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".
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 $2a^2 < 5b$
Walkthrough: a. 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 $$\frac{r_i^2+r_j^2}{2} \ge r_ir_j$$ .
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 💜
Nice
ReplyDeleteGreat!
ReplyDelete