Let $\mathcal{P}$ be a regular $2022$-gon with area $1$. Find a real number $c$ such that, if points $A$ and $B$ are chosen independently and uniformly at random on the perimeter of $\mathcal{P}$, then the probability that $AB \geq c$ is $\tfrac{1}{2}$. Espen Slettnes
Problem
Source: ELMO 2022 P3
Tags: geometry, Probabilities, ELMO 2022
CANBANKAN
24.06.2022 06:28
REDACTED: integration doesn't work
goodbear
24.06.2022 06:30
My problem!
Let $n=505$ and let the polygon be $P_1P_2\dots P_{4n+2}$. WLOG, assume that $A\in P_1P_{4n+2}$. By symmetry, we can assume that $B$ is closer to $P_1$ than to $P_{4n+2}$. Then, since the chance that $B$ is between $P_{4n+2}$ and $P_n$ is equal to the chance that $B$ is between $P_{n+2}$ and $P_{2n+2}$, and furthermore because $AB$ is always smaller in the former case than in the latter, we may assume that $B\in P_nP_{n+1}\cup P_{n+1}P_{n+2}$.
[asy][asy]
unitsize(1inch); int n=4; pair P(int i){return dir(-180*(i+n)/(2*n+1));}
for(int i=1;i<=4*n+2;++i){draw(P(i-1)--P(i),grey);}
pen p1=royalblue,p2=orange;
dot("$P_1$",P(1),P(1),p1);
dot("$P_{4n+2}$",P(4n+2),P(4n+2),p1);
draw(P(1)--P(4n+2),p1);
dot("$P_n$",P(n),P(n),p2);
dot("$P_{n+1}$",P(n+1),P(n+1),p2);
dot("$P_{n+2}$",P(n+2),P(n+2),p2);
path B=P(n)--P(n+1)--P(n+2);
draw(B,p2);
[/asy][/asy]
Now, instead of fixing the polygon and varying $A$, we fix $A$ and vary the polygon. Let $\mathcal B$ be the locus of $B$:
[asy][asy]
unitsize(1inch); int n=4; pair P(int i){return dir(-180*(i+n)/(2*n+1));}
for(int j=0;j<=1;++j)for(int i=1;i<=4*n+2;++i){draw(shift(-P(j))*(P(i-1)--P(i)),darkgrey+dotted);}
pen p1=royalblue,p2=orange;
path B=P(n)--P(n+1)--P(n+2);
filldraw(shift(-P(0))*B--shift(-P(1))*reverse(B)--cycle,0.95*white+0.05*p2,p2);
label("$\mathcal B$",P(n+1)-P(1),1.5*W,p2);
dot("$A$",(0,0),S,p1);
[/asy][/asy]
Because the areas of the top and bottom halves of $\mathcal B$ are the same, and because $B$ is uniformly distributed in each half, $B$ is uniformly distributed across $\mathcal B$. Therefore, the circle of radius $c$ centered at $A$ bisects $\mathcal B$. By symmetry, it therefore also bisects the following "banana":
[asy][asy]
unitsize(1inch); int n=4; pair P(int i){return dir(-180*(i+n)/(2*n+1));}
for(int j=0;j<=1;++j)for(int i=1;i<=4*n+2;++i){draw(shift(-P(j))*(P(i-1)--P(i)),darkgrey+dotted);}
pen p1=royalblue,p2=rgb(13/15,11/15,0);
path B=P(1)--P(2);for(int i=1;i<2n;++i)B=B--P(i+2);
filldraw(shift(-P(0))*B--shift(-P(1))*reverse(B)--cycle,0.95*white+0.05*p2,p2);
dot("$A$",(0,0),S,p1);
draw(arc((0,0),sqrt((4n+2)/pi*sin(2pi/(4n+2))),27(5-2/n),27(5+2/n)),brown+dashed);
[/asy][/asy]
Rotating this picture $4n+2$ times about $A$ gives:
[asy][asy]
unitsize(1inch); int n=4; pair P(int i){return dir(-180*(i+n)/(2*n+1));}
for(int j=0;j<=1;++j)for(int i=1;i<=4*n+2;++i){draw(shift(-P(j))*(P(i-1)--P(i)),darkgrey+dotted);}
pen p1=royalblue,p2=rgb(13/15,11/15,0);
path B=P(1)--P(2);for(int i=1;i<2n;++i)B=B--P(i+2);
for(int i=0;i<4*n+2;++i)filldraw(rotate(180i/(2n+1))*(shift(-P(0))*B--shift(-P(1))*reverse(B)--cycle),0.95*white+0.05*p2,p2);
dot((0,0),p1);
draw(circle((0,0),sqrt((4n+2)/pi*sin(2pi/(4n+2)))),brown+dashed);
[/asy][/asy]
Therefore, the circle of radius $c$ centered at $A$ has half the total area of the banana bunch, which is a regular $(4n+2)$-gon with twice the side length and therefore quadruple the area of $P_1P_2\dots P_{4n+2}$. Therefore, $\pi c^2=4/2$, so $c=\boxed{\sqrt{2/\pi}}$.
The answer is different for non-$(4n+2)$-gons, as $B$ would not be uniformly distributed in the banana. For example, the answer for a square is $\frac{2\sqrt{\pi-1}-2}{\pi-2}$. Even though I wrote this problem in 2019, it could only work three years later. (This is also a reason why integration is doomed to fail.)
This problem had the ideal score distribution in my book among MOPpers. One person got 7, everyone else got 0.
CANBANKAN
24.06.2022 06:32
In my integration, I did use the idea of changing the focus, but not in the right way darn oops What a nice problem Espen! It's a pleasure to learn from you.
samrocksnature
24.06.2022 22:51
Espen howsopro problem this is like so cute but so hard
TLP.39
25.06.2022 13:59
goodbear wrote:
The answer is different for non-$(4n+2)$-gons, as $B$ would not be uniformly distributed in the banana. For example, the answer for a square is $\frac{2\sqrt{\pi-1}-2}{\pi-2}$. Even though I wrote this problem in 2019, it could only work three years later. (This is also a reason why integration is doomed to fail.)
Unless I messed up, there is a calc solution that uses the fact that $\angle (P_{4n+2}P_1,P_nP_{n+1})+\angle (P_{4n+2}P_1,P_{n+1}P_{n+2})=\pi$ and thus also only work for $(4n+2)$-gons. MarkBcc168 wrote: Let $\mathcal{P}$ be a regular $2022$-gon with area $1$. Find a real number $c$ such that, if points $A$ and $B$ are chosen independently and uniformly at random on the perimeter of $\mathcal{P}$, then the probability that $AB \geq c$ is $\tfrac{1}{2}$. Espen Slettnes Let the vertices of $\mathcal{P}$ be $P_i$ for $1\le i \le 4n+2$. Let $X$ be the intersection of $P_{4n+2}P_1$ and $P_nP_{n+1}$ and $Y$ be the intersection of $P_{4n+2}P_1$ and $P_{n+1}P_{n+2}$. Moreover, let $a$ be the polygon's side length. It's easy to prove that if $c\le |P_1P_{n+1}|$, then $\mathbb{P}(AB\ge c) <\frac12$ for all fixed $A$. Similarly, if $c\ge |P_1P_{n+2}|$, then $\mathbb{P}(AB\ge c) >\frac12$ for all fixed $A$. Thus, $|P_1P_{n+1}| < c < |P_1P_{n+2}|$. For any point $A$ on $P_{4n+2}P_1$, there exists a unique point $B_A$ on either $P_{n}P_{n+1}$ or $P_{n+1}P_{n+2}$ such that $|AB_A|=c$. Let $x_A:=|P_{4n+2}A|$, $A_0$ be the point such that $B_{A_0}=P_{n+1}$, and \[ y_A= \begin{cases} |XB_A| &\text{if }|P_{4n+2}A|\le |P_{4n+2}A_0|\\ |YB_A| &\text{if }|P_{4n+2}A|> |P_{4n+2}A_0| \end{cases} \] From the condition $\mathbb{P}(AB\le c)=\frac12$ and the fact that $|XP_{n+1}|=|YP_{n+1}|=\frac{a}{2\sin\frac{\pi}{4n+2}}$, we have $$\int_{0}^{a}y_Adx_A=\int_{0}^{x_{A_0}}y_Adx_A+\int_{x_{A_0}}^{a}y_Adx_A=\frac{a^2}{2\sin\frac{\pi}{4n+2}}$$ Moreover, since the area of $\mathcal{P}$ is $1$, we get that $\frac{(2n+1)\cot \frac{\pi}{4n+2}}{2}a^2=1$ and thus $\frac{a^2}{2\sin\frac{\pi}{4n+2}}=\frac{1}{(2n+1)\cos\frac{\pi}{4n+2}}$ To calculate both integrals in term of $c$, we need the following lemma:
For a fixed $\theta$ and the parametric curve given by
\[
\begin{cases}
x(\gamma) = 2r\sin\left(\frac{\pi-\theta}{2}+\gamma\right)\\
y(\gamma) = 2r\sin\left(\frac{\pi-\theta}{2}-\gamma\right)
\end{cases}\]we have
$$\int_{\gamma=-\alpha}^{\gamma=\alpha}ydx = 4r^2\alpha\sin(\theta)$$for all $\alpha\in\mathbb{R}$.
\begin{align*}
\int_{\gamma=-\alpha}^{\gamma=\alpha}ydx &= \int_{\gamma=-\alpha}^{\gamma=\alpha}4r^2 \cos\left( \frac{\pi-\theta}{2}+\gamma\right) \sin\left( \frac{\pi-\theta}{2}-\gamma\right)d\gamma\\
&= \int_{\gamma=-\alpha}^{\gamma=\alpha}2r^2 \left(\sin(\pi-\theta)-\sin(2\gamma)\right)d\gamma\\
&= 4r^2\alpha\sin(\theta)
\end{align*}
For convenience, let $z_A=|XA|=|XP_{4n+2}|-x_A$. We then have $\int_{0}^{x_{A_0}}y_Adx_A=\int_{z_{A_0}}^{z_{P_{4n+2}}}y_Adz_A$. By considering the triangle $\triangle XAB_A$, we see that this integral corresponds to the integral in the lemma with $\theta = \angle P_{4n+2}XP_{n+1}=\frac{\pi}{2}+\frac{\pi}{4n+2}$, $r=\frac{c}{2\sin(\theta)}$ and $\alpha = \frac{\pi-\theta}{2}-\angle XP_{n+1}A_0$. Thus, the first integral is equal to $\frac{c^2}{\cos\frac{\pi}{4n+2}}\left(\frac{\frac{\pi}{2}-\frac{\pi}{4n+2}}{2}-\angle XP_{n+1}A_0\right)$. Similarly, if we let $w_A=|YA|=|YP_{4n+2}|-x_A$, then $\int_{x_{A_0}}^{a}y_Adx_A=\int_{w_{P_1}}^{w_{A_0}}y_Adw_A$ corresponds to the integral in the lemma with $\theta = \angle P_{4n+2}YP_{n+1}=\frac{\pi}{2}-\frac{\pi}{4n+2}$, $r=\frac{c}{2\sin(\theta)}$ and $\alpha = \angle YP_{n+1}A_0-\frac{\pi-\theta}{2}$. Thus, this integral is equal to $\frac{c^2}{\cos\frac{\pi}{4n+2}}\left(\angle YP_{n+1}A_0-\frac{\frac{\pi}{2}+\frac{\pi}{4n+2}}{2}\right)$. Since $\angle YP_{n+1}A_0-\angle XP_{n+1}A_0=\frac{\pi}{2n+1}$, we get that the sum of the two integrals is $\frac{\pi c^2}{(4n+2)\cos\frac{\pi}{4n+2}}$. As this must be equal to $\frac{1}{(2n+1)\cos\frac{\pi}{4n+2}}$, we get that $c^2=\frac{2}{\pi}\implies c=\sqrt{\frac{2}{\pi}}$. Done.
goodbear
25.06.2022 17:07
TLP.39 wrote: goodbear wrote:
The answer is different for non-$(4n+2)$-gons, as $B$ would not be uniformly distributed in the banana. For example, the answer for a square is $\frac{2\sqrt{\pi-1}-2}{\pi-2}$. Even though I wrote this problem in 2019, it could only work three years later. (This is also a reason why integration is doomed to fail.)
Unless I messed up, there is a calc solution that uses the fact that $\angle (P_{4n+2}P_1,P_nP_{n+1})+\angle (P_{4n+2}P_1,P_{n+1}P_{n+2})=\pi$ and thus also only work for $(4n+2)$-gons. Consider me highly impressed. Kudos to you for your computational fortitude!
CANBANKAN
25.06.2022 18:04
I think what is remarkable is not his computational fortitude, but the way he setup his computations. Like my bashing is way longer than his, and even if I didn't make a mistake, I still didn't avoid the arcsin.
PIartist
26.06.2022 14:52
Quote: The answer is different for non-$(4n+2)$-gons, as $B$ would not be uniformly distributed in the banana. For example, the answer for a square is $\frac{2\sqrt{\pi-1}-2}{\pi-2}$. Even though I wrote this problem in 2019, it could only work three years later. (This is also a reason why integration is doomed to fail.) I can't quite understand why the argument fails for $4n$-gons, Why isn't it uniformly distributed?
goodbear
01.07.2022 22:20
PIartist wrote: Quote: The answer is different for non-$(4n+2)$-gons, as $B$ would not be uniformly distributed in the banana. For example, the answer for a square is $\frac{2\sqrt{\pi-1}-2}{\pi-2}$. Even though I wrote this problem in 2019, it could only work three years later. (This is also a reason why integration is doomed to fail.) I can't quite understand why the argument fails for $4n$-gons, Why isn't it uniformly distributed? Good question; the reason is pretty subtle in my opinion. The banana is composed of several rhombi, and $B$ has an equal chance of appearing in each. However, the rhombi have different areas, so $B$ is not uniformly distributed in the banana — it has a disproportionally high chance of appearing in the parallelograms with smaller area near the top and bottom.