Skip to main content

Guess what? I will do class 8 geos ..

Obviously, I have become very rusted. So to unrust me ( oh god, is it unrust ?  I have become so bad in English). Okie wait.. "polish myself." I tried problems from ABJTOG 1.4. Easy ones TBH. Without further ado, here are the problems and solutions I tried. You guys can try too! I can assure you the difficulty is less than class 8. ( or class 7). I didn't try harder problems, because that would take me sometime. I did these in break :P . So yeah.. sorry for so easy  levels.

Milk and Mocha 


Problem 2 of ABJTOG:- Let $ABC$ be a triangle and let $M$ be a point on the ray $AB$ beyond $B$ such that $\overline{BM} = \overline{BC}$. Prove that $MC$ is parallel to the angle bisector of $\angle ABC$.

Solution :- Note that$$\angle BMC=\frac{1}{2} \cdot (180-B)= \frac{B}{2}=\frac{1}{2}\angle ABC.$$


Problem 1 of ABJTOG :- Let $C$ be a point on the line segment $AB$. Let $D$ be a point that doesn’t lie on the line $AB$. Let $M$ and $N$ be points on the angle bisectors of $\angle ACD$ and $\angle BCD$, respectively, such that $MN \parallel AB$. Prove that the line $CD$ bisects $MN$.

Solution:- Clearly $\angle MCN=90.$ Now, to show that line $CD$ bisects $MN$, it's enough to show that $X$ is the centre of the $\Delta CMN ,$ where $X=CD\cap MN$ or it's enough to show that $\Delta MCX,\Delta CXN$ is isosceles, which follows from the parallel line and angle bisector stuff[ $\angle XMC= \angle MCA=\angle MCX.$

Problem 14 of ABJTOG:- Let $O, I, H$ be the circumcenter, incenter and orthocenter, respectively, of $|Delta ABC.$ Prove that $B, C, O, I, H$ lie on a circle if and only if $\angle BAC = 60$

Solution:- Since $B,C,I,H$ lie on one circle, we get$$\angle BHC=BIC\implies 180-A=90+\frac A2\implies A=60.$$So we get $\angle BHC=\angle BIC=120.$ But we also have $\angle BOC=120.$ Hence Prove that $B, C, O, I, H$ are cyclic

Problem 17 of ABJTOG:- Let $ABCDEF$ be a convex hexagon with $\overline{AB} = \overline{AF}, \overline{BC} = \overline{CD}$ and $\overline{DE} = \overline{EF}$. Prove that the angle bisectors of $\angle BAF, \angle BCD$ and $\angle DEF$ are concurrent.

Solution:- Well my one was exactly Thermos. So..

Note that the angle bisectors of $\angle BAF, \angle BCD$ and $\angle DEF$ are the perpendicular bisectors of $\overline{FB}, \overline{BD}, \overline{DF}$. Thus, they are concurrent at the circumcenter of $\triangle BDF$.

Problem 15 of ABJTOG:- Let $H$ and $O$ be the orthocenter and circumcenter in a triangle $ABC$, respectively. If $\angle BAC = 60 $ , prove that $AH = AO$. Is the converse true?

Solution:- Just use the fact that $AH=2R\cdot \cos A.$ And $\cos A=\frac 12 \iff\angle A=60.$

Problem 40 of ABJTOG:- Let $D, E$ and $F$ be points on the sides $BC, CA$ and $AB,$ respectively, such that $BCEF$ is a cyclic quadrilateral. Let $P$ be the second intersection of the circumcircles of $\Delta BDF$ and $\Delta CDE.$ Prove that $A, D$ and $P$ are collinear.

Solution:- We apply radical axis theorem in circles $(BCEF),(BDF),(CDE).$ Note that $AB$ is the radical axis of $(BDF),(BFEC)$ and $AC$ is the radical axis of $(CDE),(BFEC).$ Since $AC,AB$ concur at $A$ and $DP$ is the radical axis of $(BDF),(CDE).$ Hence $A,D,P$ are collinear.

Problem 37 of ABJTOG:- Let $ABCD$ be a cyclic quadrilateral. The rays $AB$ and $DC$ intersect at $P$ and the rays $AD$ and $BC$ intersect at $Q.$ The circumcircles of $\Delta BCP$ and $\Delta CDQ$ intersect at $R.$ Prove that the points $P , Q$ and $R$ are collinear.

Solution:- Note that $\angle PBC=180-\angle PRC$ and $\angle PBC=\angle QDC=\angle QRC.$ Hence $\angle PRC+\angle QRC=180.$

Problem 38 of ABJTOG:- The diagonals of a cyclic quadrilateral $ABCD$ intersect at $S$. The circumcircle of $\Delta ABS$ intersects line $BC$ at $M$ , and the circumcircle of $ADS$ intersects line $CD$ at $N .$ Prove that $S, M$ and $N$ are collinear.

Solution:- Note that $\angle ABM= 180-\angle ASM$ and $\angle ABM=\angle ADN=\angle ASN.$ Hence $\angle ASN+\angle ASM=180.$ So $S, M$ and $N$ are collinear.



Problem 41 of ABJTOG :- Two circles are tangent to each other internally at a point $T$ . Let the chord $AB$ of the larger circle be tangent to the smaller circle at a point $P .$ Prove that $TP$ is the internal angle bisector of $\angle ATB.$

Solution:-Simple homothety.

Let $Z:= TP \cap (ATB).$ Note that $P$ is the lowest point, so by homothety, $Z$ will also be the "lowest" point i.e $Z$ is the midpoint of arc $AB.$ Hence $TZ=TP$ is the internal angle bisector of $\angle ATB.$

Problem 42 of ABJTOG:- Let $ABCD$ be a trapezoid $(AB || CD).$ Let $AC \cap BD = E$ and $AD \cap BC = F .$ Let $M, N$ be midpoints of $AB, CD,$ respectively. Prove that the points $E, F, M, N$ are collinear.

Solution:- Well clearly, $F,M,N$ are collinear. So, we will show that $E,M,N$ is collinear. Let $M'= EN\cap AB.$ By the parallel property, we get $\Delta  M'EB \sim NED$ with ratio $M'E/NE.$ So $M'B=ND \cdot \frac{M'E}{EN}.$ And we also have $\Delta  M'EA \sim NEC$ with ratio $M'E/NE .$ So $M'A=NC\cdot \frac{M'E}{EN}.\implies M'A=M'B\implies M=M'.$

----

Well yeah.. that's the problems I did. Trust me the number of Olympiad problems, I am doing has exponentially decreased :(. 

If you have time and liked the content then follow the blog.  Click the three rows thingy in the Right top which is white in colour, then follow, I have got 9 followers till now. So yayyy!!

Sunaina 💜

Labels:

Comments

  1. ArpanApril 2, 2021 at 12:12 PM

    You are on fire in writing blogs
    I checked today and had to read 4 blogs 😅

    ReplyDelete

Post a Comment

Popular posts from this blog

Geometry ( Finally!!!)

 This is just such an unfair blog.  Like if one goes through this blog, one can notice how dominated  Algebra is!! Like 6 out of 9 blog post is Algebra dominated -_- Where as I am not a fan of Algebra, compared to other genres of Olympiad Math(as of now). And this was just injustice for Synthetic Geo. So this time , go geo!!!!!!!!!!!  These problems are randomly from A Beautiful Journey through Olympiad Geometry.  Also perhaps I will post geo after March, because I am studying combi.  Problem:  Let $ABC$ 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....
3 comments
Read more

Problems I did this week [Jan8-Jan14]

Yeyy!! I am being so consistent with my posts~~ Here are a few problems I did the past week and yeah INMO going to happen soon :) All the best to everyone who is writing!  I wont be trying any new problems and will simply revise stuffs :) Some problems here are hard. Try them yourself and yeah~~Solutions (with sources) are given at the end! Problems discussed in the blog post Problem1: Let $ABC$ 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$ ...
Post a Comment
Read more

Just spam combo problems cause why not

This post is mainly for Rohan Bhaiya. He gave me/EGMO contestants a lot and lots of problems. Here are solutions to a very few of them.  To Rohan Bhaiya: I just wrote the sketch/proofs here cause why not :P. I did a few more extra problems so yeah.  I sort of sorted the problems into different sub-areas, but it's just better to try all of them! I did try some more combo problems outside this but I tried them in my tablet and worked there itself. So latexing was tough. Algorithms  "Just find the algorithm" they said and they died.  References:  Algorithms Pset by Abhay Bestrapalli Algorithms by Cody Johnson Problem1: Suppose the positive integer $n$ is odd. First Al writes the numbers $1, 2,\dots, 2n$ on the blackboard. Then he picks any two numbers $a, b$ erases them, and writes, instead, $|a - b|$. Prove that an odd number will remain at the end.  Proof: Well, we go $\mod 2$. Note that $$|a-b|\equiv a+b\mod 2\implies \text{ the final number is }1+2+\dots ...
2 comments
Read more

My experiences at EGMO, IMOTC and PROMYS experience

Yes, I know. This post should have been posted like 2 months ago. Okay okay, sorry. But yeah, I was just waiting for everything to be over and I was lazy. ( sorry ) You know, the transitioning period from high school to college is very weird. I will join CMI( Chennai Mathematical  Institue) for bsc maths and cs degree. And I am very scared. Like very very scared. No, not about making new friends and all. I don't care about that part because I know a decent amount of CMI people already.  What I am scared of is whether I will be able to handle the coursework and get good grades T_T Anyways, here's my EGMO PDC, EGMO, IMOTC and PROMYS experience. Yes, a lot of stuff. My EGMO experience is a lot and I wrote a lot of details, IMOTC and PROMYS is just a few paras. Oh to those, who don't know me or are reading for the first time. I am Sunaina Pati. I was IND2 at EGMO 2023 which was held in Slovenia. I was also invited to the IMOTC or International Mathematical Olympiad Training Cam...
11 comments
Read more

IMO 2023 P2

IMO 2023 P2 Well, IMO 2023 Day 1 problems are out and I thought of trying the geometry problem which was P2.  Problem: Let $ABC$ be an acute-angled triangle with $AB < AC$. Let $\Omega$ be the circumcircle of $ABC$. Let $S$ be the midpoint of the arc $CB$ of $\Omega$ containing $A$. The perpendicular from $A$ to $BC$ meets $BS$ at $D$ and meets $\Omega$ again at $E \neq A$. The line through $D$ parallel to $BC$ meets line $BE$ at $L$. Denote the circumcircle of triangle $BDL$ by $\omega$. Let $\omega$ meet $\Omega$ again at $P \neq B$. Prove that the line tangent to $\omega$ at $P$ meets line $BS$ on the internal angle bisector of $\angle BAC$. Well, here's my proof, but I would rather call this my rough work tbh. There are comments in the end! Proof Define $A'$ as the antipode of $A$. And redefine $P=A'D\cap (ABC)$. Define $L=SP\cap (PDB)$.  Claim1: $L-B-E$ collinear Proof: Note that $$\angle SCA=\angle SCB-\angle ACB=90-A/2-C.$$ So $$\angle SPA=90-A/2-C\implies \ang...
1 comment
Read more

Solving Random ISLs And Sharygin Solutions! And INMO happened!!

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 did...
4 comments
Read more

Challenging myself? [Jan 15-Jan 27]

Ehh INMO was trash. I think I will get 17/0/0/0-1/3-5/10-14, which is def not good enough for qualifying from 12th grade. Well, I really feel sad but let's not talk about it and focus on EGMO rather.  INMO 2023 P1 Let $S$ be a finite set of positive integers. Assume that there are precisely 2023 ordered pairs $(x,y)$ in $S\times S$ so that the product $xy$ is a perfect square. Prove that one can find at least four distinct elements in $S$ so that none of their pairwise products is a perfect square. I will use Atul's sol, cause it's the exact same as mine.  Proof: Consider the graph $G$ induced by the elements of $S$ and edges being if the products are perfect squares. Note that if $xy = a^2$ and $xz = b^2$, then $yz = \left( \frac{ab}{x} \right)^2$, since its an integer and square of a rational number its a perfect square and so $yz$ is an edge too. So the graph is a bunch of disjoint cliques, say with sizes $c_1, c_2, \cdots, c_k$. Then $\sum_{i=1}^k c_i^2 = 2023$, which ...
Post a Comment
Read more

Orders and Primitive roots

 Theory  We know what Fermat's little theorem states. If $p$ is a prime number, then for any integer $a$, the number $a^p − a$ is an integer multiple of $p$. In the notation of modular arithmetic, this is expressed as \[a^{p}\equiv a{\pmod {p}}.\] So, essentially, for every $(a,m)=1$, ${a}^{\phi (m)}\equiv 1 \pmod {m}$. But $\phi (m)$ isn't necessarily the smallest exponent. For example, we know $4^{12}\equiv 1\mod 13$ but so is $4^6$. So, we care about the "smallest" exponent $d$ such that $a^d\equiv 1\mod m$ given $(a,m)=1$.  Orders Given a prime $p$, the order of an integer $a$ modulo $p$, $p\nmid a$, is the smallest positive integer $d$, such that $a^d \equiv 1 \pmod p$. This is denoted $\text{ord}_p(a) = d$. If $p$ is a primes and $p\nmid a$, let $d$ be order of $a$ mod $p$. Then $a^n\equiv 1\pmod p\implies d|n$. Let $n=pd+r, r\ll d$. Which implies $a^r\equiv 1\pmod p.$ But $d$ is the smallest natural number. So $r=0$. So $d|n$. Show that $n$ divid...
7 comments
Read more

IMO Shortlist 2021 C1

 I am planning to do at least one ISL every day so that I do not lose my Olympiad touch (and also they are fun to think about!). Today, I tried the 2021 IMO shortlist C1.  (2021 ISL C1) Let $S$ be an infinite set of positive integers, such that there exist four pairwise distinct $a,b,c,d \in S$ with $\gcd(a,b) \neq \gcd(c,d)$. Prove that there exist three pairwise distinct $x,y,z \in S$ such that $\gcd(x,y)=\gcd(y,z) \neq \gcd(z,x)$. Suppose not. Then any $3$ elements $x,y,z\in S$ will be $(x,y)=(y,z)=(x,z)$ or $(x,y)\ne (y,z)\ne (x,z)$. There exists an infinite set $T$ such that $\forall x,y\in T,(x,y)=d,$ where $d$ is constant. Fix a random element $a$. Note that $(x,a)|a$. So $(x,a)\le a$.Since there are infinite elements and finite many possibilities for the gcd (atmost $a$). So $\exists$ set $T$ which is infinite such that $\forall b_1,b_2\in T$ $$(a,b_1)=(a,b_2)=d.$$ Note that if $(b_1,b_2)\ne d$ then we get a contradiction as we get a set satisfying the proble...
Post a Comment
Read more

Introduction

  Hey Everyone!! This is my first Blog post. So let me give a brief introduction about myself. I am Sunaina Pati. I love solving Olympiad math problems,  learning crazy astronomical facts , playing hanabi and anti-chess, listening to Kpop , love making diagrams in Geogebra and  teaching other people maths 😊 . I love geometry , number theory and Combinatorics . I am starting this blog to keep myself a bit motivated in doing studies 😎 . Right now, I am planning to write walkthroughs on some of the best problems I tried over the week which can refer for hints 'cause solutions contain some major spoilers and one learns a lot while solving the problem on his own rather than seeing solutions . Also, there will be some reviews about Kpop songs, study techniques, my day to day lifestyles,exam reviews and ofc some non-sense surprises 😂.  I am planning to  try  posting every week on Sundays or Saturdays ( most probably) ! Though there is no guarantee about when I ...
20 comments
Read more