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Triangle ABC is equilateral and inscribed in the circle with center O.

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Triangle ABC is equilateral and inscribed in the circle with center O.  [#permalink]

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New post 12 Feb 2019, 06:58
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GMATH practice exercise (Quant Class 15)

Image

Triangle ABC is equilateral and inscribed in the circle with center O. What is the value of the shaded area?

(1) The length of the radius of the circle is 2
(2) Lines PQ and BC are parallel

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Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net

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Triangle ABC is equilateral and inscribed in the circle with center O.  [#permalink]

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New post 13 Feb 2019, 07:33
fskilnik wrote:
GMATH practice exercise (Quant Class 15)

Image

Triangle ABC is equilateral and inscribed in the circle with center O. What is the value of the shaded area?

(1) The length of the radius of the circle is 2
(2) Lines PQ and BC are parallel


Image


\(? = {S_{\Delta APQ}}\)


(1) Clearly insufficient, as PROVED by the algebraic-geometric bifurcation (figure, first row):

> In the figure on the left, we have chosen P sufficiently close to B such that our focus is "as near as desired" to HALF (say 49.99%) of the area of the triangle ABC.
> In the figure on the right, we have chosen P such that lines PQ and BC are parallel, hence our focus is 4/9 (= 0.4444... , i.e., approx. 44.44%) of the area of the triangle ABC.
(The number 4/9 will be obtained in (1+2) below!)

(2) Trivial geometric bifurcation shown in the figure (second row).

(1+2) Triangles APQ and ABC are similar, hence using the "30-60-90" shortcut we get the last figure, and from it (explain):

\({{{S_{\Delta APQ}}} \over {{S_{\Delta ABC}}}} = {\left( {{2 \over {1 + 2}}} \right)^2}\,\,\,\,\, \Rightarrow \,\,\,\,\,? = {S_{\Delta APQ}} = {4 \over 9}\left( {{S_{\Delta ABC}}} \right)\,\,\,{\rm{unique}}\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{SUFF}}.\)

Important: there is no need for obtaining the area of triangle ABC explicitly: given the numerical value of the radius of the circle, triangle ABC (equilateral and inscribed in it) is unique, and its unique area COULD be calculated!


The correct answer is (C).


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio
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Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net

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Triangle ABC is equilateral and inscribed in the circle with center O.  [#permalink]

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New post 21 Feb 2019, 08:12
Rules to know:

1. In an equilateral triangle: incentre, circumcentre, centroid and orthocentre coincides.
2. In an equilateral triangle circumscribed by a circle, the centroid of the triangle and the centre of the circle coincide. (O in the diagram)
3. In a triangle, the centroid splits the median in the ratio 2:1

Lets assume that AO reaches BC at X.

From the question, we know that AO:OX = 2:1 (ref #3). Hence AO = 2x, OX = x.

(1) We know r = 2. We know the height of the shaded region to be 2.

AO = 2x, OX = x. But, we know that AO = 2. Hence, OX = 1.

BO = 2, OX = 1, Angle OXB = 90. Now, we can find BX. We know BX = CX. Hence we can find BC.

But, we cannot find PQ. Hence Insufficient.

(2) PQ || BC.

We know AO:OX = 2:1. So, we know PQ = (2/2+1) BC. We dont know BC. Hence Insufficient.

(1)+ (2)

We can find PQ with AO:OX = 2:1 and BC.

PQ = (2/2+1) * BC.

With PQ and AO, we can find the area of the shaded region.

Ans: C
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Triangle ABC is equilateral and inscribed in the circle with center O.  [#permalink]

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New post 21 Feb 2019, 11:02
fskilnik wrote:
GMATH practice exercise (Quant Class 15)

Image

Triangle ABC is equilateral and inscribed in the circle with center O. What is the value of the shaded area?

(1) The length of the radius of the circle is 2
(2) Lines PQ and BC are parallel


(1) If the shaded region is indeed an isoceles triangle, then we can split APQ into two 30-60-90 right triangles. Since ABC is equilateral, angle A must be 60 degrees. Therefore <QAO must = 30 and <AQO-60. With radius = AO =2, and PQ = 2*(2/root3) so the shaded region = 1/2 * 2/root3 (1) see figure 1

However, if the shaded region is not isoceles we cannot determine the area of the shaded region. NS

(2) PQ and BC are parallel. This gives us nothing about side lengths. Clearly NS

(1) and (2) So we know the length of the radius, and PQ || BC then we know <C = <Q = 60 and <P =<B =60, thus triangle APQ is isoceles, thus we can find the area of APQ as in (1) Sufficient
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Triangle ABC is equilateral and inscribed in the circle with center O.   [#permalink] 21 Feb 2019, 11:02
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