2 circle instersect each other. One with radius 1, another with radius r. Form $120^\circ$, outside the two cricles.
AB = $\sqrt{r^2 - r + 1}$ (matched key answer)
I find r = 0.5 (matched key answer)
So. AB = $\sqrt{\frac{3}{4}}$
- However. I have been thinking hours about 3rd problems..
I think this is what the plane looks like more or less :
But. I found that with APB = $120^\circ$, AB is not $\sqrt{\frac{3}{4}}$ Is there error with the question? Or where am i wrong?
Is there more effective way to solve it?
The situation, with $r = \frac12$, is shown in the picture below. From the fact that $\overline{AB} = \frac{\sqrt 3} 2$, $\overline{PB} = \frac12$, and $\overline{AP} = 1$, we gather that $\triangle APQ$ is equilateral.
You can then determine the area of the intersection between the two circles by summing half of the area of circle $\mathcal C_r$ with a circular segment of circle $\mathcal C$ subtended by an angle of $60°$ (that is, one sixth of the circle area minus the area of the equilateral triangle). This gives \begin{eqnarray} \mathcal A&=& \frac{\pi}{8}+\left(\frac{\pi}{6}-\frac{\sqrt 3}{4}\right)=\\ &=& \frac{7\pi}{24} - \frac{\sqrt{3}}{4}. \end{eqnarray}
EDIT To better understand the situation, you can proceed as follows.
Draw circle $\mathcal C$, with center in $A$ then draw a diameter and take $B$ on it such that $\overline{AB} = \frac{\sqrt 3}{2}$. $B$ is going to be the center of $\mathcal C_r$.
Now draw the line through $B$ perpendicular to $AB$, that intersects $\mathcal C$ in $P$ and $Q$. From Pythagorean Theorem, $\overline{PB} = \frac{1}{2}$, so $\mathcal C_r$ is the circle with center in $B$ and passing through $P$ and $Q$.
EDIT 2 The general case is shown in the figure below.
By construction we have that $\angle TPS = 120°$. From the fact that radii are perpendicular to tangent lines we have $\angle APS = 90°$ and $\angle TPB = 90°$.
Therefore we have $\angle TPA = 30°$ and $\angle APB = 60°$.
Draw from $A$ the line perpendicular to $PB$, and call $H$ its intersection with $PB$. Since $\triangle APH$ is half of an equilateral triangle, $\overline{PH} = \frac12\cdot \overline{AP} = \frac12$, and, by Pythagorean Theorem, $\overline{AH} = \frac{\sqrt 3}{2}$. $\overline{PB} = r$ by definition. Thus $\overline{HB} =r-\frac12$.
Again Pythagorean Theorem, on $\triangle AHB$, gives, in conclusion \begin{eqnarray} \overline{AB}^2 &=& \frac34 + \left(r-\frac12\right)^2=\tag{1}\label{eq1}\\ &=& r^2-r+1. \end{eqnarray}
From \eqref{eq1}, $$\overline{AB}^2 \geq \frac34,$$ with the minimum attained when $r=\frac12$ and thus $H\equiv B$, as in the figure above.