(a) To solve the problem, we need to determine the region inside the equilateral triangle \(ABC\) where a point \(P\) is closer to the center \(O\) than to any of the vertices \(A\), \(B\), or \(C\).
First, let's find the medians of the equilateral triangle. Each median will divide the triangle into two equal areas, and all three medians intersect at point \(O\), the centroid of the triangle. We denote the centroid-to-vertex distance as \(R\).
Because the triangle is equilateral, each median also acts as an altitude. The ratio of centroid-to-vertex distance to the centroid-to-side midpoint distance is \(2:1\), giving the relation:
\[
\text{Centroid-to-midpoint distance} = \frac{R}{2}
\]
Now, consider drawing circles centered at \(O\) with radius \(R\). Each vertex of the triangle is at distance \(R\) from \(O\):
- Circle with radius \(R\) centered at \(O\): All points on this circle are exactly at distance \(R\) from \(O\), meaning they are equidistant to one of the vertices.
- Inside this circle, all points are closer to \(O\) than to any vertices.
The smaller inscribed circle inside the equilateral triangle, mentioned above with radius equal to the distance from the centroid to a midpoint of a side, occupies a part of the triangle. Thus, the desired region is the interior of this inscribed circle.
Now, we calculate the fraction of the area:
The area of an equilateral triangle is \(A = \frac{\sqrt{3}}{4} s^2\), where \(s\) is the side length. The area of the inscribed circle (radius \(\frac{s \sqrt{3}}{6}\)) is:
\[
\text{Area of inscribed circle} = \pi \left(\frac{s \sqrt{3}}{6}\right)^2 = \frac{\pi s^2}{12}
\]
Next, to find the probability:
\[
P = \frac{\text{Area of inscribed circle}}{\text{Area of triangle}} = \frac{\frac{\pi s^2}{12}}{\frac{\sqrt{3}}{4} s^2} = \frac{\pi s^2/12}{\sqrt{3} s^2 / 4} = \frac{\pi/12}{\sqrt{3}/4} = \frac{\pi}{12} \times \frac{4}{\sqrt{3}} = \frac{\pi}{3\sqrt{3}} = \frac{\pi \sqrt{3}}{9}
\]
Thus, the required probability of \(P\) being closer to \(O\) than to any vertex is:
\[
\boxed{\frac{\pi \sqrt{3}}{9}}
\]
(b) The problem here is computationally simpler: Let \(ABCD\) be the square and point \(P\) be inside the square. Calculate the condition where the area of triangle \(PAB\) is less than half of the area of the square.
We start by letting the area of the square be \(k\). Thus, each side of the square has length \(\sqrt{k}\). The condition \( \frac{1}{2} \text{Area of square}\) is half of \(k\), indicating our constraint deals with finding if:
\[
\text{Area of} \triangle PAB < \frac{k}{2}
\]
The typical area expression for \( \triangle PAB \) when \(A\) and \(B\) are vertices of the square is determined by the coordinates, setting vertices at \((0,0)\), \((\sqrt{k},0)\), and \((0,\sqrt{k})\). Using any potential point \(P\) inside, integrating over the potential coordinates verifies that less than similar equal-bisect partition applies.
Fortunately, straightforward expectation over the weighted uniform density ensures at half-partion space are alequivalent, giving thus:
\[
P = 0.5
\]
So the probability that triangle \(PAB\) will be less than halfway area in that context translates to:
\[
\boxed{\frac{1}{2}}
\]
(a) Let $ABC$ be an equilateral triangle, centered at $O.$ A point $P$ is chosen at random inside the triangle. Find the probability that $P$ is closer to $O$ than to any of the vertices. (In other words, find the probability that $OP$ is shorter than $AP,$ $BP,$ and $CP.$)
(b) Let $O$ be the center of square $ABCD.$ A point $P$ is chosen at random inside the square. Find the probability that the area of triangle $PAB$ is less than half the area of the square.
1 answer
1 answer