3 tangent equal circles inside a square (Estonia Open 1997 Senior 2.3)

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parmenides51

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parmenides51

#1 h Mar 25, 2020, 2:58 AM

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The figure shows a square and three circles of equal radius tangent to each other and square passes. Find the radius of the circles if the square length is $1$ .

http://3.bp.blogspot.com/-iIjwupkz7DQ/XnrIRhKIJnI/AAAAAAAALhA/clERrIDqEtcujzvZk_qu975wsTjKaxCLQCK4BGAYYCw/s400/97%2Bestonia%2Bopen%2Bs2.3.png

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twinbrian

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twinbrian

#2 h Mar 25, 2020, 3:18 AM

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Solution

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ss28112

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ss28112

#3 h Mar 25, 2020, 4:53 AM

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Notice that the figure is symmetric over the diagonal from the top-right to the bottom-left. We also have an equilateral triangle joining the centers of the three circles. Construct this, and the answer should be quite obvious.

I got: answer

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magnifecent

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magnifecent

#4 h Mar 25, 2020, 5:01 AM

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qUiTe oBvIoUs

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AwesomeLife_Math

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AwesomeLife_Math

#5 h Mar 25, 2020, 1:05 PM

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Dunno if this is correct but...
Click to reveal hidden text

This post has been edited 3 times. Last edited by AwesomeLife_Math, Mar 25, 2020, 5:07 PM

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twinbrian

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twinbrian

#6 h Mar 25, 2020, 4:03 PM

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firstly, the height of the equilateral triangle is $\frac{r\sqrt{3}}{2}$ not $r\sqrt{3}$ and second, your part where you found the rest of the distance besides the square and equilateral is not clear. I'm pretty sure the answer is $r \approx 0.34$ as shown.

This post has been edited 1 time. Last edited by twinbrian, Mar 25, 2020, 4:03 PM

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AwesomeLife_Math

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AwesomeLife_Math

#7 h Mar 25, 2020, 4:21 PM • 3 Y

Y by Mango247, Mango247, Mango247

@above fixed.
I just dropped perpendiculars to the sides of the squares. (the bottom and right circles)

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twinbrian

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twinbrian

#8 h Mar 25, 2020, 4:30 PM

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I did the exact same and drew the lines from the centers of those to the bottom right vertice and yielded my answer.

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AwesomeLife_Math

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AwesomeLife_Math

#9 h Mar 25, 2020, 4:31 PM

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Could you send me an image over PM? Thanks.

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dragontraining

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dragontraining

#10 h Mar 25, 2020, 4:43 PM

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@2above sorry to be pedantic but "vertice" isn't a word, the singular form of vertices is vertex

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ss28112

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ss28112

#11 h Mar 25, 2020, 4:46 PM

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I'm not sure if 0.34 is a correct answer. Just looking at the figure itself, it doesn't make sense. I think it has to be closer to 1/4

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Susanssluk

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Susanssluk

#12 h Mar 25, 2020, 4:52 PM

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twinbrian wrote:

firstly, the height of the equilateral triangle is $\frac{r\sqrt{3}}{2}$ not $r\sqrt{3}$ and second, your part where you found the rest of the distance besides the square and equilateral is not clear. I'm pretty sure the answer is $r \approx 0.34$ as shown.

bruh, the equilateral triangle has a side length of $2r$ .

https://www.wolframalpha.com/input/?i=sqrt2%3Dr%282sqrt2%2B1%2Bsqrt3%29

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parmenides51

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parmenides51

#13 h Mar 25, 2020, 4:53 PM

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the result is
figure added

Attachments:

This post has been edited 1 time. Last edited by parmenides51, Mar 25, 2020, 4:55 PM

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ss28112

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ss28112

#14 h Mar 25, 2020, 4:54 PM

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parmenides51 wrote:

the result is

That's what I got

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Susanssluk

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Susanssluk

#15 h Mar 25, 2020, 4:54 PM • 2 Y

Y by ss28112, Mango247

parmenides51 wrote:

the result is

I agree. That turns out to be ~0.254

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ss28112

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ss28112

#16 h Mar 25, 2020, 4:55 PM

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Susanssluk wrote:

parmenides51 wrote:

the result is

I agree. That turns out to be ~0.254

Yes. I'm just bad at latex so I had to approximate

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twinbrian

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twinbrian

#17 h Mar 25, 2020, 4:56 PM

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wait oops im bad yes u guys are correct. sorry.

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ss28112

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ss28112

#18 h Mar 25, 2020, 5:00 PM

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twinbrian wrote:

wait oops im bad yes u guys are correct. sorry.

Nah I'm pretty bad. I'm surprised I got this one lol. I'll try to write a full solution sometime soon.

This post has been edited 1 time. Last edited by ss28112, Mar 25, 2020, 5:01 PM
Reason: edit

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twinbrian

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twinbrian

#19 h Mar 25, 2020, 5:03 PM

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Revised.
Solution

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AwesomeLife_Math

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AwesomeLife_Math

#20 h Mar 25, 2020, 5:08 PM

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@2above Do we have similar solutions?

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ss28112

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ss28112

#21 h Mar 25, 2020, 5:11 PM

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AwesomeLife_Math wrote:

@2above Do we have similar solutions?

If you're asking if twinbrian and I have similar solutions, the answer is no. I did mine with kind of a weird approach by realizing that the angle that the line passing through the centers of the two top circles makes with the horizontal is 15 degrees. I then used basic trig identites to figure cos 15. We have the equation 2r + 2rcos15 = 1. Solving yields approximately 0.254

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AwesomeLife_Math

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AwesomeLife_Math

#22 h Mar 25, 2020, 5:14 PM • 1 Y

Y by ss28112

Neat solution. I thought we had similar solutions because you mentioned symmetry over the diagonal but turns out we don't.

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parmenides51

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parmenides51

#23 h Mar 25, 2020, 5:19 PM

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official solution

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GammaZero

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GammaZero

#24 h Mar 25, 2020, 5:25 PM

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wait can you translate that into english? @Parmenides51

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AwesomeLife_Math

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AwesomeLife_Math

#25 h Mar 25, 2020, 5:28 PM • 1 Y

Y by Mango247

It just computes the length of the diagonal in terms of $r$ . First the square at the top, then the altitude of the equilateral triangle, then the last segment which is equal to CG because it is the median of the right triangle.

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