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The question is pretty self-explanatory from this image.

Note: This image is not to scale. Please look at the lengths of the lines.

The question is rather self explanatory from this image and the title, but to reiterate: I would like to find the exact area of this circle in terms of $\pi$. I thought of maybe estimating the circumference, but that led me nowhere. This is from a problem set which I was given 15 minutes to do, so there should be some trick to finding the solution. Thank you in advance!

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4 Answers 4

up vote 4 down vote accepted

I get the same answer as Ross Millikan by a slightly different route.

Draw the six radii from the center of the circle to the vertices of the inscribed hexagon. You'll have three triangles with sides of length $r$, $r$, and $2$, and three triangles with sides of length $r$, $r$, and $1$. Let the angles of these two sets at the center be $\theta$ and $\phi$. Then by the law of cosines we have

$$\begin{align} 4&=2r^2(1-\cos\theta)\\ 1&=2r^2(1-\cos\phi)\\ \end{align}$$

with $3\theta+3\phi=2\pi$, hence

$$1-\cos\theta=4(1-\cos\phi)=4(1-\cos{2\pi\over3}\cos\theta-\sin{2\pi\over3}\sin\theta)$$

which simplifies to

$$2\sin\theta=\sqrt3(1+\cos\theta)$$

Squaring both sides and using $\sin^2\theta=1-\cos^2\theta$ leads to the quadratic

$$7\cos^2\theta+6\cos\theta-1=(7\cos\theta-1)(\cos\theta+1)=0$$

from which we can concluse that $\cos\theta=1/7$, which plugs in to give

$$4=2r^2\left(1-{1\over7}\right)={12\over7}r^2$$

which means the requested area of the circle is

$$A=\pi r^2={7\over3}\pi$$

Added later: The OP has asked if there is a simpler solution that doesn't use any trig. What I'm about to offer is not exactly simple, but it does avoid trig.

As above, drawing the six radii produces two sets of three identical triangles. If we rearrange these triangles so that the sides of length $1$ and $2$ alternate instead of being grouped together, the hexagon produced is still inscribed in a circle of the same radius. So let's look at that figure, and label the points $A$, $B$, $C$, $D$, $E$, and $F$, with $|AB|=|CD|=|EF|=1$ and $|BC|=|DE|=|FA|=2$.

The first thing to notice, if you draw the figure, is that $$|AC|=|CE|=|EA|=\sqrt3r$$

The second thing to notice is that, by symmetry, the angles at all the vertices of the hexagon are equal, and thus all equal $120^\circ$.

Now consider the triangle $\triangle ABC$. Extend the line $BC$ and drop a perpendicular to it from $A$. Let $P$ be the point at which the perpendicular meets the line. Note that $\angle ABC=120^\circ$ implies $\angle ABP=60^\circ$, hence $|AB|=1$ implies $|BP|={1\over2}$ and $|AP|={\sqrt3\over2}$. But we now have $|CP|=|BC|+|BP|=2+{1\over2}={5\over2}$ at which point the Pythagorean Theorem tells us

$$3r^2=|AC|^2=|AP|^2+|CP|^2={3\over4}+{25\over4}=7$$

The key idea here was the notion that you can rearrange the sides to alternate without changing the radius of the circle. Once you accept that, the rest really is fairly simple. But I rather doubt I would have come up with this as a seventh grader on a 15-minute problem set.

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Thank you both for your answers, but are you sure there's not something a bit more clever? We weren't even given calculators for this set, and a very small amount of time. Also, I am in seventh grade, so I doubted that the answer would even use trig. Don't get me wrong, I appreciate that you solved it and understand your answer, I'm just wondering if there's some easy trick that I overlooked and can be employed on a problem like this. –  recursive recursion Feb 12 at 22:04
    
I think this is a cleaner approach than mine. –  Ross Millikan Feb 12 at 22:11
    
@recursiverecursion, I'll try giving it more thought. In the meantime, you might edit your question with an update acknowledging the two trig-based answers but emphasizing that you'd like to know if there is a truly simple approach. (I might note, it took me more than 15 minutes to get the correct solution. I'm impressed you're doing this as a seventh grader.) –  Barry Cipra Feb 12 at 22:25
    
I think this is simpler. If we arrange the triangles in this alternating manner and call the vertexes of the hexagon $A,B,\ldots,F$ then $A,C,E$ is an equilateral triangle with $AC=\sqrt{3}r$. $M$ is the center of the circle and $\angle(A,M,C)=\frac{2 \pi}{3}$ and so $\angle(A,B,C)=\frac{2 \pi}{3}$. Using the law of cosine we get ${AC}^2=1^2+2^2-2 \cdot 1 \cdot 2 \cdot \cos{\frac{2 \pi}{3}}=7$. So $3r^2=7$ –  miracle173 Feb 13 at 0:56
    
@miracle173, nice observation. But the OP wanted to avoid using trig, so I had taken the law of cosines off the table. –  Barry Cipra Feb 13 at 1:02

Let $\alpha$ (respectively, $\beta$) be the measure of any inscribed angle subtending a chord of length $1$ (respectively, $2$). Let $\overline{PQ}$ be the "middle" chord of length $1$, and $\overline{RS}$ the "middle" chord of length $2$, as shown. Let $X$ be the intersection of $\overline{PS}$ and $\overline{QR}$, and let $\gamma$ be the measure of angle $\angle RXS$ (and $\angle PXQ$). Let $O$ (not shown) be the center of the circle.

enter image description here

By the Inscribed Angle Theorem, $$\alpha = \frac{1}{2}\angle POQ \qquad \text{and} \qquad \beta = \frac{1}{2}\angle ROS$$ By a related theorem, the measure of an angle formed by two chords is the average of the measure of opposing subtended arcs: $$\gamma = \frac{1}{2}\left( \; \angle POQ + \angle ROS \; \right) = \alpha + \beta$$

Consequently, $\triangle XRS$ (also $\triangle POQ$) is equiangular, hence equilateral.

Now, in $\triangle PQS$, we have $$|\overline{PQ}| = 1 \qquad |\overline{PS}| = |\overline{PX}|+|\overline{XS}| = 1 + 2 = 3 \qquad \angle QPS = \frac{\pi}{3}$$ so that, by the Law of Cosines, $$|\overline{QS}|^2 = 1^2 + 3^2 - 2\cdot 1 \cdot 3 \cos\frac{\pi}{3} = 10 - 3 = 7$$ and by the Law of Sines, $$ d = \frac{|\overline{QS}|}{\sin\angle QPS} = \frac{\sqrt{7}}{\sqrt{3}/2} = \frac{2\sqrt{7}}{\sqrt{3}}$$ where $d$ is the diameter of the circle. Thus, the radius of the circle is $\sqrt{\frac{7}{3}}$, and the area $\frac{7}{3}\pi$. $\square$

(I believe that there are quicker ways to get from the equilateral triangles to the circle's area.)

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Let the circle have radius $r$. If you draw the radii to each point on the circle and through the midpoints of each side you get six $1,\sqrt{r^2-1},r$ right triangles and six $\frac 12, \sqrt{r^2-\frac 14},r$ right triangles. Using the angles at the center gives $$6\arcsin \frac 1r + 6 \arcsin \frac 1{2r}=2\pi\\ \arcsin \frac 1r + \arcsin \frac 1{2r}=\frac \pi 3\\$$ Take the sine of this, use the angle-sum formula and the fact that $\cos \arcsin x=\sqrt {1-x^2}$ should get you to the Alpha solution $r=\sqrt{\frac 73}$

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Thank you both for your answers, but are you sure there's not something a bit more clever? We weren't even given calculators for this set, and a very small amount of time. Also, I am in seventh grade, so I doubted that the answer would even use trig. Don't get me wrong, I appreciate that you solved it and understand your answer, I'm just wondering if there's some easy trick that I overlooked and can be employed on a problem like this. –  recursive recursion Feb 12 at 22:04
    
I think neither Barry Cipra's answer nor mine need a calculator (and in fact most calculators will not get the exact answer) and are doable in 15 minutes for someone who is comfortable with trig. I don't see a more clever approach, but maybe there is one. –  Ross Millikan Feb 12 at 22:11
    
@RossMillikan, I solved the problem in 15 minutes, but it took me another 15 minutes to solve it correctly.... –  Barry Cipra Feb 12 at 22:29
    
Not only that, but there were 7 other problems that I needed to solve. –  recursive recursion Feb 13 at 3:27

We name the left upper vertex with the sides $2$ and $1$ as $(x_0,y_0)$ and continue clockwise This six expressions must be $0$ because the vertexes lie on the circle. $$-r^2+y_{0}^2+x_{0}^2$$ $$-r^2+y_{1}^2+x_{1}^2$$ $$-r^2+y_{2}^2+x_{2}^2$$ $$-r^2+y_{3}^2+x_{3}^2$$ $$-r^2+y_{4}^2+x_{4}^2$$ $$-r^2+y_{5}^2+x_{5}^2$$ These expressions must be $0$ because of the given length of the sides of the polygon. $$\left(y_{0}-y_{1}\right)^2+\left(x_{0}-x_{1}\right)^2-1$$ $$\left(y_{1}-y_{2}\right)^2+\left(x_{1}-x_{2}\right)^2-1$$ $$\left(y_{2}-y_{3}\right)^2+\left(x_{2}-x_{3}\right)^2-1$$ $$\left(y_{3}-y_{4}\right)^2+\left(x_{3}-x_{4}\right)^2-4$$ $$\left(y_{4}-y_{5}\right)^2+\left(x_{4}-x_{5}\right)^2-4$$ $$\left(y_{5}-y_{0}\right)^2+\left(x_{5}-x_{0}\right)^2-4$$ So we have 12 polynomail equations with 13 variables. We can use symmetries to reduce the number of polynomials and the number of variables. I use Maxima to solve the system and get 48 solutions. All but one can be ignored because of symmetries, rotations and the fact, that the radius must be larger $0$. The radius of this solution is ${{\sqrt{7}}\over{\sqrt{3}}}$ and the area ${{7\,\pi}\over{3}}$

Here is a Maxima program to solve the problem

display2d:false$
load(grobner)$
/* define the sqare of the length of the sides of the polygon*/
squaredist:fillarray(make_array (fixnum, 6),[1,1,1,4,4,4]);
/* generate all equations */
/* 6 equations for the point lying on the circle */
equations1:append(makelist(x[i]^2+y[i]^2-r^2,i,0,5),
/* 6 equations for the sides of the polygon */
makelist((x[mod(i,6)]-x[mod(i+1,6)])^2
+(y[mod(i,6)]-y[mod(i+1,6)])^2-squaredist[i],i,0,5))$
/* define the symmetries and eliminate some variables 
and equationy by using these symmetries*/
symmetries:[x[3]=-x[0],y[3]=y[0],x[2]=-x[1],y[2]=y[1],x[5]=-x[4],y[5]=y[4]]$
equations2:ev(equations1,symmetries,ratexpand)$
equations2:listify(setify(equations2))$
/* extract the list of variables of these equations */
lov:listofvars(equations2);
/* calculate a groebner base of the equations and
solve these instead of the original equations 
Maxima got stuck when trying to solve the equations directly
*/
gb:poly_reduced_grobner(equations2,lov)$
solutions1:solve(gb,lov)$
/* there are more than one solutions 
but we can filter out most of them */
length(solutions1);
solutions2:sublist(solutions1,lambda([t],
is(subst(t,x[0])y[0]) 
and subst(t,y[0]>y[4]) 
and subst(t,r>0)))$
/* only one solution remains */
length(solutions2);
/* this is the requested area */
area:r^2*%pi,solutions2;
/*these are the radius and the vertexes of the polygon */
ss:makelist(ev([x[i],y[i]],symmetries,solutions2),i,0,5)$
rr:r,solutions2;
for ll in ss do print(ll);

This is the output of the run of the program

(%i2) display2d : false
(%i3) load(grobner)
Loading maxima-grobner $Revision: 1.6 $ $Date: 2009/06/02 07:49:49 $
(%i4) squaredist:fillarray(make_array(fixnum,6),[1,1,1,4,4,4])
(%o4) ?\#\(1\ 1\ 1\ 4\ 4\ 4\)
(%i5) equations1:append(makelist(-r^2+y[i]^2+x[i]^2,i,0,5),
  makelist(-squaredist[i]+(y[mod(i,6)]-y[mod(1+i,6)])^2
  +(x[mod(i,6)]-x[mod(1+i,6)])^2,i,0,5))
(%i6) symmetries:[x[3] = -x[0],y[3] = y[0],x[2] = -x[1],
  y[2] = y[1],x[5] = -x[4],y[5] = y[4]]
(%i7) equations2:ev(equations1,symmetries,ratexpand)
(%i8) equations2:listify(setify(equations2))
(%i9) lov:listofvars(equations2)
(%o9) [x[1],x[0],y[0],y[1],x[4],y[4],r]
(%i10) gb:poly_reduced_grobner(equations2,lov)
(%i11) solutions1:solve(gb,lov)
(%i12) length(solutions1)
(%o12) 40
(%i13) solutions2:sublist(solutions1,
  lambda([t],
  is(subst(t,x[0])  y[0]) and subst(t,y[0] > y[4])
  and subst(t,r > 0)))
(%i14) length(solutions2)
(%o14) 1
(%i15) ev(area:r^2*%pi,solutions2)
(%o15) 7*%pi/3
(%i16) ss:makelist(ev([x[i],y[i]],symmetries,solutions2),i,0,5)
(%i17) ev(rr:r,solutions2)
(%o17) sqrt(7)/sqrt(3)
(%i18) for ll in ss do print(ll)
[-9/7,10/(7*sqrt(3))] 
[-1/2,5/(2*sqrt(3))] 
[1/2,5/(2*sqrt(3))] 
[9/7,10/(7*sqrt(3))] 
[1,-2/sqrt(3)] 
[-1,-2/sqrt(3)] 
(%o18) done
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