Weird Theorems

Welcome back to this blog.

 Oki it's a new day and I decided to prove some weird theorems/lemmas with weird names. So, you can try proving it too! 

In my list we have, 

• Sawayama theorem
• Butterfly Theorem
• Iran lemma
• Simon's Trick
• Chicken McNugget Theorem
• Pick's theorem
• Fermat's Christmas Theorem

Sawayama Thebault Theorem: Let $ABC$ be a triangle with circumcircle $\Gamma$ and incenter I. Let $D \in BC.$ Let $\Omega_1$ be the circle tangent to the line segments $DA$ and $DB$ and to the circle $\Gamma$, and let $\Omega_2$ be the circle tangent to the line segments $DA$ and $DC$ and to the circle $\Gamma$. If $O_1$ and $O_2$ are the centres of  $\Omega_1$ and $\Omega_2$, respectively, prove that $O_1 -I -O_2$ are collinear.

Walkthrough: 

• Note $I\in EF,~~I\in GH$
• Prove $O_1DO_2=90$
• $DO_1|| GH$ and $DO_2 || EF.$
• Since we want to prove that $I \in O_1 O_2 $, let $EF \cap O_1 O_2 = I '$ and let $GH \cap O_1 O_2 = I".$ 
• $$\frac {O_1I'}{I'O_2}=\frac{O_1X}{DX}, \frac {O_1I''}{O_2I"}=\frac{DY}{O_2Y}$$
• Note that $$\frac{O_1X}{DX}=\frac{DY}{O_2Y}$$

Butterfly's theorem: Let $M$ be the midpoint of a chord $PQ$ of a circle $\omega,$ through which two other chords $AB$ and $CD$ are drawn. Let $AD \cap PQ = X$ and $BC \cap PQ = Y.$ Prove that $M $ is also the midpoint of $XY.$

Walkthrough: Introduce the midpoint and then drop perps.

Introduce the midpoints of $AD$ and $CB.$

• Note that $$\frac{AD}{AM}=\frac{CB}{CM}$$
• So $$\frac{AR}{AM}=\frac{CS}{CM}$$
• So we have $AMR$ similar to $CMS.$
• So by cycilcity stuff $\angle MOX=\angle MOY.$

Incenter perpendicularity: The incircle of $\Delta ABC$ is tangent to $BC, CA, AB$ at $D, E, F$ respectively. Let $M_{BC}$ and $M_{AC}$ be the midpoints of $BC$ and $AC.$ If $K$ is the intersection of lines $BI$ and $EF, $ then $BK \perp KC.$ In addition, $K$ lies on line $MN.$

Walkthrough: 

• Define $K$ as a point on $BI$ such that $BI\cap CK.$ So $I, D, E,K, C$ is cyclic. Now, note that $M_{BC}$ is the centre of the right angle triangle $BKC.$ So $$\angle KM_{BC}C=2\angle IBC=\angle B\implies M_{BC}-M_{AC}-K$$
• Note that $$\angle EKI=\angle ECI=C/2=\angle (FE,BI)\implies F-E-K$$

Simon's trick: It's pretty famous and the origin of the name is discussed here. Also, Simon is Akshay Venkatesh's student :O

$$ mn+m+n+1= (m+1)(n+1)$$

Yeah, that's it.

The Chicken McNugget Theorem: For any two relatively prime positive integers $m,n$, the greatest integer that cannot be written in the form $am + bn$ for nonnegative integers $a, b$ is $mn-m-n$.

Moreover, exactly $\frac{(m - 1)(n - 1)}{2}$ positive integers which cannot be expressed in the form $am + bn$.

Walkthrough: 

I think the picture is enough :P

Pick's theorem: The area of a lattice  polygon is

$$A = I + \frac{1}{2}B - 1$$

where $I$ is the number of lattice points in the interior and $B$ is the number of lattice points on the boundary.

Proof: Define nice triangles as triangles with no interior points. Clearly nice triangles satisfy the theorem.

 Now, for any arbitrary triangle, we kinda use induction. And consider a point inside the arbitrary triangle ( there exist, else the triangle is a nice triangle). And then find the area of new triangles ( where we can again divide and so on, till we get nice triangles).

Now $$[ABC]=[AOC]+[BOC]+[AOB]$$ 

$$ [AOB]=I_1+\frac{B_1+B_2+B_4}{2}$$

$$ [AOC]=I_2+\frac{B_1+B_3+B_5}{2}$$

$$ [OBC]=I_3+\frac{B_3+B_2+B_6}{2}$$

Now, the boundary points $B_1,B_2,B_3$ are actually the interior point of $ABC.$

So $$[AOC]+[BOC]+[AOB]= I_1+\frac{B_1+B_2+B_4}{2}+I_2+\frac{B_1+B_3+B_5}{2} +I_3+\frac{B_3+B_2+B_6}{2}$$ 

$$= [I_1+I_2+I_3+ B_1+B_2+B_3]+\frac{B_4+B_5+B_6}{2}$$

And then we can divide the polygon into triangles and proceed the same.

Fermat's Christmas Theorem: An odd prime can be written as the sum of two squares iff the prime is $1 \pmod 4.$

Direction1: If the odd prime is $3\pmod 4$ then it can not be written as a sum of two squares.

Proof: Consider $\mod 4,$ an odd square is always $0,1,\mod 4.$

The other direction uses Gaussian integers and more fancy stuff..so I will probably dedicate a whole blogpost on it :P

Moreover, we also have, 

Corollary: If $p$ is an odd prime $p\equiv 3  $ mod $4$ and $p|(x^2+y^2)$ then prove that $p|x$ and $p|y$

Proof: Let $p=4k+3,$ and $p\not\mid x, p\not\mid y,$

$$x^2\equiv -y^2\mod p $$

$$(x^2)^{2k+1}\equiv (-y^2)^{2k+1}\mod p $$

$$(x^2)^{p-1}\equiv -(y^2)^{p-1}\mod p $$

By FLT, we get a contradiction. 

So yeee! That's it for today's blog post! I think the names are very funny :P and proving them was equally fun! Summary of 2021 blog post coming soon :P

See ya soon! ( thanks to Anand, now I too say "ya" instead of "you")

Sunaina💜