Bunuel wrote:

If ∆ PQR and ∆ PRS are equilateral, what fraction of PQRS is shaded?

(A) 1/3

(B) 1/4

(C) 1/6

(D) 1/9

(E) 1/12

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Finding the answer to this question did not take a ton of time, but explaining that answer without a diagram is hard

1) SIX CONGRUENT SMALL TRIANGLES

Consider just ∆ PRS in figure PQRS

An equilateral triangle's altitude is also its median, angle bisector, and perpendicular bisector of base

All medians of an equilateral triangle are the same length:

Medians bisect the 60-degree angles to form two 30-degree angles

Medians bisect the bases into equal lengths and create right angles

Medians, therefore, create six identical 30-60-90 triangles

2) ANGLE MEASURES AND SIDE LENGTHS OF SIX TRIANGLES ARE CONGRUENT

Each triangle has:

60-degree angle at circumcenter, 90-degree angle at base bisector point, 30-degree angle at vertex

30-60-90 triangles have sides in ratio: \(x: x\sqrt{3}: 2x\)

See angles and side lengths in diagram of each small triangle. They are identical.

3) SHADED PORTION = \(\frac{1}{3}\)OF ONE TRIANGLE and \(\frac{1}{6}\) of figure PQRS

Because the six small triangles are equal,

one shaded portion equals \(\frac{1}{6}\) of area of ∆ PRS

There are two shaded portions. \(\frac{1}{6} + \frac{1}{6} = \frac{1}{3}\) of area of ∆ PRS

Given: ∆ PQR and ∆ PRS are equilateral

Length of PR = length of both PQ and QR

Areas, therefore, of ∆ PQR and ∆ PRS are equal

∆ PRS is \(\frac{1}{2}\) of the area of figure PQRS

Find shaded area's fraction of ∆ PRS and double the fraction

Shaded portion is therefore

\(\frac{1}{3} * \frac{1}{2} = \frac{1}{6}\)of the area of PQRS

Fraction of PQRS that is shaded = \(\frac{1}{6}\)

ANSWER C

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that within me there lay an invincible summer.

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