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# A puzzle (for those who are bored!)

+22
2171
13 An equilateral triangle of side length 2 is inscribed in a circle.  A rectangle, with side lengths x and y, is inscribed in another circle of the same size as the first one.  If the area of the rectangle is the same as that of the triangle, what are the sizes of x and y?

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May 11, 2015

#3
+19

1: Cos(30) = 1/r therefore r = 2/sqrt3 and r^2 = 4/3

2:Area of equil Triang = (1/2)*(2)*sqrt((2^2)-(1^2))=sqrt3 therefore Area of rectangle = xy = sqrt3

3 Draw lines to the opposite vertices of the rectangle then add the area of the 4 triangles you just made inside the rectangle = total area:

2*area Tri A= 2*[(1/2)(x)sqrt(4/3 - (x^2)/4)

2*area Tri B= 2*[(1/2)(y)(x/2)]

4, Solve simultaneous equation to get x and y:

xy = sqrt3

(x)sqrt(4/3 - (x^2)/4) + (1/2)xy = sqrt3

--> If doing manually you substitute the first equation into the second then solve for x, followed by y, but since I'm lazy I used Wolfram alpha to do this for me and got

x≈2.16662 and y≈0.799424

May 12, 2015

#1
0

Oh, cool! Thanks!

May 11, 2015
#2
0

I thought this was a really cool question.  Thanks for providing it Alan.

Isn't any one of you high school students going to show us what you are made of and answer it.

You could at least give it a try :/ May 12, 2015
#3
+19

1: Cos(30) = 1/r therefore r = 2/sqrt3 and r^2 = 4/3

2:Area of equil Triang = (1/2)*(2)*sqrt((2^2)-(1^2))=sqrt3 therefore Area of rectangle = xy = sqrt3

3 Draw lines to the opposite vertices of the rectangle then add the area of the 4 triangles you just made inside the rectangle = total area:

2*area Tri A= 2*[(1/2)(x)sqrt(4/3 - (x^2)/4)

2*area Tri B= 2*[(1/2)(y)(x/2)]

4, Solve simultaneous equation to get x and y:

xy = sqrt3

(x)sqrt(4/3 - (x^2)/4) + (1/2)xy = sqrt3

--> If doing manually you substitute the first equation into the second then solve for x, followed by y, but since I'm lazy I used Wolfram alpha to do this for me and got

x≈2.16662 and y≈0.799424

Brodudedoodebrodude May 12, 2015
#4
0

That was nice!!

May 12, 2015
#5
+5

Correct numerical answer Bro, though it would be nice to see your full expressions for x and y!  The reason for this is that there are several ways to solve this problem, each of which can result in expressions that look different, but are, of course, equivalent.

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May 12, 2015
#6
+13

I did like Brodudedoodebrodude in principle (even if I´m a little late.)

x:

$${\frac{{\sqrt{{\mathtt{3}}}}}{{\sqrt{{\frac{\left({\mathtt{8}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{37}}}}\right)}{{\mathtt{3}}}}}}}}$$

y:

$${\sqrt{{\frac{\left({\mathtt{8}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{37}}}}\right)}{{\mathtt{3}}}}}}$$

May 12, 2015
#7
+13

This has probably been done to death but too bad.

The area of the triangle is

$$\\3\times \frac{1}{2}\times r^2*sin \frac{2\pi}{3}\\\\ =\frac{3r^2}{2}sin\frac{\pi}{3}\\\\ =\frac{3r^2}{2}\times\frac{\sqrt{3}}{2}\\\\ =\frac{3\sqrt{3}r^2}{4}\\\\$$ $$\\Area\\ =4\times\frac{1}{2}\times r^2sin\theta\\\\ =2r^2sin\theta\\\\ so\\\\ 2r^2sin\theta=\frac{3\sqrt3*r^2}{4}\\\\ sin\theta=\frac{3\sqrt3}{8}\\\\ If\;opp=3\sqrt3\;\;and\;\;hyp=8\;\;then\;\;adj=\sqrt{64-27}=\sqrt{37}\\\\ so \;\;cos\theta=\frac{\sqrt{37}}{8}\\\\ -----\\ x^2=2r^2-2r^2cos\theta\\ x^2=2r^2-2r^2\times\frac{\sqrt{37}}{8}\\ x^2=\frac{8r^2}{4}-\frac{r^2\sqrt{37}}{4}\\ x^2=\frac{r^2}{4}\left(8-\sqrt{37}\right)\\\\ x=\frac{r\sqrt{\left(8-\sqrt{37}\right)}}{2}\;\;unit\;thingies$$

$$\\xy=\frac{3\sqrt3r^2}{4}\\\\ y=\frac{3\sqrt3r^2}{4}\div x\\\\ y=\frac{3\sqrt3r^2}{4}\times \frac{1}{x} \\\\ y=\frac{3\sqrt3r^2}{4}\times \frac{2}{r\sqrt{8-\sqrt{37}}} \\\\ y=\frac{3\sqrt3r}{2}\times \frac{1}{\sqrt{8-\sqrt{37}}} \\\\ y= \frac{3r\sqrt3}{2\sqrt{8-\sqrt{37}}} \;unit\;thingies\\\\$$

Is this answer the same as the others?

May 12, 2015
#8
+10

This is fine as far as it goes Melody, and if you calculate r you can take it all the way!

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May 13, 2015
#9
+13

Yes I forgot about that Alan.   Maybe this is better $$\\4=2r^2-2r^2cos120\\ 4=2r^2-2r^2*-0.5\\ 4=2r^2+r^2\\ 4=3r^2\\ \frac{4}{3}=r^2\\ r=\frac{2}{\sqrt3}\\\\ y= \frac{3r\sqrt3}{2\sqrt{8-\sqrt{37}}} \;unit\;thingies\\\\ y= \frac{3*\frac{2}{\sqrt3}\sqrt3}{2\sqrt{8-\sqrt{37}}} \;unit\;thingies\\\\ y= \frac{6}{2\sqrt{8-\sqrt{37}}} \;unit\;thingies\\\\ y= \frac{3}{\sqrt{8-\sqrt{37}}} \;unit\;thingies\\\\\\ x=\frac{r\sqrt{\left(8-\sqrt{37}\right)}}{2}\;\;unit\;thingies\\\\ x=\frac{\frac{2}{\sqrt3}*\sqrt{\left(8-\sqrt{37}\right)}}{2}\;\;unit\;thingies\\\\ x=\frac{\sqrt{3\left(8-\sqrt{37}\right)}}{3}\;\;unit\;thingies$$

May 13, 2015
#10
+10

The area of the equilateral triangle = (1/2) 4 sin120 = 2(√3)/ 2  = √3

The radius of both circles is given by  √[ 1^2 + (1/√3)^2 ]  =   √[1 + 1/3]   =  2/ √3

So....the diagonal of the rectangle = 2 ( 2/ √3 )  =  4 / √3

And, using the Pythagorean Thorem, "y" in the rectangular figure = √ [( 4/ √3) ^2 - x^2 ] = √[16/3 - x^2]

So......since the area of the triangle and the rectangle is equal, we have...

y * x  = √3

√[16/3 - x^2] * x =  √3    square   both sides

(16/3)x^2 - x^4  = 3        multiply through by 3

16x^2 - 3x^4  = 9            rearrange

3x^4 - 16x^2  + 9 = 0         let m= x^2   ....  so we have

3m^2 - 16m + 9 = 0

Using the quadratic formula [and an assist from the on-site solver], we have

$${\mathtt{3}}\left[{{m}}^{{\mathtt{2}}}\right]{\mathtt{\,-\,}}{\mathtt{16}}{\mathtt{\,\times\,}}{\mathtt{m}}{\mathtt{\,\small\textbf+\,}}{\mathtt{9}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{m}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{37}}}}{\mathtt{\,-\,}}{\mathtt{8}}\right)}{{\mathtt{3}}}}\\ {\mathtt{m}} = {\frac{\left({\sqrt{{\mathtt{37}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{8}}\right)}{{\mathtt{3}}}}\\ \end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{m}} = {\mathtt{0.639\: \!079\: \!156\: \!567\: \!260\: \!1}}\\ {\mathtt{m}} = {\mathtt{4.694\: \!254\: \!176\: \!766\: \!073\: \!2}}\\ \end{array} \right\}$$

If x = .799, y = √[16/3 - .7994^2] = .2166    ...but this is impossible....since y is smaller than x

If x = 2.1666, y = √[16/3 - 2.1666^2]  = about . 7994   May 14, 2015
#11
+5

Nice answer Chris Yours looks a bit easier than mine.

I suppose I should include my approx answers so that the answersc an be compared.

May 14, 2015
#12
0

This was a really good question Alan, thanks for posting it :) May 14, 2015
#13
+5

Thanks, Melody.....I realized that making that quadratic substitution  would prevent some "messy" radicals...!!!!   May 14, 2015