision as of 21:01, 29 July 2018

Problem

Form a pentagon by taking a square of side length 1 and an equilateral triangle of side length 1, and placing the triangle so that one of its sides coincides with a side of the square. Then "circumscribe" a circle around the pentagon, passing through three of its vertices, so that the circle passes through exactly one of the vertices of the equilateral triangle, and through exactly two vertices of the square. What is the radius of the circle?

$\text{(A) }\frac{2}{3}\qquad \text{(B) }\frac{3}{4}\qquad \text{(C) }1\qquad \text{(D) }\frac{5}{4}\qquad \text{(E) }\frac{4}{3}\qquad \text{(F) }\frac{\sqrt{2}}{2}\qquad \text{(G) }\frac{\sqrt{3}}{2}\qquad \text{(H) }\sqrt{2}\qquad$

$\text{(I) }\sqrt{3}\qquad \text{(J) }\frac{1+\sqrt{3}}{2}\qquad \text{(K) }\frac{2+\sqrt{6}}{2}\qquad \text{(L) }\frac{7}{6}\qquad \text{(M) }\frac{2+\sqrt{6}}{4}\qquad \text{(N) }\frac{4}{5}\qquad \text{(O) }2007\qquad$

Solutions

Solution 1

[asy] draw(circle((0,0),100)); pair O=(0,0), A=(0,100), B=(-50,13.397), C=(50,13.397), D=(50,-86.603), E=(-50,-86.603); draw(B--E--D--C--A--B); dot(A); label("$A$",A,N); dot(B); label("$B$",B,NW); dot(C); label("$C$",C,NE); dot(E); label("$E$",E,SW); dot(D); label("$D$",D,SE); dot((0,0)); label("$O$",(0,-5),S);  draw(O--A); draw(O--B); draw(O--C); draw(O--D); draw(O--E);  [/asy]

Let $A,B,C$ be points of the equilateral triangle, $B,C,D,E$ be points of the square, and $O$ be the center of the circle.


By the Base Angle Theorem, $\angle OED = \angle ODE$, so $\angle BEO = \angle CDO$. Since $OD = OE$, by SAS Congruency, $\triangle BEO \cong \triangle CDO$, so $OB = OC$.


By SSS Congruency, $\triangle ABO \cong \triangle ACO$, so $\angle BAO = \angle CAO = 30^\circ$.


Finally, by SSS Congruency, $\triangle ABO \cong \triangle EBO$, so $\angle BEO = 30^\circ$. Thus, $\angle OED = 60^\circ$, making $\triangle OED$ equilateral, so the radius of the circle is $\boxed{\textbf{(C) }1}$.

Solution 2

[asy] draw(circle((0,0),100)); draw((-50,-86.603)--(50,-86.603)--(50,13.397)--(50,13.397)--(0,100)--(-50,13.397)--(-50,-86.603)); draw((-50,-86.603)--(0,100)--(50,-86.603)); [/asy]

Draw lines connecting the top vertex of the equilateral triangle to the bottom two vertices of the square (as seen in the diagram). This means that the circle is the circumcircle of the triangle shown above.


From the Law of Cosines, the distance of each of the remaining sides is $\sqrt{1 + 1 - 2\cdot \cos(150^\circ)} = \sqrt{2 + \sqrt{3}}$.


Now we need to find the length of the circumradius. It can be done in a number of ways, but we’re going to be using the area formula $A = \tfrac{abc}{4R}$.


The area of the triangle is $(1 + \tfrac{\sqrt{3}}{4}) - (2 \cdot \tfrac 12 \cdot \sin{150^\circ}) = \tfrac{2 + \sqrt{3}}{4}$. The product of the sides of the triangle is $1 \cdot (\sqrt{2 + \sqrt{3}})^2 = 2 + \sqrt{3}$. Plugging the values in the area formula yields \[\frac{2 + \sqrt{3}}{4} = \frac{2 + \sqrt{3}}{4R}\] \[R = 1\] The length of the radius of the circle is $\boxed{\textbf{(C) }1}$.

See Also

2007 iTest (Problems)
Preceded by:
Problem 14
Followed by:
Problem 16
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