Some of the ISLs I did before INMO :P
[2005 G3]: Let 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 \angle KCL=360-(\angle ACB+\angle DKC+\angle BCK)=\angle DAB/2 +180-\angle DAB=180-\angle DAB/2
Motivation: I took a hint on this. I had other angles but I didn't realise that \Delta CBK\sim \Delta LDK.
Probably I should have ratio chased a bit better. Will try that next time!
[2018 G4]: A point T is chosen inside a triangle ABC. Let A_1, B_1, and C_1 be the reflections of T in BC, CA, and AB, respectively. Let \Omega be the circumcircle of the triangle A_1B_1C_1. The lines A_1T, B_1T, and C_1T meet \Omega again at A_2, B_2, and C_2, respectively. Prove that the lines AA_2, BB_2, and CC_2 are concurrent on \Omega.
Solution: Oh god this was so much a ggb problem. Not really but fineeeee...
Rename the point T as D.
Claim: The three circles (AC_1B),(AB_1C),(A_1BC) concur
Proof: We can simply ignore A_2,B_2,C_2. Also, there might be configuration issues.
Define H=(AC_1B)\cap (AB_1C).
Note that \angle BHC=\angle AHB-\angle AHC
=\angle AC_1B-180+\angle AB_1C
=\angle ADB-180+\angle ADC
Now, we have to show that \angle BHC=180-\angle BA_1C=180-\angle BDC
But \angle BHC=\angle ADB+\angle ADC-180
So enough to show \angle ADB+\angle ADC-180=190-\angle BDC
Or show \angle ADB+\angle ADC+\angle BDC=360.
Claim: H\in(A_1B_1C_1)
Proof: Note that \angle A_1HC_1=\angle A_1HB+\angle BHC_1=\angle A_1CB+180-\angle C_1AB=\angle A_1B_1D+180-\angle C_1B_1D.
Where we used the fact that A is centre of B_1C_1D and C is centre of DB_1A_1.
Now, we claim that H is the concurrency point. So if we show C_2,C,H collinear, then we are done as other collinearities follows similarly.
Enough to show that A_1HC=\angle A_1HC_2
or show that \angle A_1BC=\angle A_1HC_2
or show that \angle A_1C_1D=\angle A_1HC_2
But \angle A_1C_1H=\angle A_1HC_2
And done!
[2016 G4]: Let ABC be a triangle with AB = AC \neq BC and let I be its incentre. The line BI meets AC at D, and the line through D perpendicular to AC meets AI at E. Prove that the reflection of I in AC lies on the circumcircle of triangle BDE.
Here's the diagram!
Solution: Let the reflected point be I'. Let \angle ACB=C. We express all the angles in the form of \angle C.
By simple angle chase, we get ADI=C+C/2.
And so \angle I'DE=90+C+C/2.
Moreover, we have \angle I'CM=C+C/2.
Define X=EM\cap I'C, we get that \angle 90-C-C/2\implies I'DEX\text{ is cyclic }.
Also, note that as II'||ED, \angle BDE=\angle EDI=\angle DII'
So \angle BDE=90-C-C/2=\angle BXI\text { as } BX=XC\implies BEDX \text{ is cyclic }.
Hence BDEI' is cyclic.
Remark: It was pretty easy for G4. I think constructing X wasn't that easy, but once you notice that angles are 90-C-C/2 then everything breaks down..
[2001 N1]: Determine all positive integers n\geq 2 that satisfy the following condition: for all a and b relatively prime to n we havea \equiv b \pmod n\qquad\text{if and only if}\qquad ab\equiv 1 \pmod n.
Motivation: Everything was simply motivated with the fact that 3|2^2-1 where as other primes are just too big. CRT motivation was also the same, like I really wanted to construct a.
Case 2 still works because of the uniqueness of CRT..
[2011 C1]: Let n > 0 be an integer. We are given a balance and n weights of weight 2^0, 2^1, \cdots, 2^{n-1}. We are to place each of the n weights on the balance, one after another, in such a way that the right pan is never heavier than the left pan. At each step we choose one of the weights that has not yet been placed on the balance, and place it on either the left pan or the right pan, until all of the weights have been placed.
Determine the number of ways in which this can be done.
Solution: We will use induction. Denote f(n) the number of distinct process.
We claim that the answer is f(n)=(2n-1)!!. And by induction we will prove that f(n)=(2n-1)f(n-1). Base case is true clearly.
Let the blocks be 2^0,2^1,2^2,\dots 2^{n-2},2^{n-1}
Claim: At any step, the weight can never be equal.
Proof: Say \text{ left pan:} 2^{x_1}+\dots+2^{x_k}, \text{ Right pan:}2^{y_1}+\dots+2^{y_l}
And x_1<\dots<x_k and y_1<\dots<y_l.
We compare both the sides with v_2. Note that v_2(\text{ left pan})=x_1 and v_2(\text{ right pan})=y_1.
So sum of weight in left pan is never equal to sum of weight in right pan, as x_1\ne y_1.
Hence we can place the block 2^0 in any of the pans and it will satisfy the condition. Since adding 1, we will have still have at least equality of both the pans.
Now consider the blocks 2^1,2^2,\dots 2^{n-2},2^{n-1}
Then the no of distinct process to arrange these blocks in (2n-3)!!. Note that everything is same but multiplied by 2.
Now consider any process, we have 2\times n-1 ways to include the 2^0 block( As if we are including 2^0 in the first step then it must be on the left side).
Motivation: The answer was easy to guess. I think 2^0 was very special and once you notice that you can place 2^0 in both of the pans, induction then follows.
[2000 N4]: Find all triplets of positive integers (a,m,n) such that a^m + 1 \mid (a + 1)^n.
Motivation: The main motivation was that a+1|a^m+1|(a+1)^n.
But then by zsigmondy theorem, which states,
There’s a prime dividing a^n + b^n and not a^k + b^k for k < n with one
exception (a, b, n) = (2, 1, 3), a>b>0
Solution: By the second form of Zsigmondy theorem, we get that there exists a "new" prime factor of a^m+1, p such that (p,a+1)=1, but then a^m+1| (a+1)^n\implies p|(a+1)^n.
Now, let's see the exception case. Note that there is three exception case.
Case1: a=1\implies (1,m,n)\text{ works}
Case2: m=1\implies (a,1,n)\text{ works}
Case3: 3,2 case. We get that (2,3,n)\implies 2+1|2^3+1=9|(2+1)^n.
Our solutions are \boxed{(a,1,n),(1,m,n),(2,3,n);n>1}.
[2018 C1]:Let n\geqslant 3 be an integer. Prove that there exists a set S of 2n positive integers satisfying the following property: For every m=2,3,...,n the set S can be partitioned into two subsets with equal sums of elements, with one of subsets of cardinality m.
Solution:Let F_1+F_2+\dots+F_n=S_n. Then consider set \{F_1,F_2,\dots,F_{2n-2},F_{2n-1},S_{2n-3}+F_{2n-1}-F_{2n-2}=S_{2n-3}+F_{2n-3}\}.
Then consider A_2=(S_{2n-3}+F_{2n-1}-F_{2n-2})+(F_{2n-2})=S_{2n-3}+F_{2n-1}.
And B_{2n-2}=F_1+F_2+\dots+F_{2n-3}+F_{2n-1}.
Then A_3=(S_{2n-3}+F_{2n-1}-F_{2n-2})+(F_{2n-3})+(F_{2n-4})=S_{2n-3}+F_{2n-1}
And B_{2n-1}=F_1+F_2+\dots+F_{2n-5}+F_{2n-2}+F_{2n-1}
Similarly, to get A_4 we replace F_{2n-4} with F_{2n-5}+F_{2n-6}.
And so on.
Our general A_m construction is,
A_m=(S_{2n-3}+F_{2n-1}-F_{2n-2})+F_{2n-3}+F_{2n-5}+\dots+F_{2n-2m+1}+F_{2n-2m}
Motivation: It's kind of like "process thing." The main motivation was like F_{n+1}=F_n+F_{n-1}.
So, I can change F_{k+1}\rightarrow F_k+F_{k-1} increasing m but the sum is constant.
[2003 G3]:Let ABC be a triangle and let P be a point in its interior. Denote by D, E, F the feet of the perpendiculars from P to the lines BC, CA, AB, respectively. Suppose thatAP^2 + PD^2 = BP^2 + PE^2 = CP^2 + PF^2.Denote by I_A, I_B, I_C the excenters of the triangle ABC. Prove that P is the circumcenter of the triangle I_AI_BI_C.
Now, let's get into my Sharygin 2022 Solutions!! I did 12 complete :P.
Sharygin P13: Eight points in a general position are given in the plane. The areas of all 56 triangles with vertices at these points are written in a row. Prove that it is possible to insert the symbols "+" and "−" between them in such a way
that the obtained sum is equal to zero.
Proof: We begin with a lemma.
Lemma: Given ABCD a quadrilateral, we consider areas of the triangle. Then it is possible to insert the symbols + and - between them in such a way that the obtained sum is equal to zero.
Proof: Note that [ABD]+[ABC]=[ACD]+[ABC]. So assign [ABD],[ABC] as positive and [ACD],[ABC] negative.
Claim: Given 8 points ABCDEFGH, we can choose 17 tuples of four points, such no two tuples share three points.
Proof: Just consider the following contruction.
ABCD,ABEF,ABGH,ACEG,ACEF,ACFH,ADEH,ADFGBCEH,BCFG,BDEG,BDHF,CDEF,CDGH,EFGH
Let the eight points be A,B,C,D,E,F,G,H. Then consider the following quadrilateralsABCD,ABEF,ABGH,ACEG,ACEF,ACFH,ADEH,ADFGBCEH,BCFG,BDEG,BDHF,CDEF,CDGH,EFGH. Then note that by Lemma we can assign sign to the areas of each of the triangles of any of quadrilateral such that sum is 0. Now note since no to quadrilateral share three vertices, we get that[ABCD]+[ABEF]+[ABGH]+[ACEG]+[ACEF]+[ACFH]+[ADEH]+[ADFG]+[BCEH]+[BCFG]+[BDEG]+[BDHF]+[CDEF]+[CDGH]+[EFGH]=\text{ Sum of all } 56 \text{ triangles.}Let the 56 triangles arranged in a row be T_1,T_2,\dots T_{56}. So the sign of T_1 is positive, then we can proceed by noticing where T_1 belongs to from the 8 quadrilateral i.e ABCD,ABEF,ABGH,ACEG,ACEF,ACFH,ADEH,ADFG
BCEH,BCFG,BDEG,BDHF,CDEF,CDGH,EFGH.
And then proceed with assigning the signs ( using the two above lemmas).
Proof: Note that \angle ATB=\angle CTA=120\implies \angle CAT=60-\angle TAB=\angle TBA\implies \Delta ATB\sim\Delta CTA.
Proof: Define M_A:= AI\cap (ABC).
Proof: Since AB\times CD=BC\times AD\implies \frac{AB}{BC}\times \frac{CD}{AD}=1\implies (A,C;B,D)=-1.So ABCD is a harmonic quadrilateral.
Proof: To prove that \angle BAX=\angle CAY, it is enough to show that \angle XAI=\angle YAI as \angle BAI=\angle CAI.
Then the no of distinct process to arrange these blocks in (2n−3)!!. Note that everything is same but multiplied by 2.
ReplyDeleteAre you notating double factorial or the other one is just an exclamation mark?
Yes! It's very common notation :)
Deleteloved it!
ReplyDeletehi
ReplyDeletehow do you get better at geo?
thx