A geometric hors d'oeuvre ......

Geometry Level 3

Suppose within a unit square $ABCD$ line segments are drawn from $A$ to the midpoint of $BC$ , from $B$ to the midpoint of $CD$ , from $C$ to the midpoint of $DA$ , and from $D$ to the midpoint of $AB$ .

The resulting $4$ points of intersection of these line segments within $ABCD$ serve as the corners of a square of area $\dfrac{a}{b}$ , where $a,b$ are positive coprime integers. Find $a + b$ .

The answer is 6.

1 solution

Brian Charlesworth
Nov 10, 2014

Let $\theta$ be the angle that the line segment joining $A$ to the midpoint of $BC$ makes with side $AB$ . Also, let $x$ be the length of the side of the square defined by the $4$ points of intersection.

Then $x = \cos(\theta) - \sin(\theta)$ . Now $\cos(\theta) = \frac{2}{\sqrt{5}}$ and $\sin(\theta) = \frac{1}{\sqrt{5}}$ , so $x = \frac{1}{\sqrt{5}}$ .

The area of the square is thus $x^{2} = \frac{1}{5}$ , making $a + b = \boxed{6}$ .

how do you say that x= costheta - sintheta.

Raven Herd - 6 years, 5 months ago

Let points $E$ and $F$ be the vertices of the interior square lying on the line segment joining $A$ with the midpoint of $BC$ that lie, respectively, closest to and furthest from $A$ .

Now $\angle BAF = \theta$ . So $\cos(\theta) = \frac{AF}{AB} = AF$ and $\sin(\theta) = \frac{BF}{AB} = BF$ . But by symmetry $BF = AE$ , and so

$x = EF = AF - AE = AF - BF = \cos(\theta) - \sin(\theta)$ .

Brian Charlesworth - 6 years, 5 months ago

A geometric hors d'oeuvre ......

The answer is 6.

1 solution

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