Consider two points: $A=(0,0)$ which is the center of a unit circle, and $B=(0,1)$ which lies on that unit circle. Now, you choose a third point $C$ inside the circle uniformly at random.

What is the probability that you will be able to draw a square such that all three points $A, B, C$ lie on two adjacent sides of the square?

Please submit your answer to two decimal places.

The answer is 0.75.

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Could you provide theoretical proof on how you determined that the third point would have to be in the pink region?

Utsav Garg
- 4 years ago

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For the bottom half of the whole circle, you can draw a line perpendicular to the y-axis, connecting the third point with the first and second point. For the circle above the x axis, knowing that the angle formed between the 2 opposite ends of the diameter and a third point on the circumference of the circle is a right angle, we know that any point along the circumference is possible. Now, if we were to extend the point beyond the circle, the angle formed between the three points would become acute, and attempting to make the line shorter to reform a right angle would disconnect the third point. In contrast, we can move the third point freely inside the circle along the lines connecting its original location on the circumference of the circle, therefore, any point in the inner part of that circle is possible for the third point to reside. You can see my explanation to this problem further down in the discussions.

Kevin Tong
- 4 years ago

How u determined, the upper half area that the 3rd point shd lie within it, must be a circle?

Rajendra Prasad
- 4 years ago

We claim three points $A, B, C$ can be located on two adjacent sides of a square iff they do not make an acute triangle.

The forward implication is easy. If they are collinear, it's obvious (draw a square using that line). Otherwise, suppose $AB$ is on one side of the square and $C$ is on another; let they meet at $D$ , and $B$ is closer to $D$ than $A$ . Then $\angle DBC \le 90^\circ$ (because $\angle BDC = 90^\circ$ ), and $\angle ABC$ is complementary to $\angle DBC$ , so $\angle ABC \ge 90^\circ$ .

The backward implication is also simple. Suppose $\angle ABC \ge 90^\circ$ . Draw line $l$ passing $AB$ , and draw line $m$ perpendicular to $l$ passing $C$ . Let $l,m$ intersect on $D$ . Let $r$ be larger than $AD, CD$ and draw a circle $c$ centered on $D$ . It intersects $l$ at $E$ such that $A,B$ are in between $D,E$ , and intersects $m$ at $G$ such that $C$ is between $D,G$ . Now $D,E,G$ can be completed to a square $DEFG$ ; it contains $A,B,C$ on two adjacent sides ( $DE$ , $DG$ ), as desired.

Thus the question is, if we pick the third point $C$ randomly, what is the probability that $A(0,0), B(0,1)$ , and $C$ do not form an acute triangle? We'll look at where the non-acute angle is.

Clearly it's not $\angle B$ ; $C$ must be on the halfplane $y \ge 1$ for that to happen. So either $\angle A$ or $\angle C$ is the non-acute angle. If $\angle A$ is the non-acute angle, then $C$ lies on the halfplane $y \le 0$ . If $\angle C$ is the non-acute angle, then $C$ lies inside the circle of diameter $AB$ . You can look at @Geoff Pilling 's solution for where $C$ can be. The halfplane has area $\frac{\pi}{2}$ (we only consider the portion inside the unit circle), and the circle with diameter $AB$ has area $\frac{\pi}{4}$ . They don't intersect, so the total area is $\frac{3\pi}{4}$ . This is out of $\pi$ , the area of the unit circle, so the probability is $\frac{3}{4} = \boxed{0.75}$ .

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I'm sorry but I really don't understand your solution, first thing I don't get is how are you supposing angle ABC to be greater than or equal to 90°, which I'm pretty sure is not possible if C is placed inside the circle? I would really appreciate your help, thanks

Utsav Garg
- 4 years ago

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The first portion of the proof is a lemma; I just used the same letters.

Ivan Koswara
- 4 years ago

Notice how any point under the x-axis and inside the circle is a possible point since we can draw a line through the point from the y-axis. Therefore, we can include, the bottom half of the circle is usable for the point C. Now we check above the x-axis.

Knowing that a point on a circle forms a right angle between 2 points on opposite sides of the diameter of the circle, we can start to figure out the bounds of C above the x-axis. We draw a circle of diameter 1, with A and B as the opposite ends of the diameter. Notice, if we were to extend the line AC further out of the circle, the line formed by the end point becomes an acute angle, making it impossible to reform a right angle adjacent side connecting A, B and C. Also, notice you can still form a square anywhere inside of the circle with diameter one (You can simply move the point C along the original side of the square). Therefore, we can conclude that the entire circle of diameter one above the x-axis is possible for point C to reside in.

Finally, using what we've discovered, we find the probability to be $\frac{\frac{\pi}{2} + {\pi \cdot (\frac{1}{2})^2}}{\pi} \\ \frac{\frac{3\pi}{4}}{\pi} \\ \frac{3}{4} = \boxed{0.75}$

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The point can be chosen either from upper half or lower half of the unit circle.

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Case(1)
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When third point is selected from lower half.Let position of selected point is
$(h,k)$

First construct a straight line by connecting $(0,0)$ and $(h,k)$ .Then drop a perpendicular form $(0,1)$ on this line.Then we have to select the side of the square in such a way that the chosen point will be within or on the side.Then we can get our required side length of the square.So, we can complete the square.

There for every points in the lower half all points will satisfy the condition.

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Case(2)
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When the third point will be chosen from upper half.Let the point is
$(h,k)$

Construct the line by connecting $(0,0)$ and $(h,k)$ .The equation of line $L_1 : y=\dfrac{kx}{h}$

Then drop a perpendicular from $(0,1)$ on $L_1$ .The equation of these line is $L_2 : y=-\dfrac{hx}{k} +1$

By solving $L_1,L_2$ we can get the intersection point which is the vertex of the square.Let intersection point is $(a,b)$

$a=\dfrac{hk}{h^2+k^2} \\ b=\dfrac{k^2}{h^2+k^2}$

To make a square the co-ordinate of the vertex should be greater than the chosen point (Now consider only first quadrant)

So, $\dfrac{hk}{h^2+k^2} > h \implies h >0 ; h^2+k^2-k<0 \implies h>0 ; h^2+(k-1/2)^2 < (1/2)^2$

Same also for $k$ .If $k>0$ then $h^2+(k-1/2)^2 < (1/2)^2$

So, for every points in first quadrant the square possible if the third point lies within the right half circle whose center at $(0,1/2)$ with radius $1/2$ .For 2nd quadrant as there is a symmetry then these point will lie within the left half of the circle.

So, for upper half all points should lie within the circle.

The required probability is $\dfrac{\pi (1)^2/2 + \pi .(1/2)^2}{\pi(1)^2}=0.75$

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The locus of points for which this is possible is the pink region in this figure:

A partial argument is as follows...

The third point needs to either be in the top half of the unit circle (the larger circle in the diagram above) or the bottom half.

If it is in the bottom half, then a square can be constructed by extending the line segment from that point to the origin until it is perpendicular to a second line segment that connects to $(0,1)$ , and this will be the corner of your square. And this is possible for all points in the bottom half.

If it is in the top half, then a corner of the square can be constructed the same way as above, but that is only possible within the smaller circle shown above, since otherwise the intersection will be between the origin and the new point which won't form a corner of a square.

(Will try to come up with a more rigorous proof soon)

So, the probability will be given by:

$P = \dfrac{\text{pink area}}{\text{area of unit circle}}$

$P = \dfrac{\frac{\pi}{2}+ \frac{\pi}{4}}{\pi}$

$P = \boxed{0.75}$