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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 3905 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 \(AC=2DC=2r\sqrt{3}\).

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 \(AC=2DC=2r\sqrt{3}\).

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.

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