kyotsu-test 2018 QCourse1-IV
View
Let the quadrangle ABCD be a rhombus where the length of the sides is $\sqrt { 2 }$ and $\angle \mathrm { ABC } = 30 ^ { \circ }$.
(1) We have
$$\mathrm { AC } ^ { 2 } = \mathbf { A } - \mathbf { B } \sqrt { \mathbf { C } } , \quad \mathrm { BD } ^ { 2 } = \mathbf { E } + \mathbf{F} \sqrt{\mathbf{E}} .$$
Now, for any positive numbers $a$ and $b$, we have
$$( \sqrt { a } \pm \sqrt { b } ) ^ { 2 } = a + b \pm 2 \sqrt { a b } \quad \text { (double-sign correspondence). }$$
Using this formula, we obtain
$$\mathrm { AC } = \sqrt { \mathbf { G } } - \mathbf { H } , \quad \mathrm { BD } = \sqrt { \mathbf { I } } + \mathbf{I} . \mathbf { J } .$$
(2) Let us draw four circles, each centered on one vertex of rhombus ABCD, with the following conditions:
The radii of the circles centered on vertices A and C are of length $r$, and those centered on vertices B and D are of length $\sqrt { 2 } - r$.
Circles centered on opposite vertices (A and C, B and D) may touch each other but may not intersect.
Let us denote the area of the region common to rhombus ABCD and these four circles by $S$. We have
$$S = \pi \left( r ^ { 2 } - \frac { \sqrt { \mathbf { K } } } { \mathbf { L } } r \right)$$
where the range of $r$ is
$$\sqrt { \mathbf { O } } - \frac { \sqrt { \mathbf { P } } + \mathbf { Q } } { \mathbf { R } } \leqq r \leqq \frac { \sqrt { \mathbf { S } } - \square \mathbf { T } } { \square \mathbf { U } }$$
Hence $S$ is minimized when $r = \frac { \sqrt { \mathbf { V } } } { \mathbf { W } }$, and the value of $S$ then is $\frac { \mathbf{X} } { \mathbf { Y } } \pi$.