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In the diagram above, the triangle is equilateral with a side of 2... [#permalink]

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26 Jul 2016, 05:46

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In the diagram above, the triangle is equilateral with a side of 2. All three circles are of equal size and all are tangent to each other and to two sides of the triangle. Which of the following is the radius of one of the circles?

A) 1+ √3 B) 2 - √3 C) √6/4 D) 1 - √3/2 E) (√3-1)/2)

Re: In the diagram above, the triangle is equilateral with a side of 2... [#permalink]

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26 Jul 2016, 07:57

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Here's how I'd solve it.

As in the first 30secs I didn't reckon any pattern (there are, I just didn't see them), I estimated.

We can find the height of the equilateral triangle which is √3, and we can see that r has to be between √3/4 and √3/5, so approximately 0,43<r<0,34.

The solutions are: A) 1+ √3 ≈ 2,7 B) 2 - √3 ≈ 0,29 C) √6/4 ≈ 0,6 D) 1 - √3/2 ≈ 0,25 E) (√3-1)/2) ≈ 0,35

The only one included in the range is E), so the answer has to be E). (And in fact it is).

There are other ways to approach it, if enough of you guys like the problem I'll post Magoosh official solution (although recopying everything is a pain :D).

In the diagram above, the triangle is equilateral with a side of 2... [#permalink]

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06 Aug 2016, 08:36

That was hard. It actually took me a good time.

I thought in dividing the figure in more geometric figures:

If you think of the bottom of the triangle, you can make 2 (30-60-90) triangles drawing a line dividing the vertex to the center of each circle and from the center to the bottom of the triangle.

If smaller side is \(r\) - of the new (30-60-90) triangle - than its hypotenuse is \(2r\) and its longer leg is \(r\sqrt{3}\).

Now that is the same for the both sides of the bottom of the original triangle. This will make a lot of more sense if you draw.

Now you have \(r\sqrt{3}\), then \(2r\) between the 2 (30-60-90) triangles and another \(r\sqrt{3}\).

These measurements summed up must be equal to the side, which is 2.

\(r\sqrt{3}+r+r+r\sqrt{3} = 2\)

\(r = \frac{sqrt(3)-1}{2}\)

ps: I tried to use square root inside the fraction, but it didn't work :/

Last edited by matsuda on 07 Aug 2016, 05:45, edited 2 times in total.

In the diagram above, the triangle is equilateral with a side of 2... [#permalink]

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06 Aug 2016, 11:34

matsuda wrote:

That was hard. It actually took me a good time.

I thought in dividing the figure in more geometric figures:

If you think of the bottom of the triangle, you can make 2 (30-60-90) triangles drawing a line dividing the vertex to the center of each circle and from the center to the bottom of the triangle.

If smaller side is \(r\)- of the new (30-60-90) triangle - than its hypotenuse is \(2r\)and its longer leg is \(2\sqrt{3}\).

Now that is the same for the both sides of the bottom of the original triangle. This will make a lot of more sense if you draw.

Now you have \(2\sqrt{3}\), then \(2r\)between the 2 (30-60-90) triangles and another \(2\sqrt{3}\).

These measurements summed up must be equal to the side, which is 2.

\(2\sqrt{3}+r+r+2\sqrt{3} = 2\)

\(r = \frac{sqrt(3)-1}{2}\)

ps: I tried to use square root inside the fraction, but it didn't work :/

I am getting a 30-60-90 triangle with radius r. So, hypotenuse 2r and longest tail "sqrt(3)r". So in total: 2sqrt(3)r+2r=2.

Re: In the diagram above, the triangle is equilateral with a side of 2... [#permalink]

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07 Aug 2016, 05:36

mahfuz_fahim wrote:

matsuda wrote:

That was hard. It actually took me a good time.

I thought in dividing the figure in more geometric figures:

If you think of the bottom of the triangle, you can make 2 (30-60-90) triangles drawing a line dividing the vertex to the center of each circle and from the center to the bottom of the triangle.

If smaller side is \(r\)- of the new (30-60-90) triangle - than its hypotenuse is \(2r\)and its longer leg is \(2\sqrt{3}\).

Now that is the same for the both sides of the bottom of the original triangle. This will make a lot of more sense if you draw.

Now you have \(2\sqrt{3}\), then \(2r\)between the 2 (30-60-90) triangles and another \(2\sqrt{3}\).

These measurements summed up must be equal to the side, which is 2. \(2\sqrt{3}+r+r+2\sqrt{3} = 2\)

\(r = \frac{sqrt(3)-1}{2}\)

ps: I tried to use square root inside the fraction, but it didn't work :/

I am getting a 30-60-90 triangle with radius r. So, hypotenuse 2r and longest tail "sqrt(3)r". So in total: 2sqrt(3)r+2r=2.

Please help me to clarify my understandings

The red part is my mistake. It should be the way you did:

\(r\sqrt{3}+r+r+r\sqrt{3} = 2\) \(2r\sqrt{3}+2r = 2\) -------------- add similar factors \(2r(\sqrt{3}+1) = 2\) ------------- take 2r out \(r(\sqrt{3}+1) = 1\) ------------- divide by 2 \(r = \frac{1}{(\sqrt{3}+1)}\) ------------- take the square root out of the denominator \(r = \frac{1}{(\sqrt{3}+1)}*\frac{sqrt(3)-1}{sqrt(3)-1)} = \frac{sqrt(3)-1}{2}\)

Hope is clear.

ps: for some reason my sqrt formula stopped workings :/ ps2: corrected the mistake

Re: In the diagram above, the triangle is equilateral with a side of 2... [#permalink]

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07 Aug 2016, 06:13

matsuda wrote:

mahfuz_fahim wrote:

matsuda wrote:

That was hard. It actually took me a good time.

I thought in dividing the figure in more geometric figures:

If you think of the bottom of the triangle, you can make 2 (30-60-90) triangles drawing a line dividing the vertex to the center of each circle and from the center to the bottom of the triangle.

If smaller side is \(r\)- of the new (30-60-90) triangle - than its hypotenuse is \(2r\)and its longer leg is \(2\sqrt{3}\).

Now that is the same for the both sides of the bottom of the original triangle. This will make a lot of more sense if you draw.

Now you have \(2\sqrt{3}\), then \(2r\)between the 2 (30-60-90) triangles and another \(2\sqrt{3}\).

These measurements summed up must be equal to the side, which is 2. \(2\sqrt{3}+r+r+2\sqrt{3} = 2\)

\(r = \frac{sqrt(3)-1}{2}\)

ps: I tried to use square root inside the fraction, but it didn't work :/

I am getting a 30-60-90 triangle with radius r. So, hypotenuse 2r and longest tail "sqrt(3)r". So in total: 2sqrt(3)r+2r=2.

Please help me to clarify my understandings

The red part is my mistake. It should be the way you did:

\(r\sqrt{3}+r+r+r\sqrt{3} = 2\) \(2r\sqrt{3}+2r = 2\) -------------- add similar factors \(2r(\sqrt{3}+1) = 2\) ------------- take 2r out \(r(\sqrt{3}+1) = 1\) ------------- divide by 2 \(r = \frac{1}{(\sqrt{3}+1)}\) ------------- take the square root out of the denominator \(r = \frac{1}{(\sqrt{3}+1)}*\frac{sqrt(3)-1}{sqrt(3)-1)} = \frac{sqrt(3)-1}{2}\)

Hope is clear.

ps: for some reason my sqrt formula stopped workings :/ ps2: corrected the mistake

Can you please explain, how you got this step? \(2\sqrt{3}+r+r+2\sqrt{3} = 2\)[/color]

Re: In the diagram above, the triangle is equilateral with a side of 2... [#permalink]

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07 Aug 2016, 06:37

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acegmat123 wrote:

Can you please explain, how you got this step? \(2\sqrt{3}+r+r+2\sqrt{3} = 2\)[/color]

Like I said, the red part was wrong. I corrected the post before.

I'm attaching a image - poorly made in ms paint - that I think it can help.

Attachment:

triangle.jpg [ 24.05 KiB | Viewed 2371 times ]

But essentially, the side of the original triangle must be equal to two longer leg of the 30-60-90 triangle summed with two \(r\). You get a 30-60-90 triangle if you draw a bisector of any vertex of the original triangle to the center of the closest circle and a radius from the center of the circle to the side of the triangle. You know that it will make a 90 degrees angle because the circle is tangent to the triangle. Since the triangle is equilateral, inside angle is 60 and divided by 2 when divided by the bisector, so 30. The other angle must be 60. Distance between the centers of 2 circles is \(2r\).

30-60-90 ratio is \(1:\sqrt{3}:2\), in this case, \(r:r\sqrt{3}:2r\)

Then, sum of longer leg of the similar smaller triangle, the \(2r\), and the other leg of the other triangle must be equal to 2 or

Re: In the diagram above, the triangle is equilateral with a side of 2... [#permalink]

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07 Aug 2016, 08:55

matsuda wrote:

acegmat123 wrote:

Can you please explain, how you got this step? \(2\sqrt{3}+r+r+2\sqrt{3} = 2\)[/color]

Like I said, the red part was wrong. I corrected the post before.

I'm attaching a image - poorly made in ms paint - that I think it can help.

Attachment:

triangle.jpg

But essentially, the side of the original triangle must be equal to two longer leg of the 30-60-90 triangle summed with two \(r\). You get a 30-60-90 triangle if you draw a bisector of any vertex of the original triangle to the center of the closest circle and a radius from the center of the circle to the side of the triangle. You know that it will make a 90 degrees angle because the circle is tangent to the triangle. Since the triangle is equilateral, inside angle is 60 and divided by 2 when divided by the bisector, so 30. The other angle must be 60. Distance between the centers of 2 circles is \(2r\).

30-60-90 ratio is \(1:\sqrt{3}:2\), in this case, \(r:r\sqrt{3}:2r\)

Then, sum of longer leg of the similar smaller triangle, the \(2r\), and the other leg of the other triangle must be equal to 2 or

Re: In the diagram above, the triangle is equilateral with a side of 2... [#permalink]

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20 Aug 2017, 00:02

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I dont get the AB/BE = \(\sqrt{3}\) /1 Could someone please explain this?

The big triangle is equilateral, thus each angle is 60 degrees. BAE is half of that, so 30 degrees, which makes triangle BAE 30-60-90 right triangle. This is one of the 'standard' triangles you should be able recognize on sight. A fact you should commit to memory is: The sides are always in the ratio \(1 : \sqrt{3}: 2\). Notice that the smallest side (1) is opposite the smallest angle (30°), and the longest side (2) is opposite the largest angle (90°).Therefore \(\frac{AB}{BE} = \frac{\sqrt{3}}{1}\)

Re: In the diagram above, the triangle is equilateral with a side of 2... [#permalink]

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23 Aug 2017, 08:59

sarathgopinath wrote:

Bunuel wrote:

The big triangle is equilateral, thus each angle is 60 degrees. BAE is half of that, so 30 degrees, which makes triangle BAE 30-60-90 right triangle.

Hi Bunuel. Why is BAE half of 60? Is it due to symmetry ?

it is property,

when from any point you draw 2 tangents to a circle and a line is drawn joining the centre of circle and that point then that line will be angle bisector of the angle made. like here the angle was 60 degrees and half of that will be 30 since it is angle bisector i.e. the line from centre of circle to that point will bisect 60 degrees in two angles of 30 degrees.

hope its clear.
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Re: In the diagram above, the triangle is equilateral with a side of 2... [#permalink]

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28 Aug 2017, 03:39

can't draw the fig on my scratch-book and therefore wrong calculation..I was able to think of 30-60-90 triangle but drawing messed everything..hope we don't get such question

Re: In the diagram above, the triangle is equilateral with a side of 2... [#permalink]

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28 Aug 2017, 03:55

r19 wrote:

can't draw the fig on my scratch-book and therefore wrong calculation..I was able to think of 30-60-90 triangle but drawing messed everything..hope we don't get such question

see the concept involved, learn from your mistakes and move ahead, if this type comes, you will definitely not do wrong, and hope for the best.. all the best
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