Question

A circle is inscribed into a rhombus $ABCD$ with one angle $60°$. The distance from the 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 the rhombus be $ABCD$ with side length $a$. Let the angle at vertex $B$ be $60°$, so the adjacent angle at $A$ is $120°$. The diagonals bisect each other at right angles at the center $O$. Let $OA = d_1/2$ and $OB = d_2/2$. In the right-angled triangle $\triangle OAB$, $\angle OAB = 120°/2 = 60°$ and $\angle OBA = 60°/2 = 30°$. We have $OA = a \cos(60°) = a/2$ and $OB = a \sin(60°) = a\sqrt{3}/2$. The radius $r$ of the inscribed circle is the altitude from $O$ to the side $AB$. The area of $\triangle OAB = \frac{1}{2} OA \cdot OB = \frac{1}{2} (a/2)(a\sqrt{3}/2) = \frac{a^2\sqrt{3}}{8}$. Also, the area of $\triangle OAB = \frac{1}{2} AB \cdot r = \frac{1}{2} a \cdot r$. Equating the areas: $\frac{a^2\sqrt{3}}{8} = \frac{ar}{2} \implies r = \frac{a\sqrt{3}}{4}$.

The problem states "The distance from the vertex is equal to 1". Given the integer options, it is highly probable that this phrase implies the distance from the center $O$ to one of the vertices is 1. Let's test this interpretation.

Case 1: $OA = 1$ If $OA = 1$, then $a/2 = 1 \implies a = 2$. Then $OB = a\sqrt{3}/2 = 2\sqrt{3}/2 = \sqrt{3}$. The radius $r = \frac{a\sqrt{3}}{4} = \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2}$. The sum of squared distances from any point $P$ on the circle to the vertices is given by $S = 4r^2 + 2(OA^2 + OB^2)$. $S = 4(\frac{\sqrt{3}}{2})^2 + 2(1^2 + (\sqrt{3})^2) = 4(\frac{3}{4}) + 2(1 + 3) = 3 + 2(4) = 3 + 8 = 11$.

Case 2: $OB = 1$ If $OB = 1$, then $a\sqrt{3}/2 = 1 \implies a = 2/\sqrt{3}$. Then $OA = a/2 = (2/\sqrt{3})/2 = 1/\sqrt{3}$. The radius $r = \frac{a\sqrt{3}}{4} = \frac{(2/\sqrt{3})\sqrt{3}}{4} = \frac{2}{4} = \frac{1}{2}$. $S = 4r^2 + 2(OA^2 + OB^2) = 4(\frac{1}{2})^2 + 2((\frac{1}{\sqrt{3}})^2 + 1^2) = 4(\frac{1}{4}) + 2(\frac{1}{3} + 1) = 1 + 2(\frac{4}{3}) = 1 + \frac{8}{3} = \frac{11}{3}$.

The phrasing "The distance from the vertex is equal to 1" is ambiguous. If it means the distance from the vertex to the inscribed circle, the calculation leads to $44/3$ or $308+176\sqrt{3}$, neither of which matches the options. However, if we interpret "The distance from the vertex is equal to 1" as the distance from the center of the rhombus to one of its vertices, and specifically $OA=1$ (where $A$ is adjacent to the $60^\circ$ angle), we get $S=11$, which is option (B). This interpretation aligns with the presence of integer options.

Let's verify this interpretation. If $OA=1$, then $a=2$. The vertices are $A(1,0)$, $C(-1,0)$, $B(0, \sqrt{3})$, $D(0, -\sqrt{3})$. The circle is centered at $O(0,0)$ with radius $r=\sqrt{3}/2$. Let $P(x,y)$ be a point on the circle, so $x^2+y^2 = r^2 = 3/4$. $|PA|^2 = (x-1)^2 + y^2 = x^2 - 2x + 1 + y^2 = (x^2+y^2) - 2x + 1 = 3/4 - 2x + 1 = 7/4 - 2x$. $|PC|^2 = (x+1)^2 + y^2 = x^2 + 2x + 1 + y^2 = (x^2+y^2) + 2x + 1 = 3/4 + 2x + 1 = 7/4 + 2x$. $|PB|^2 = x^2 + (y-\sqrt{3})^2 = x^2 + y^2 - 2\sqrt{3}y + 3 = (x^2+y^2) - 2\sqrt{3}y + 3 = 3/4 - 2\sqrt{3}y + 3 = 15/4 - 2\sqrt{3}y$. $|PD|^2 = x^2 + (y+\sqrt{3})^2 = x^2 + y^2 + 2\sqrt{3}y + 3 = (x^2+y^2) + 2\sqrt{3}y + 3 = 3/4 + 2\sqrt{3}y + 3 = 15/4 + 2\sqrt{3}y$.

Summing these: $|PA|^2 + |PB|^2 + |PC|^2 + |PD|^2 = (7/4 - 2x) + (15/4 - 2\sqrt{3}y) + (7/4 + 2x) + (15/4 + 2\sqrt{3}y)$ $= (7/4 + 7/4) + (15/4 + 15/4) + (-2x+2x) + (-2\sqrt{3}y+2\sqrt{3}y)$ $= 14/4 + 30/4 = 44/4 = 11$.

This confirms that if the distance from the center to the vertex adjacent to the $60^\circ$ angle is 1, the sum is 11.