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 $1$.
Problem
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Tags: geometry, equal circles, circles, square, tangent circles
twinbrian
25.03.2020 06:18
Let the radius be $r$ and we obtain the base of the triangle connected by the upper two circles as $1-2r$. The hypotenuse is $2r$ so we have the height as $\sqrt{4r-1}$. Thus, we obtain the hypotenuse of the triangle created by the center of the leftmost $2$ circles with the lower right vertice of the square as $\sqrt{(1-r-\sqrt{4r-1})^2+r^2}=\sqrt{2r^2+2r+(r-1)(\sqrt{4r-1})}$. We have the height of it as $\sqrt{r^2+2r+(r-1)(\sqrt{4r-1})}$, and thus we obtain the equation $\sqrt{r^2+2r+(r-1)(\sqrt{4r-1})}+\frac{r\sqrt{3}}{2}+r\sqrt{2}=\sqrt{2}$ so $r \approx 0.34$
ss28112
25.03.2020 07:53
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.
0.254
magnifecent
25.03.2020 08:01
qUiTe oBvIoUs
AwesomeLife_Math
25.03.2020 16:05
Dunno if this is correct but...
First you drop perpendiculars from the leftmost circle's center ($A$) to the top and left sides of the square. You will form a square with side length $r$ ($r$ is the radius of the circles). Then you drop a perpendicular from the rightmost circle's center ($B$) to the bottom side of the square and do the same with ($C$), except to the right side of the square. Since we know that the circles $B$ & $C$ are symmetric around the diagonal, we know that all the shapes we formed are bisected by the diagonal. Now we find the values for all the lengths and set them equal to $\sqrt{2}$ and then solve.
We first have a square of radius $r$, so its diagonal is $r\sqrt{2}$. Then you have an equilateral triangle, which obviously has a height of $r\sqrt{3}$. The two perpendicular segments that we dropped form a right triangle, two rectangles and a square. And this right triangle, in fact, is a 45-45-90 (this can be shown through angle chasing or by intuition: cause the figure is symmetric about the diagonal). The hypotenuse of this right triangle is equal to $2r$, so the median must be equal to $r$. Finally, at the lower right corner we have another square, with side lengths $r$ (we dropped perpendiculars from the centers of the circles). Thus our final equatioin is $r\sqrt{2}+r\sqrt{3}+r+r\sqrt{2}=\sqrt{2}$. Solving, the answer is approximately 0.25433.
twinbrian
25.03.2020 19:03
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.
AwesomeLife_Math
25.03.2020 19:21
@above fixed. I just dropped perpendiculars to the sides of the squares. (the bottom and right circles)
twinbrian
25.03.2020 19:30
I did the exact same and drew the lines from the centers of those to the bottom right vertice and yielded my answer.
AwesomeLife_Math
25.03.2020 19:31
Could you send me an image over PM? Thanks.
dragontraining
25.03.2020 19:43
@2above sorry to be pedantic but "vertice" isn't a word, the singular form of vertices is vertex
ss28112
25.03.2020 19:46
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
Susanssluk
25.03.2020 19:52
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
Susanssluk
25.03.2020 19:54
parmenides51 wrote:
$ \frac{\sqrt2}{1+ 2\sqrt2 +\sqrt3}$
I agree. That turns out to be ~0.254
ss28112
25.03.2020 19:55
Susanssluk wrote: parmenides51 wrote:
$ \frac{\sqrt2}{1+ 2\sqrt2 +\sqrt3}$
I agree. That turns out to be ~0.254 Yes. I'm just bad at latex so I had to approximate
twinbrian
25.03.2020 19:56
wait oops im bad yes u guys are correct. sorry.
ss28112
25.03.2020 20:00
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.
twinbrian
25.03.2020 20:03
Revised.
Let the radius be $r$ and we obtain the base of the triangle connected by the upper two circles as $1-2r$. The hypotenuse is $2r$ so we have the height as $\sqrt{4r-1}$. Thus, we obtain the hypotenuse of the triangle created by the center of the leftmost $2$ circles with the lower right vertice of the square as $\sqrt{(1-r-\sqrt{4r-1})^2+r^2}$. We have the height of it as $1-r-\sqrt{4r-1}$, and thus we obtain the equation $1-r-\sqrt{4r-1}+r\sqrt{3}+r\sqrt{2}=\sqrt{2}$ so $r \approx 0.254$
AwesomeLife_Math
25.03.2020 20:08
@2above Do we have similar solutions?
ss28112
25.03.2020 20:11
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
AwesomeLife_Math
25.03.2020 20:14
Neat solution. I thought we had similar solutions because you mentioned symmetry over the diagonal but turns out we don't.
GammaZero
25.03.2020 20:25
wait can you translate that into english? @Parmenides51
AwesomeLife_Math
25.03.2020 20:28
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.