1
$\begingroup$

In the picture, ABCD is a square of side 1 and the semicircles have centers on A, B, C and D. What is the length of PQ? picture

The solution says that the triangle AQD is equilateral, then extends PQ and ... But I dont know how ADQ is equilateral .

2 Answers 2

0

The line $(PQ)$ cuts $(BC)$ at $I$, and $(AD)$ at $H$. Since $ADQ$ is equilateral, $HQ=a\frac{\sqrt3}{2}$ (use trigonometry or the Pythagorean theorem), where $a$ is the side of the square (here it's $1$ but it doesn't matter). Then $IQ=PH$ by symmetry, and $IQ=a-HQ=a\left(1-\frac{\sqrt3}{2}\right)$, and finally

$$PQ=a-2IQ=a(\sqrt3-1)$$

  • 0
    Why is ADQ equilateral? – 2017-02-05
  • 0
    @ArshiaMoniri $Q$ is both on the circle with center $A$ and radius $a$, and the circle with center $D$ and radius $a$, hence $AQ=DQ=a$, and of course, $AD=a$ by definition of $a$. – 2017-02-05
0

Remark that by construction you have AQ=DQ. Hence AQD is equilateral. The height QH of this triangle is therefore $\sqrt{3}/2$ (H is where the extension of segment PQ touches the AD segment). Now you can prove similarly that PH' is also $\sqrt{3}/2$ (H' is the point where the extension of PQ touches BC).

Now you know that

$1=DC=PQ+2PH$

and that

$\sqrt{3}/2=PQ+PH$

So you have a system of two equations and two variables.

Solving you find:

$PH=\frac{2-\sqrt 3}{2}$

$PQ=\sqrt 3 -1$