Week #5 ( a bit different?)

Well.. IOQM is too close, so I haven't been doing any Olympiad type problems lately :( . And continuously attempting mocks. I encountered a lot of nice problems, easy but cute , a bit hard too.. but then typing them was really not feasible for me right now!

But then blog without regular updates is bad. Anyways I completed Alexander Remorov's projective geo Part 1 handout last week with JIB's, Rohan Bhaiya ,Sanjana's and Serena's help. :P . And I did 2004 G8 ( it's in egmo too ). But still I found it cute.

So here fully motivating harmonic G8. I won't say it's easy, but yes killed by harmonic.

Problem(2004 G/8): Given a cyclic quadrilateral $ABCD$ , let $M$ be the midpoint of the side $CD$ , and let $N$ be a point on the circumcircle of triangle $ABM$ . Assume that the point $N$ is different from the point $M$ and satisfies $\frac{AN}{BN}=\frac{AM}{BM}$ . Prove that the points $E$ , $F$ , $N$ are collinear, where $E=AC\cap BD$ and $F=BC\cap DA$ .

Here's the walkthrough: First of all. Draw a nice diagram ( not necessarily accurate , but don't use ggb )

a. Note that $AMNB$ is harmonic.

b. Define $T=CD\cap EF$ and show $T\in (AMB)$ . For this we have to define $AB\cap CD=I$ . Find a harmonic bundle and use lemma 9.17 from EGMO ( which is "Points $A, X, B, P$ lie on a line in that order and $(A, B; X, P ) = −1$ . Let $M$ be the midpoint of $AB$ . Then $P X \cdot P M = P A \cdot PB$ " ). So basically POP.

c. Projecting through $T$ on $AB$ and use "cevian induces harmonic lemma".

Yeah!! So this was the walkthrough for 2004 G8 :P. Also try this problem too..

Problem: (USAMO 1983/2): Prove that the roots of $x^5 + ax^4 + bx^3 + cx^2 + dx + e = 0$ cannot all be real if $2a^2 < 5b$

Let $r_1, r_2,r_3,r_4,r_5$ be the roots. Now suppose $2a^2\ge 5b$ , we will show that atleast one of $r_1, r_2,r_3,r_4,r_5$ is non real.

b. Note that by vieta, $a=-(r_1+r_2+r_3+r_4+r_5)$ and $b=\sum_{sym}r_1\cdot r_2$ . So simplify it and get $2a^2=2\sum {a_i}^2+ 4\sum_{sym}r_1\cdot r_2 \le 5b=5\cdot \sum_{sym}r_1\cdot r_2 \Rightarrow 2\sum {a_i}^2\le \sum_{sym}r_1\cdot r_2.$

c. Now if $r_1, r_2,r_3,r_4,r_5$ were real then by AM-GM, we get $\frac{r_i^2+r_j^2}{2} \ge r_ir_j$ .

oh also... I compiled the solutions of Alexander Remorov's projective geo Part 1 handout . Actually true story is I was trying asymptote in Evan.sty and it's my really awkward habit to write every solution I got , irrespective of how much time it takes. So wrote the solutions, and let me say, the asymptote looks so nice. Still learning though. See this.

Do write in the comments section about what was the most seemingly hard problem, you realised was pretty nice.(at least write something ! I will be happy to hear your comments ). Follow this blog if you want to see more contest math problems, till then happy problem solving! Also all the best for IOQM everyone!

Sunaina 💜

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$ABCD$ be a parallelogram. A variable line

$g$ through the vertex

$A$ intersects the rays

$BC$ and

$DC$ at the points

$X$ and

$Y$ , respectively. Let

$K$ and

$L$ be the

$A$ -excenters of the triangles

$ABX$ and

$ADY$ . Show that the angle

$\measuredangle KCL$ is independent of the line

$g$ Solution: Note that

$\Delta LDK \sim \Delta XBK$ and

$\Delta ADY\sim \Delta XCY.$ So we have

$\frac{BK}{DY}=\frac{XK}{LY}$ and

$\frac{DY}{CY}=\frac{AD}{XC}=\frac{AY}{XY}.$ Hence

$\frac{BK}{CY}=\frac{AD}{XC}\times \frac{XK}{LY}\implies \frac{BK}{BC}=\frac{CY}{XC}\times \frac{XK}{LY}=\frac{AB}{BC}\times \frac{XK}{LY}$

$\frac{AB}{LY}\times \frac{XK}{BK}=\frac{AB}{LY}\times \frac{LY}{DY}=\frac{AB}{DL}$

$\implies \Delta CBK\sim \Delta LDK$ And we are done. We get that

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$G$ is said to be

$k$ -vertex-connected (or

$k$ -connected) if it has more than

$k$ vertices and remains connected whenever fewer than

$k$ vertices are removed. Show that every

$k$ -connected graph of order atleast

$2k$ contains a cycle of length at least

$2k$ . We begin with a lemma. Prove that a graph

$G$ of order

$n \geq 2k$ is

$k$ connected then every 2 disjoint set

$V_1$ and

$V_2$ of

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$ABC$ be an acute triangle where

$\angle BAC = 60^{\circ}$ . Prove that if the Euler’s line of

$\triangle ABC$ intersects

$AB$ and

$AC$ at

$D$ and

$E$ , respectively, then

$\triangle ADE$ is equilateral. Solution: Since

$\angle A=60^{\circ}$ , we get

$AH=2R\cos A=R=AO$ . So

$\angle EHA=\angle DOA.$ Also it's well known that

$H$ and

$O$ isogonal conjugates.

$\angle OAD =\angle EAH.$ By

$ASA$ congruence, we get

$AE=AD.$ Hence

$\triangle ADE$ is equilateral....

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$z=a+ib,$ where

$a$ and

$b$ are real numbers and

$i^2=-1.$ 2. In polar form,

$z=r(\cos \theta+~~i\sin\theta)=~~re^{i\theta},$ where

$r=~~|z|=~~\sqrt{a^2+b^2},$ which is called the magnitude. 3. Here we used euler's formula i.e

$\cos \theta+~~i\sin\theta=~~e^{i\theta}.$ 4. The

$\theta$ is called the argument of

$z,$ denored

$\arg z.$ (

$\theta$ can be considered in

$\mod 360$ and it is measured anti-clockwise). 5. The complex conjugate of

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$ABC$ be a triangle whose incircle

$\omega$ touches sides

$BC, CA, AB$ at

$D,E,F$ respectively. Let

$H$ be the orthocenter of

$DEF$ and let altitude

$DH$ intersect

$\omega$ again at

$P$ and

$EF$ intersect

$BC$ at

$L$ . Let the circumcircle of

$BPC$ intersect

$\omega$ again at

$X$ . Prove that points

$L,D,H,X$ are concyclic. Problem 2: Let

$ABCD$ be a convex quadrangle,

$P$ the intersection of lines

$AB$ and

$CD$ ,

$Q$ the intersection of lines

$AD$ and

$BC$ and

$O$ the intersection of diagonals

$AC$ and

$BD$ . Show that if

$\angle POQ= 90^\circ$ then

$PO$ is the bisector of

$\angle AOD$ ...

Sunaina thinks absurd

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