TOP 10 problems of Week#1

This week was full Geo and NT 😊 .

Do try all problems first!! And if you guys get any nice solutions , do post in the comments section!

Here are the walkthroughs of this week's top 5 geo problems!

5th position (PUMac 2009 G8): Consider $\Delta ABC$ and a point $M$ in its interior so that $\angle MAB = 10^{\circ}, \angle MBA = 20^{\circ}, \angle MCA =30^{\circ}$ and $\angle MAC = 40^{\circ}$. What is $\angle MBC$? 

Walkthrough: a. Take $D$ as a point on $CM$ such that $\angle DAC=30^{\circ}$, and define $BD\cap AC=E$ . So $\Delta DAC$ is isosceles .

b. Show M is the incentre of $\Delta ABD$

c. Show $\angle EDC=60^{\circ}$

d. Show $\Delta BAC$ is isosceles .

e. So $\boxed{\angle MBC=60^{\circ}}$

4th position (IMO SL 2000 G4): Let $ A_1A_2 \ldots A_n$ be a convex polygon, $ n \geq 4.$ Prove that $ A_1A_2 \ldots A_n$ is cyclic if and only if to each vertex $ A_j$ one can assign a pair $ (b_j, c_j)$ of real numbers, $ j = 1, 2, \ldots, n,$ so that $ A_iA_j = b_jc_i - b_ic_j$ for all $ i, j$ with $ 1 \leq i < j \leq n.$

Walkthrough : Thanks to crystal1011 :)

a. Ptolemy is OP for both the cases

b. For the case where real numbers exists; prove it by ptolemy !

c. For the other case , we need to find one construction, find one!

d. $b_2=A_1A_2, c_1=1,b_1=0,c_2=0 $ . What can you say about $b_j$ and $c_j$?

e. Find about $b_j$ using formula on $A_1A_j$ and $c_j$ using formula on $A_2A_j$.

f. verify by ptolemy!

3rd position (AIME 2010 I P15):In $ \triangle{ABC}$ with $ AB = 12$, $ BC = 13$, and $ AC = 15$, let $ M$ be a point on $ \overline{AC}$ such that the incircles of $ \triangle{ABM}$ and $ \triangle{BCM}$ have equal radii. Let $ p$ and $ q$ be positive relatively prime integers such that $ \tfrac{AM}{CM} = \tfrac{p}{q}$. Find $ p + q$.

Walkthrough:  I kept it in 3rd position, not because it's cute or anything, I just want the people to go through the lethal pain I went through while solving this!

a. Denote $I_1,I_2$ as the centres and $D_1, D_2$ as the touch points. Let $AM=2x, CM=15-2x, BM=2y$

b. Use formula $ r = \sqrt {\frac {(s - a)(s - b)(s - c)}{s}}$ to get 2 equations. $(6+y-x)(x+6-y)= (-x+y+1)(x+y+6)$

c. The equations should be $r^2=\frac {(x+y-6)(6+y-x)(x+6-y)}{x+y+c}$ and $r^2=\frac{(x+y-1)(-x+14-y)(y+1-x)}{-x+y+14}$

d. Show $r^2=MD_1\cdot MD_2$ using the fact $\Delta MD_1I_1\sim \Delta ID_2M$.

e. Now we need to solve these equations, which I suffered 'cause I did a lot of sillies, anyways we get $(6+y-x)(x+6-y)= (-x+y+1)(x+y+6)$ 

$\implies -x^2+2xy-y^2+36=-x^2-5x+y^2+7y+6$ 

$ \implies 2y^2+7y-2xy-5x-30=0$ .

 Similarly $(x+y-1)(-x+14-y)=(x+y-6)(-x+y+14)$ 

$\implies -x^2-2xy+15x-y^2+15y-14=-x^2+20x+y^2+8y-84$ 

$\implies 2y^2 -7y+2xy+5x-70=0 $

f. Now it's trivial , and we get $\boxed{y=5}$. Find rest on your own :P.

2nd position (AIME II 2016/10): Triangle $ABC$ is inscribed in circle $\omega$. Points $P$ and $Q$ are on side $\overline{AB}$ with $AP<AQ$. Rays $CP$ and $CQ$ meet $\omega$ again at $S$ and $T$ (other than $C$), respectively. If $AP=4,PQ=3,QB=6,BT=5,$ and $AS=7$, then $ST=\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$ 

Walkthrough: Thanks to Crystal1011

a. Project through $C$ and notice  $ (A,Q;P,B)=(A,T;S,B) $ . Done!

1st Position (USAJMO 2016 P5): Let $\triangle ABC$ be an acute triangle, with $O$ as its circumcenter. Point $H$ is the foot of the perpendicular from $A$ to line $\overleftrightarrow{BC}$, and points $P$ and $Q$ are the feet of the perpendiculars from $H$ to the lines $\overleftrightarrow{AB}$ and $\overleftrightarrow{AC}$, respectively.

Given that$$AH^2=2\cdot AO^2,$$prove that the points $O,P,$ and $Q$ are collinear.

Walkthrough: a. $2AO^2$ looks nice.. introduce antipode of $A$(say $A'$)

b. invert wrt $A$ with radius $AH$.

c. Note that $O\rightarrow A', P\rightarrow B, Q\rightarrow C$. Conclude!

Next are the walkthroughs of this week's top 5 Number theory problems!

5th position(IMO 2009/1): Let $ n$ be a positive integer and let $ a_1,a_2,a_3,\ldots,a_k$ $ ( k\ge 2)$ be distinct integers in the set $ { 1,2,\ldots,n}$ such that $ n$ divides $ a_i(a_{i + 1} - 1)$ for $ i = 1,2,\ldots,k - 1$. Prove that $ n$ does not divide $ a_k(a_1 - 1).$

Walkthrough: a. For the sake of contradiction assume $n|a_k(a_1 - 1).$ 

b. Note that $a_ia_{i+1} \equiv a_i \pmod n\text{ for }i=1,2, \dots , k-1.$

c. Show $ a_1 \equiv a_1a_2a_3\cdots a_k \pmod n. $

d. So $a_1\equiv a_2 \pmod n$. contradiction

4th position (HMMT Feb 2017 NT): Find all pairs of positive integers $(a, b)$ for which $ab$

divides $a^{2017} + b.$

Walkthrough: a. Since $ab|a^{2017} + b \implies a|b $ . So let $b=b_1a$

b. Again we get $a^2b_1|a^{2017} + ab_1 \implies a|b_1 $. So let $b_1=b_2a$ , and the process continues .

c. Finally show $ab_{2017}|1+b_{2017}$

d. Hence $a=1,2 $ . 

e. Conclude using the fact that $b|a^{2017}$

3rd position (IMO 2013 N1): Let $\mathbb{Z} _{>0}$ be the set of positive integers. Find all functions $f: \mathbb{Z} _{>0}\rightarrow \mathbb{Z} _{>0}$ such that

$$ m^2 + f(n) \mid mf(m) +n $$

for all positive integers $m$ and $n$.

Walkthrough: Part b and c are useless TBH.

 a. take $P(n,n)$ and show $n\leq f(n)$ (for $n>1$)

b. with $P(x,1)$, where $f(1)=x$ ,show that $1=f(1)$ 

c. with $P(2,2)$, show that $f(2)=2$.

d. with $P(2,x)$, show that $f(x)\le x$, so $f(x)=x$

2nd position (IMO ShortList 2004, number theory problem 3):Find all functions $ f: \mathbb{N^{*}}\to \mathbb{N^{*}}$ satisfying

$ \left(f^{2}\left(m\right)+f\left(n\right)\right) \mid \left(m^{2}+n\right)^{2}$

for any two positive integers $ m$ and $ n$.

Walkthrough: a. take $P(1,1)$ and show $f(1)=1$

b. take $P(1,p-1)$ for prime $p$ , show $f(p-1)=p-1$ or $p(p-1) $

c.take $P(p-1,1)$ , show that if $f(p-1)=p(p-1)$ then $(p(p-1))^2 +1 \leq (p^2 - 2p + 2)^2 $ . which is not possible for large $p$. So $f(p-1)=p-1$.

d. take $P(x,n)$ , where x is a  very large number of the form $p-1 $. 

e. Show that $x^2+f(n)|(f(n)-n)^2 \implies f(n)=n$

1st position(APMO 2009/P4): Prove that for any positive integer $ k$, there exists an arithmetic sequence $ \frac{a_1}{b_1}, \frac{a_2}{b_2}, \frac{a_3}{b_3}, ... ,\frac{a_k}{b_k}$ of rational numbers, where $ a_i, b_i$ are relatively prime positive integers for each $ i = 1,2,...,k$ such that the positive integers $ a_1, b_1, a_2, b_2, ...,  a_k, b_k$ are all distinct.

Walkthrough: Credits to anser and SnowPanda . This very intuitive problem, and in my opinion, walkthrough won't be that good.

a. try to introduce $k!$ as denominator.  What about $\frac{1}{k!}, \frac{2}{k!},\dots \frac{k}{k!}$ ? This does form an AM sequence and we can reduce it to lowest form too , but this doesn't ensure that the  numerators and denominators will be different. 

b. One way to ensure this is multiply some large $p$ ( should be greater than $k$). 

c. Still $\frac{p}{k!},\frac{ p}{(k!/2)}, ..., \frac{p }{(k!/k)}$ doesn't satisfy all the a_i's to be different 

d. So what about  $\frac{p(k! + 1)}{k!},\frac{ p(k!/2 + 1)}{(k!/2)}, ..., \frac{p(k!/k + 1)}{(k!/k)}$ for some very large prime $p$ ? Show that this works!

So these were my top 10 ! I personally loved the 1st position geo problem 😊. It was first giving so bad computational vibes, but it turned out to be so good! 

What are your top 10s, do write in the comments section (at least write something ! I will be happy to hear your comments ). Follow this blog if you want to see more contest math problems! See you all soon 😊.

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I also compiled them in a pdf here https://drive.google.com/file/d/1OB-lFxxPDP_SbaK4yy0k8lTGYiE1_y9G/view?usp=sharing

Sunaina 💜