Question

A circle is inscribed into a rhombus $ABCD$ with one angle $60°$. The distance from the centre of the circle to the nearest vertex is equal to 1. If $P$ is any point of the circle, then $|PA|^2 + |PB|^2 + |PC|^2 + |PD|^2$ is equal to :

• A. 12
• B. 11
• C. 9
• D. None of these

Answer 1

Answer: (B)

11

Answer 2

Let $O$ be the center of the inscribed circle and the rhombus. Let the vertices of the rhombus be $A, B, C, D$ such that $\angle A = 60°$. Then $\angle B = 120°$, $\angle C = 60°$, $\angle D = 120°$. The diagonals of the rhombus bisect each other at right angles at $O$.

Consider the right-angled triangle $\triangle OAB$. The angles are $\angle OAB = 60°/2 = 30°$ and $\angle OBA = 120°/2 = 60°$. Let $OA = d_1$ and $OB = d_2$. In $\triangle OAB$: $\tan(30°) = \frac{OB}{OA} \implies \frac{1}{\sqrt{3}} = \frac{d_2}{d_1} \implies d_1 = \sqrt{3} d_2$. The distance to the nearest vertex is given as 1. Since $d_1 = \sqrt{3} d_2$, $d_2$ is the shorter distance. Thus, $OB = OD = d_2 = 1$. Consequently, $OA = OC = d_1 = \sqrt{3} \times 1 = \sqrt{3}$.

The radius $r$ of the inscribed circle is the altitude from $O$ to the hypotenuse $AB$ in $\triangle OAB$. The area of $\triangle OAB$ can be calculated in two ways: Area $= \frac{1}{2} \times OA \times OB = \frac{1}{2} \times \sqrt{3} \times 1 = \frac{\sqrt{3}}{2}$. Also, Area $= \frac{1}{2} \times AB \times r$. The length of the side $AB = \sqrt{OA^2 + OB^2} = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3+1} = 2$. So, Area $= \frac{1}{2} \times 2 \times r = r$. Equating the areas, $r = \frac{\sqrt{3}}{2}$.

Let $P$ be any point on the inscribed circle. We want to find $|PA|^2 + |PB|^2 + |PC|^2 + |PD|^2$. We can use the property that for any point $P$ and a set of points $V_1, V_2, \dots, V_n$, if $O$ is the centroid of these points, then $\sum_{i=1}^n |PV_i|^2 = n|PO|^2 + \sum_{i=1}^n |OV_i|^2$. In our case, $n=4$ and $O$ is the centroid of the vertices $A, B, C, D$ of the rhombus. $P$ is on the circle with center $O$, so $|PO|^2 = r^2$. The sum is $S = 4|PO|^2 + |OA|^2 + |OB|^2 + |OC|^2 + |OD|^2$. $S = 4r^2 + OA^2 + OB^2 + OC^2 + OD^2$. Substitute the values: $r^2 = (\frac{\sqrt{3}}{2})^2 = \frac{3}{4}$. $OA^2 = (\sqrt{3})^2 = 3$. $OB^2 = 1^2 = 1$. $OC^2 = (\sqrt{3})^2 = 3$. $OD^2 = 1^2 = 1$.

$S = 4\left(\frac{3}{4}\right) + 3 + 1 + 3 + 1$ $S = 3 + 8$ $S = 11$.

Alternatively, using coordinates: Let $O$ be the origin $(0,0)$. The vertices are $A=(\sqrt{3}, 0)$, $B=(0, 1)$, $C=(-\sqrt{3}, 0)$, $D=(0, -1)$. Let $P=(x,y)$ be a point on the circle $x^2+y^2 = r^2 = \frac{3}{4}$. $|PA|^2 = (x-\sqrt{3})^2 + y^2 = x^2 - 2\sqrt{3}x + 3 + y^2$. $|PB|^2 = x^2 + (y-1)^2 = x^2 + y^2 - 2y + 1$. $|PC|^2 = (x+\sqrt{3})^2 + y^2 = x^2 + 2\sqrt{3}x + 3 + y^2$. $|PD|^2 = x^2 + (y+1)^2 = x^2 + y^2 + 2y + 1$. Summing these: $|PA|^2 + |PB|^2 + |PC|^2 + |PD|^2 = (x^2+y^2 - 2\sqrt{3}x + 3) + (x^2+y^2 - 2y + 1) + (x^2+y^2 + 2\sqrt{3}x + 3) + (x^2+y^2 + 2y + 1)$ $= 4(x^2+y^2) + (-2\sqrt{3}x + 2\sqrt{3}x) + (-2y + 2y) + (3+1+3+1)$ $= 4(x^2+y^2) + 8$. Since $x^2+y^2 = \frac{3}{4}$, Sum $= 4(\frac{3}{4}) + 8 = 3 + 8 = 11$.