Welcome back! So today I will be sharing a few problems which I did last week and some ISLs. Easy ones I guess.
Happy September 2021!
Problem[APMO 2018 P1]: Let ABC be a triangle with orthocenter H and let M and N denote the midpoints of {AB} and {AC}. Assume H lies inside quadrilateral BMNC, and the
circumcircles of \triangle BMH and \triangle CNH are tangent. The line through H parallel to {BC} intersects (BMH) and (CNH) again at K, L respectively.
Let F = {MK} \cap {NL}, and let J denote the incenter of \triangle MHN.
Prove that FJ = FA.
Proof:
- By angle chase, we get \angle FKL=\angle FLK.
- Hence KL||MN\implies FM=FN.
- And we get \angle MFN=2A\implies F is circumcentre if (AMN)\implies FA=FM=FN.
- And we get \angle MHN=180-2A
- Hence MFHN is cyclic.
- By fact 5, ME=FJ=FN\implies FJ=FA.
Problem[Shortlist 2007 G3]:
Let ABCD be a trapezoid whose diagonals meet at P. Point Q lies between parallel lines BC and AD, and line CD separates points P and Q.
Given that \angle AQD = \angle CQB, prove that \angle BQP = \angle DAQ.
Proof: (With Erijon)
- consider the homothety centered at P which takes B to D and C to A
- Note that BL||QD, L is the image of Q We also have AL||CQ
- And we get \angle DLA=\angle CLB=\angle CDB=\angle AQB
- Hence we get QCBL,QADL cyclic.
- So we have BQ||DL then we get
\angle BQL=\angle QLD =\angle QAD
and we are done.
Problem[EGMO 2013/1]:
The side BC of the triangle ABC is extended beyond C to D so that CD = BC. The side CA is extended beyond A to E so that AE = 2CA. Prove that if AD=BE then the triangle ABC is right-angled.
Proof:
- So introduce the midpoint of AE,AB as M,J.
- Note that JC=\frac{1}{2}AD
- And we have JM=\frac{1}{2}AB
- AM=AC\implies JA\perp MC\implies BAC=90.
Problem[APMO 2013 P5 ]:
Let ABCD be a quadrilateral inscribed in a circle \omega, and let P be a point on the extension of AC such that PB and PD are tangent to \omega. The tangent at C intersects PD at Q and the line AD at R. Let E be the second point of intersection between AQ and \omega. Prove that B, E, R are collinear.
Proof: Note that -1=(A,E;C,D)=(A,RE\cap \omega; C,D) But we know (A,C;B,D)=-1\implies RE\cap \omega=B
Problem[2008 G1]:
Let H be the orthocenter of an acute-angled triangle ABC. The circle \Gamma_{A} centered at the midpoint of BC and passing through H intersects the sideline BC at points A_{1} and A_{2}. Similarly, define the points B_{1}, B_{2}, C_{1} and C_{2}.
Prove that the six points A_{1}, A_{2}, B_{1}, B_{2}, C_{1} and C_{2} are concyclic.
Proof:
- Let (M_c)\cap (M_B)=X,H.
- Note that XH is the radical axis, so we have XH\perp M_CM_B. But we have AH\perp BC||M_CM_B\implies A-X-H
- So A lies on the radical axis \implies Ac_1\cdot AC_2=AB_2\cdot AB_1\implies (C_1B_2B_1C_2) is cyclic, similarly we get C_1C_2B_1B_2 cyclic, B_1A_2A_1B_2 is cyclic.
- Now note that the pairwise radical axis doesn't concur.
Problem 6: Find all integers n for which both 4n + 1 and 9n + 1 are perfect squares
Solution: I am not gonna lie this question felt really hard to me :(
The idea took time, but it's simple.
So let 4n+1=a^2 then note b^2=9n+1= 2\cdot (4n+1)+n-1= 2\cdot a^2+ \frac{a^2-1}{4} -1= \frac{9a^2-5}{4}\implies 9a^2-5\text{~~is a square}
So both a^2,9a^2-5 is a square. Let 9a^2-5=k^2, 9a^2=l^2\implies l^2-k^2= (l+k)(l-k)=5\implies l=2,k=\implies a^2=1=4n+1\implies n=0.
Problem 7: Find all n \in \mathbb{N} for which \exists x,y,k \in \mathbb{N} such that (x,y)=1 , k>1 and 3^{n}=x^k+y^k
Solution: The second form of Zigmondy Theorem states that : there’s a prime dividing a^k + b^k and not a^l + b^l for l < k ( The exception is (a,b,k)=(2,1,3)).
So now if (a,b,k)\ne(2,1,3) then it is not possible since we will have two distinct primes dividing x^k+y^k=3^n. Not possible.
If we have (a,b,k)=(2,1,3) we get 3^n=2^k+1. Using Mihăilescu theorem ( which states that the only consecutive perfect powers are
8 and 9)
Hence \boxed{n=2, k=3, x=2, y=1} is the only solution.
Problem 8: Let P (x) be a polynomial with integer coefficients and P (n) =m for some integers n, m (m \ne 10). Prove that P (n + km) is divisible by m for any integer k.
Proof: We know that a-b|P(a)-P(b)\implies km|P(n+km)-P(n)\implies m|P(n+km)-P(n)\implies m|P(n+km).
Problem 9: The sequence of integers a_n is given by a_0 = 0, a_n= p(a_{n-1}), where p(x) is a polynomial whose coefficients are all positive integers. Show that for any two positive integers m, k with greatest common divisor d, the greatest common divisor of a_m and a_k is a_d.
Proof : We assume m\le k.
Note that a_{m-1}-a_{k-1}|P(a_{m-1})-P(a_{k-1})
Similarly we get a_{m-2}-a_{k-2}|P(a_{m-2})-P(a_{k-2})
And we continue we get a_0-a_{k-m}| P(a_{m-1})-P(a_{k-1}).
So we get gcd(a_m,a_k)=gcd(a_{k-m},a_k). And this continues.
By the euclidean algorithm, we are done.
Problem 10:Given a polynomial P(x) with natural coefficients;
Let us denote with a_n the sum of the digits of P(n) value.
Prove that there is a number encountered in the sequence a_1, a_2, ... , a_n, ... infinite times.
Proof : Let (b_n,b_{n-1},\dots, b_0) be the coefficients of the polynomial P(x)=b_n\cdot x^n+\dots+x_0.
Then surely there exist a natural number k such that P(10^k) is of the form \overline{b_n0\dots 0b_{n-1}0\dots 0 b_0}.
Then consider 10^k,10^{k+1},\dots .
Yep, this is it for this week's blog! Autumn is about to come! I didn't even realise it but 2021 is about to end in 4 months!
Also, I have exported a few contacts from Feedburner to Mailchimp and I am in the process of removing FeedBurner completely from this blog. And people will be getting mails once a week only!
P.S. Do comment on how you liked the problems, which one was easy and what was your favourite!
P.P.S. Since IOQM is close, would you all like it if I post 10 IOQM problems each week? That would be great I think? It would probably be a series of collaborative posts.
See you soon
Sunaina 💜
Love this blog a lot. Hopefully it never dies :D
ReplyDeleteThanks :D, will try my best to not let it die!
DeleteBeautiful problems! 2008 G1 was very nice. Yeah, IOQM is close. So posting IOQM problems would be great.
ReplyDeleteGlad you liked it :). And yeah, would be sending those problems in every wed/thursday.
DeleteWoahh nice type ups, I like when u give concise and nice solutions to past international contest problems :p.
ReplyDeleteLooking forward to future posts 😁😁😁
Woah thanks! Will try more problems!
DeleteAlso, 7000 views coming soon 🤩
ReplyDeleteOh no, nothing special for this :P ( maybe Isl n1s? IDK :P)
DeletePranav pro gawd solves N5s now :orz: big phan pranav sar
DeleteBro what ? I solved only 1 N5, that too not completely on my own, I needed a hint, so I in some sense can't classify as a solve completely, but yeah I did learn but didn't solve myself -__-
Deletelol pranav. How many people did you tell that u solved N5 :joy:?
DeleteI didn't tell anyone else except you and anand ;-;.
DeleteI was actually solving the problem in LT, like writing my progress there in a thread and asked bora for hint meanwhile. And when I was done I wrote "yay! Finally I am done."
And then prabh -_____-
As his Pf name states, a problem :rotfl:
DeleteNoice. Do try USA TSTST 2012 P7. My proudest solve ever (coz most solutions use something except basic angle chase (like projective inversion miquel points et al) which idk)
ReplyDeletewoah sure!
DeleteFor your problem on 3^n=x^k+y^k, Zsigmondy is overkill. We can see that if k is not a power of 2, using the lifting the exponent lemma + size is an easier finish. If k is a power of 2, we win by taking mod 3 to find 3|x, 3|y and then we can consider a solution (x/3,y/3) and win with infinite descent.
ReplyDeleteI recommend using basic tools to solve easy problems because it allows you to be able to recognize patterns and is generally more flexible.
HOLYY.. Hiii. I am happyy that you commented! Also yep, LTE and Infinite descent would be better! Although Zsig is really nice. I was like
Delete" Oh it's doable by zsig" and once I think a problem is solved.. why find the long route :P JK. But yes, nice solution. Will try to cut down the use!