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# In the figure below, if the radius of circle O is r and tria

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In the figure below, if the radius of circle O is r and tria [#permalink]

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31 Mar 2012, 03:10
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In the figure below, if the radius of circle O is r and triangle ABC is equilateral, what is the length of AB, in terms of r?
Attachment:

Triangle.GIF [ 3.47 KiB | Viewed 4049 times ]

(A) r $$\sqrt{2}$$
(B) r $$\sqrt{3}$$
(C) 2r $$\sqrt{3}$$
(D) $$\frac{3}{2}$$ r
(E) 2r
[Reveal] Spoiler: OA

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Re: Inscribed circle in an Equilateral triangle [#permalink]

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31 Mar 2012, 05:01
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enigma123 wrote:
In the figure below, if the radius of circle O is r and triangle ABC is equilateral, what is the length of AB, in terms of r?

(A) r $$\sqrt{2}$$
(B) r $$\sqrt{3}$$
(C) 2r $$\sqrt{3}$$
(D) $$\frac{3}{2}$$ r
(E) 2r

Any idea how to solve?

There are many ways to solve this question including using the direct formula saying that the radius of the inscribed circle in an equilateral triangle is $$r=a*\frac{\sqrt{3}}{6}$$.

One can also do the following, consider the diagram below:
Attachment:

Circle.gif [ 6.02 KiB | Viewed 4033 times ]
Angle DCO is 60/2=30 degrees, hence triangle DOC is 30°-60°-90° right triangle (angle DOC is 60°). Next, in a right triangle where the angles are 30°, 60°, and 90° the sides are always in the ratio $$1 : \sqrt{3}: 2$$ and the leg opposite 30° (OD=r) corresponds with $$1$$ and the leg opposite 60° (DC) corresponds with $$\sqrt{3}$$, so $$\frac{r}{DC}=\frac{1}{\sqrt{3}}$$ --> $$DC=r\sqrt{3}$$. Now, since DC=AC/2 then $$AC=2DC=2r\sqrt{3}$$.

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Re: In the figure below, if the radius of circle O is r and tria [#permalink]

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04 Dec 2015, 08:18
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Expert's post
rahulkashyap wrote:
how does the line bisect the angle 60?

thanks

It a special property of incircles of equilateral triangles. Look below for the proof.

Triangle ABC is an equilateral traingle with O as the center of the incircle. Radii of the incircle (OD and OF) are perpendicular to the sides AC and BC (by property of a circle wherein radius is perpendicular to the tangent at the point of tangency, CD and CF are tangents to the incircle at points of tangencies D and F respectively).

Additionally, for a circle, 2 tangents (CD and CF) drawn from the same external point (C) are of same length --> CD = CF

Thus, in right triangles CDO and CFO,

CD=CF
OD=OF = radius of the incircle
OC=OC (common side)

Thus Triangle CDO is congruent to triangle CFO (by SSS property of congruence)--->$$\angle { DCO} = \angle{FCO}$$ and thus $$\angle { DCO} = \angle{FCO} = \frac{\angle {ACB}}{2}$$

Dont need to learn the proof but do remember that the line drawn from the center of an incircle of an equilateral triangle to one of the veritces will always bisect that angle.

Hope this helps.
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12-04-15 9-53-29 AM.jpg [ 10.92 KiB | Viewed 1014 times ]

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Re: Inscribed circle in an Equilateral triangle [#permalink]

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31 Mar 2012, 04:37
Ok so as in the diagram.
the ratio of green line to red line will be 2:1.
Pink line will be perpendicular to triangle edge.
therefore

(2a)^2 = (x/2)^2 + a^2

Upon solving you will get x = 2(3)^1/2 a

Hope it helps.
Attachments

Untitled.png [ 5.14 KiB | Viewed 3985 times ]

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Re: In the figure below, if the radius of circle O is r and [#permalink]

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31 Mar 2012, 05:11
Bunuel - Thanks. But how can (angle DOC is 60°)?
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Re: In the figure below, if the radius of circle O is r and [#permalink]

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31 Mar 2012, 05:17
enigma123 wrote:
Bunuel - Thanks. But how can (angle DOC is 60°)?

DOC is 30°-60°-90° right triangle. Angle D=90°, angle DCO=30° and angle DOC=60°.
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Re: In the figure below, if the radius of circle O is r and [#permalink]

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31 Mar 2012, 17:00
Sorry Bunuel - what I meant to ask was - How come OC will bisect the angle DCO?
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Re: In the figure below, if the radius of circle O is r and [#permalink]

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31 Mar 2012, 17:16
enigma123 wrote:
Sorry Bunuel - what I meant to ask was - How come OC will bisect the angle DCO?

Ask yourself, why should any angle from ACO and BCO be greater than another?
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Re: In the figure below, if the radius of circle O is r and tria [#permalink]

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06 Sep 2013, 03:48
Bumping for review and further discussion.
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Re: In the figure below, if the radius of circle O is r and tria [#permalink]

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10 Sep 2013, 22:59
Hi Bunuel, can we say that in equilateral triangle median = angle bisector = altitude = perpendicular bisector?
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Re: In the figure below, if the radius of circle O is r and tria [#permalink]

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11 Sep 2013, 01:45
b2bt wrote:
Hi Bunuel, can we say that in equilateral triangle median = angle bisector = altitude = perpendicular bisector?

Yes, that's true. For more check here: math-triangles-87197.html
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Re: In the figure below, if the radius of circle O is r and tria [#permalink]

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07 Jan 2014, 02:57
Hi Bunnel,

How we solve this problem by using the direct formula for inradius of equilateral triangle
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Re: In the figure below, if the radius of circle O is r and tria [#permalink]

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04 Dec 2015, 07:06
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Re: In the figure below, if the radius of circle O is r and tria [#permalink]

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04 Dec 2015, 07:45
Bunuel wrote:
enigma123 wrote:
In the figure below, if the radius of circle O is r and triangle ABC is equilateral, what is the length of AB, in terms of r?

(A) r $$\sqrt{2}$$
(B) r $$\sqrt{3}$$
(C) 2r $$\sqrt{3}$$
(D) $$\frac{3}{2}$$ r
(E) 2r

Any idea how to solve?

There are many ways to solve this question including using the direct formula saying that the radius of the inscribed circle in an equilateral triangle is $$r=a*\frac{\sqrt{3}}{6}$$.

One can also do the following, consider the diagram below:
Attachment:
Circle.gif
Angle DCO is 60/2=30 degrees, hence triangle DOC is 30°-60°-90° right triangle (angle DOC is 60°). Next, in a right triangle where the angles are 30°, 60°, and 90° the sides are always in the ratio $$1 : \sqrt{3}: 2$$ and the leg opposite 30° (OD=r) corresponds with $$1$$ and the leg opposite 60° (DC) corresponds with $$\sqrt{3}$$, so $$\frac{r}{DC}=\frac{1}{\sqrt{3}}$$ --> $$DC=r\sqrt{3}$$. Now, since DC=AC/2 then $$AC=2DC=2r\sqrt{3}$$.

how does the line bisect the angle 60?

thanks
Re: In the figure below, if the radius of circle O is r and tria   [#permalink] 04 Dec 2015, 07:45
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