Problems done in July

 July was LIOG month. Just LIOG was done, and then we had MMC ( till July 14). SO from the last two weeks, I was doing LIOG only. And I did some non-geo too! Also, I am not adding all the problems I did. A few only in fact. One can go through my apps post ( Jelena_ivanchic)

Here are a few problems and solutions/walkthroughs:

Problem: Let $A_1$ be the intersection of tangent at $A$ to the circumcircle of a triangle $\Delta ABC$ with sideline $BC$. Similarly define $B_1,C_1$. Show, that $A_1,B_1$ and $C_1$ are collinear

Walkthrough: 

• Use this $\Delta ACA_1\sim \Delta BAA_1.$
• So we get $$\frac{AC}{BA}=\frac{CA_1}{AA_1}=\frac{AA_1}{BA_1}$$ $$\implies \frac{CA_1}{AA_1}\cdot \frac{AA_1}{BA_1} =\frac{AC^2}{BA^2}$$ $$\implies  \frac{AC^2}{BA^2}=\frac{CA_1}{BA_1}.$$ Now, apply menelaus on $ABC.$

Problem: Let $\Gamma$ be a circle and let $B$ be a point on a line that is tangent to $\Gamma$ at the point $A$. The line segment $AB$ is rotated about the center of the circle through some angle to the line segment $A’B’$. Prove that $AA’$ passes through the midpoint of $BB’$.

Walkthrough:

Solved with PC and Krutarth.

• Let $D$ be the midpoint of $BB',$ we get $\angle XAB=\angle XDB,$ hence $XADB$ cyclic.
• Now by similarity (spiral but let's not bring fancy words), we get $XBB'\sim XAA'$ ( actually isosceles but we don't need that fact).
• So $\angle  XBB'=\angle  XAA',$ but $\angle XBB'=\angle XAF,$ where $F$ is a point on ray $DA.$

Problem: Given an infinite sequence of numbers $a_1, a_2, a_3,...$ . For each positive integer $k$ there exists a positive integer $t = t(k)$ such that $a_k = a_{k+t} = a_{k+2t} =...$. Is this sequence necessarily periodic? That is, does a positive integer $T$ exist such that $a_k = a_{k+T}$ for each positive integer k?

Walkthrough: No, it’s not necessary.

Here’s the construction:

Let $X$ denote the place of the sequence, for example in sequence $1,2,3,4,\dots$ $1st$ place is $1.$

Then let $$1\rightarrow v_2(x)=0$$ $$2\rightarrow v_2(X)=1$$ $$3\rightarrow v_2(X)=2$$ $$4\rightarrow v_2(X)=3$$ $$5\rightarrow v_2(X)=4$$ $$\vdots$$

So the sequence looks like $$1213121412131215 \dots.$$ Clearly, it’s not periodic.

Problem: Find all pairs of natural numbers $(m,n)$ bigger than $1$ for which $2^m+3^n$ is the square of the whole number.

Walkthrough: Note that $\mod 3$ gives $m$ is even and $\mod 4$ gives $n$ is even. We get $$2^{2k}= (X^2-3^l)(X^2+3^l)\implies X^2-3^l= 2^a,~~ X^2+3^l=2^b \implies 2^a| 2^b \implies 2^a|2\cdot 3^l| \implies a=0 \text{or} 1$$ So $X^2-3^l=1, 2$ but squares are $0,1\mod 3.$

So $$X^2-3^l=1 \mod 3$$ Now we will try to solve $$X^2-3^l=1$$

Note that $$X^2-1=(X-1)(X+1)=3^l\implies X-1=3^w,X+1=3^v\implies 3^w|2\implies w=0 \implies X-1=1\implies 2.$$

Clearly, $X=2$ doesn't satisfy. So we don't have any solutions.

Problem: Let $BD$ be a bisector of triangle $ABC$. Points $I_a$, $I_c$ are the incenters of triangles $ABD$, $CBD$ respectively. The line $I_aI_c$ meets $AC$ in point $Q$. Prove that $\angle DBQ = 90^\circ$.

Walkthrough: 

• Let $F:= I_aI_c \cap BD,~~ J:=BI_a\cap AC,~~ K:=BI_c\cap AC.$
• Now, we have $$\angle I_aDI_c=90,~~\angle FDI_a=\angle I_aDA.$$ Then we have $$-1=(E,F;I_A,I_C)$$ projecting through $B$ on $AC.$ We get $$-1=(E,F;I_A,I_C)=(E,D;J,K).$$ We already have $$\angle JBD=\angle KBD\implies \angle DBE=90.$$

 

Problem: Let $ABC$ be a triangle with incenter $I$, let $,I_{1}, I_{2}, I_{3}$ be the incenters of triangles $BIC, CIA, AIB$ respectively. Prove that lines $AI_{1}, BI_{2}, CI_{3}$ are concurrent.

Walkthrough: 

• Note that $\angle ABI_3= B/2=\angle I_1BC.$ Similarly we get $\angle I_1CB=C/2=\angle I_2CA,~~\angle BAI_3=\angle CAI_2.$
• Now apply jacobi and we are done

Apart from that CAMP started! We had Hanabi night, where we got 20/25, which I feel is quite good at least for beginners. 

And then LIMIT results came out. In subjective I got 21/120. I was sad since I thought I could have got more marks. I might send my sols here too, and maybe someone can open grade here :O.

Sad note: My mom decreased my net time to 9pm :(, earlier it was 11pm.

Anyways, hope you enjoyed this post!

Sunaina💜