Inspired by IMO 1984

by sqing, Mar 26, 2025, 3:01 AM

Let $a,b,c\geq 0$ and $a+b+c=1$ . Prove that
$a^2+b^2+ ab +24abc\leq\frac{81}{64}$ Equality holds when $a=b=\frac{3}{8},c=\frac{1}{4}.$
$a^2+b^2+ ab +18abc\leq\frac{343}{324}$ Equality holds when $a=b=\frac{7}{18},c=\frac{2}{9}.$

inequalities proposed

algebra

equal angles

by jhz, Mar 26, 2025, 12:56 AM

In convex quadrilateral $ABCD, AB \perp AD, AD = DC$ . Let $E$ be a point on side $BC$ , and $F$ be a point on the extension of $DE$ such that $\angle ABF = \angle DEC>90^{\circ}$ . Let $O$ be the circumcenter of $\triangle CDE$ , and $P$ be a point on the side extension of $FO$ satisfying $FB =FP$ . Line BP intersects AC at point Q. Prove that $\angle AQB =\angle DPF.$

geometry

2025 Caucasus MO Seniors P1

by BR1F1SZ, Mar 26, 2025, 12:37 AM

For given positive integers $a$ and $b$ , let us consider the equation $a + \gcd(b, x) = b + \gcd(a, x).$

For $a = 20$ and $b = 25$ , find the least positive integer $x$ satisfying this equation.
Prove that for any positive integers $a$ and $b$ , there exist infinitely many positive integers $x$ satisfying this equation.

(Here, $\gcd(m, n)$ denotes the greatest common divisor of positive integers $m$ and $n$ .)

number theory

Inspired by IMO 1984

by sqing, Mar 25, 2025, 3:04 PM

Let $a,b,c\geq 0$ and $a+b+c=1$ . Prove that
$a^2+b^2+ ab +17abc\leq\frac{8000}{7803}$ $a^2+b^2+ ab +\frac{163}{10}abc\leq\frac{7189057}{7173630}$ $a^2+b^2+ ab +16.23442238abc\le1$

inequalities

inequalities proposed

algebra

Long condition for the beginning

by wassupevery1, Mar 25, 2025, 1:49 PM

Find all functions $f: \mathbb{Q}^+ \to \mathbb{Q}^+$ such that $\dfrac{f(x)f(y)}{f(xy)} = \dfrac{\left( \sqrt{f(x)} + \sqrt{f(y)} \right)^2}{f(x+y)}$ holds for all positive rational numbers $x, y$ .

function

algebra

functional equation

Smallest value of |253^m - 40^n|

by MS_Kekas, Jan 28, 2024, 9:35 PM

Find the smallest value of the expression $|253^m - 40^n|$ over all pairs of positive integers $(m, n)$ .

Proposed by Oleksii Masalitin

number theory

Diophantine equation

A Powerful Ratio Theorem

by math_pi_rate, Sep 27, 2018, 12:43 PM

So this topic is for a wonderful theorem about isogonal lines, namely Steiner's Ratio Theorem. Here's the theorem:

THEOREM (Steiner) Let $D$ and $E$ be points on $BC$ , so that $AD$ and $AE$ are isogonal with respect to $\angle BAC$ . Then we have the following equality:- $\left (\frac{AB}{AC} \right)^2=\frac{BD \cdot BE}{CD \cdot CE}$
REMARK: The converse of Steiner's Theorem is also true.

PROOF 1 (Trig Bash

) Let $\angle BAD=\angle CAE=x$ and $\angle BAE=\angle CAD=y$ . Apply sine law in $\triangle ADB, \triangle ADC, \triangle AEB, \triangle AEC$ . Then we have $\frac{BD}{AB}=\frac{\sin x}{\sin \angle BDA},\frac{BE}{AB}=\frac{\sin y}{\sin \angle BEA},\frac{CD}{AC}=\frac{\sin y}{\sin \angle CDA},\frac{CE}{AC}=\frac{\sin x}{\sin \angle CEA}$ Multiplying the first two fractions, dividing them by the next two fractions, and using the fact that $\sin(180^{\circ}-\theta)=\sin \theta$ , one gets the required equality. The converse can be proved in a similar way.

PROOF 2 (Inversion) Let $\omega$ be the circumcircle of $\triangle ABC$ , and let $AD \cap \omega=P, AE \cap \omega =Q$ . Invert about $A$ with radius $\sqrt{AB \cdot AC}$ followed by reflection in the internal angle bisector of $\angle BAC$ . Then $D \rightarrow Q$ and $E \rightarrow P$ . Thus it suffices to show that $\frac{BP}{CQ}=\frac{CP}{BQ} \Leftrightarrow \frac{BP}{CP}=\frac{CQ}{BQ}$ As $\angle BPC=\angle BQC$ , this is equivalent to proving that $\triangle BPC \sim \triangle CQB$ . But this is obviously true, cause we have $PQ \parallel BC$ , i.e. $BPQC$ is an isosceles trapezoid. The converse can be proved in a similar fashion.

So now let's move on to some questions which actually use this theorem. Remember that this theorem is just a part of these solutions (albeit an important one), and so most solutions using this theorem might require other lemmas and observations too. Having said that, let's move forward.

PROBLEM 1 (Source=Balamatda)
Let $ABC$ be a triangle inscribed in a circle $(O)$ . Suppose that $AH$ is the altitude and the line $AO$ intersects $BC$ at $D$ . The circumcircle of $ADC$ intersects the circumcircle of $AHB$ at $E$ . The tangent at $A$ of $(O)$ meets $BC$ at $I$ . Prove that $IA=IE$ .

https://scontent.fdad3-3.fna.fbcdn.net/v/t1.0-9/42614647_2256121971285643_3789226474465132544_n.jpg?_nc_cat=100&oh=dddf8d8f46971fe51fc3e49a6899879e&oe=5C61361D

SOLUTION: We have $\angle BEH=\angle BAH=\angle CAO=\angle CED \Rightarrow ED$ and $EH$ are isogonal wrt $\angle BEC$ . By Steiner's Ratio Theorem, we get that $\left (\frac{AB}{AC} \right)^2=\frac{BH \cdot BD}{CH \cdot CD}=\left (\frac{BE}{CE} \right)^2 \Rightarrow \frac{AB}{AC}=\frac{BE}{CE}$ This means that $E$ lies on the $A$ -Apollonius circle, giving that $IA=IE$ (as $I$ is the center of the $A$ -Apollonius circle). $\blacksquare$

REMARK: The above solution also shows that the problem is true for any two isogonal lines.

PROBLEM 2 (ELMO 2016 P6)
Elmo is now learning olympiad geometry. In triangle $ABC$ with $AB\neq AC$ , let its incircle be tangent to sides $BC$ , $CA$ , and $AB$ at $D$ , $E$ , and $F$ , respectively. The internal angle bisector of $\angle BAC$ intersects lines $DE$ and $DF$ at $X$ and $Y$ , respectively. Let $S$ and $T$ be distinct points on side $BC$ such that $\angle XSY=\angle XTY=90^\circ$ . Finally, let $\gamma$ be the circumcircle of $\triangle AST$ .

(a) Help Elmo show that $\gamma$ is tangent to the circumcircle of $\triangle ABC$ .

(b) Help Elmo show that $\gamma$ is tangent to the incircle of $\triangle ABC$ .

James Lin

SOLUTION: WLOG assume $AB \leq AC$ . Let $I$ be the incenter of $\triangle ABC$ .

Let $\omega$ be the circumcircle of $XSYT$ , and $Z$ be the midpoint of $XY$ , i.e. center of $\omega$ . Also, Let $AI \cap BC = P$ .

Let $DA$ meet the incircle again at $K$ . Then $DEKF$ is harmonic $\Rightarrow -1=(K,D;E,F) \overset{D}{=} (A,P;X,Y)$ .

This means that $A$ and $P$ are inverses w.r.t. $\omega$ , i.e. $ZX^2=ZP \cdot ZA = PZ(PZ+PA) =PZ^2+PZ \cdot PA$

$\Rightarrow PZ \cdot PA = ZX^2-ZP^2 = (ZX-ZP)(ZX+ZP) = (ZX-ZP)(ZY+ZP) = PX \cdot PY = PS \cdot PT \Rightarrow ASZT$ is cyclic.

As $ZS = ZT$ , $AZ$ is the internal angle bisector of $AS$ and $AT$ $\Rightarrow AS$ and $AT$ are isogonal.

(a) By Steiner's Ratio Theorem, $\frac{BT}{CT} \cdot \frac{BS}{CS} = \left(\frac{AB}{AC} \right)^2 \Rightarrow AC \cdot \sqrt{BS \cdot BT} - AB \cdot \sqrt{CS \cdot CT} = 0$ . By the Converse of Casey's Theorem on point circles $\odot (A), \odot (B), \odot (C)$ and $\odot (AST)$ , we get that $\odot (AST)$ and $\odot (ABC)$ are tangent to each other at $A$ . $\blacksquare$

(b) Let $AI \cap EF = T$ , then $\angle FTY = 90^{\circ} \Rightarrow \angle DYX = \angle FYT = 90^{\circ}-\angle EFD = \angle IDE = \angle IDX$

$\Rightarrow ID$ is tangent to $\odot (DXY) \Rightarrow ID^2 = IX \cdot IY \Rightarrow X$ and $Y$ are inverses w.r.t. the incircle.

Thus, $\omega$ and the incircle are orthogonal $\Rightarrow$ Length of tangent from $Z$ to $\omega$ , i.e. $\ell$ , is equal to $ZS$ .

Now, $ZS \cdot ST = ZS(SD+TD) = ZS \cdot TD+ZT \cdot SD \Rightarrow ZS \cdot TD +ZT \cdot SD - ST \cdot \ell = 0$ . By the Converse of Casey's Theorem on point circles $\odot (Z), \odot (S), \odot (T)$ and the incircle, and using the fact that $Z$ lies on $\odot (AST)$ , we get that $\odot (AST)$ and the incircle are tangent to each other. $\blacksquare$

REMARK: The first part gives an important result, which has been stated more clearly below.

RESULT Let $D$ and $E$ be points on $BC$ , so that $AD$ and $AE$ are isogonal with respect to $\angle BAC$ . Then $\odot (ADE)$ is tangent to $\odot (ABC)$ at $A$ . The converse is also true.

PROOF: See the solution to the first part of ELMO 2016 P6 given above.

PROBLEM 3 (Sharygin 2018 Correspondence Round Problem 16)
Let $ABC$ be a triangle with $AB < BC$ . The bisector of angle $C$ meets the line parallel to $AC$ and passing through $B$ , at point $P$ . The tangent at $B$ to the circumcircle of $ABC$ meets this bisector at point $R$ . Let $R'$ be the reflection of $R$ with respect to $AB$ . Prove that $\angle R'PB = \angle RPA$ .

SOLUTION: (Bary bash

) The attached solution is the one that I submitted during the actual contest.

Attachments:

Solution_Q 16.docx (72kb)

This post has been edited 3 times. Last edited by math_pi_rate, Feb 13, 2019, 4:18 PM

IMO 2018 Problem 2

by juckter, Jul 9, 2018, 11:26 AM

Find all integers $n \geq 3$ for which there exist real numbers $a_1, a_2, \dots a_{n + 2}$ satisfying $a_{n + 1} = a_1$ , $a_{n + 2} = a_2$ and
$a_ia_{i + 1} + 1 = a_{i + 2},$ for $i = 1, 2, \dots, n$ .

Proposed by Patrik Bak, Slovakia

This post has been edited 2 times. Last edited by djmathman, Jun 16, 2020, 4:02 AM
Reason: problem author

Prime-related integers [CMO 2018 - P3]

by Amir Hossein, Mar 31, 2018, 2:16 AM

Two positive integers $a$ and $b$ are prime-related if $a = pb$ or $b = pa$ for some prime $p$ . Find all positive integers $n$ , such that $n$ has at least three divisors, and all the divisors can be arranged without repetition in a circle so that any two adjacent divisors are prime-related.

Note that $1$ and $n$ are included as divisors.

number theory

graph theory

Operating on lamps in a circle

by anantmudgal09, Dec 9, 2017, 1:08 PM

There are $n$ lamps $L_1, L_2, \dots, L_n$ arranged in a circle in that order. At any given time, each lamp is either on or off. Every second, each lamp undergoes a change according to the following rule:

(a) For each lamp $L_i$ , if $L_{i-1}, L_i, L_{i+1}$ have the same state in the previous second, then $L_i$ is off right now. (Indices taken mod $n$ .)

(b) Otherwise, $L_i$ is on right now.

Initially, all the lamps are off, except for $L_1$ which is on. Prove that for infinitely many integers $n$ all the lamps will be off eventually, after a finite amount of time.

This post has been edited 1 time. Last edited by anantmudgal09, Dec 9, 2017, 1:09 PM

combinatorics

Flee Jumping on Number Line

by utkarshgupta, Dec 11, 2015, 3:59 PM

An immortal flea jumps on whole points of the number line, beginning with $0$ . The length of the first jump is $3$ , the second $5$ , the third $9$ , and so on. The length of $k^{\text{th}}$ jump is equal to $2^k + 1$ . The flea decides whether to jump left or right on its own. Is it possible that sooner or later the flee will have been on every natural point, perhaps having visited some of the points more than once?

This post has been edited 2 times. Last edited by djmathman, Apr 15, 2016, 5:59 PM

combinatorics

number theory

All Naturals

Submit

First comment of January 26, 2025!
by Yiyj1, Jan 27, 2025, 1:02 AM
And here's the first of 2025!
by CrimsonBlade273, Jan 9, 2025, 6:16 AM
First comment of 2024!
by mannshah1211, Jan 14, 2024, 3:01 PM
Wowowooo my man came backkkk
by HoRI_DA_GRe8, Nov 11, 2023, 4:21 PM
Aah the thrill of coming back to a dead blog every year once!
by math_pi_rate, Jul 28, 2023, 8:43 PM
kukuku 1st comment of 2023
by kamatadu, Jan 3, 2023, 7:32 PM
1st comment of 2022
by HoRI_DA_GRe8, May 25, 2022, 8:29 PM
duh i found this masterpiece after the owner went inactive noooooooooooo :sadge:
by Project_Donkey_into_M4, Nov 23, 2021, 2:56 PM
Hi everyone! Sorry but I doubt if I'll be reviving this anytime soon

I might post something if I find anything interesting, but they'll mostly be short posts, unlike the previous ones!
by math_pi_rate, Nov 7, 2020, 7:39 PM
@below be patient. He must be busy with some other work.
by amar_04, Oct 22, 2020, 10:23 AM
Advanced is over now, please revive this please!
by Geronimo_1501, Oct 12, 2020, 5:10 AM
REVIVE please
by Gaussian_cyber, Aug 28, 2020, 2:19 PM
re-vi-ve
by nprime06, Jun 12, 2020, 2:17 AM
re-vi-ve
by fukano_2, May 30, 2020, 2:09 AM
Try this: Prove that a two colored cow has an odd number of Agis Phesis
by Synthetic_Potato, Apr 18, 2020, 5:29 PM
Nice blog!!
by enthusiast101, Nov 9, 2018, 10:44 AM
Updates required
by TheDarkPrince, Oct 29, 2018, 10:07 AM
All hail to geo god math_pi_rate -- who solved 2011 G8 in kindergarten
by Kayak, Oct 24, 2018, 6:46 AM
Thanks. Will probably post after RMO only now.
by math_pi_rate, Sep 30, 2018, 7:22 AM
Nice blog!!! Keep posting!
by ayan.nmath, Sep 28, 2018, 9:01 PM
Thanks. I'll make sure that I include my thought process in the next post.
by math_pi_rate, Sep 27, 2018, 1:01 PM
Can you post some advice on how to get better at synthetic geometry on your blog or say your thought process while solving geo problems (like your post on FE)? Nice post on FE though !
by Kayak, Sep 27, 2018, 11:38 AM
I am sorry, but what do u mean by tangents?
by math_pi_rate, Sep 27, 2018, 9:23 AM
Can you post something on Tangents?
by PhysicsMonster_01, Sep 27, 2018, 6:42 AM
Thanks @below and @2below.
@below Posted smth apart from geo just for you. Jai Mahismati.
by math_pi_rate, Sep 17, 2018, 4:47 PM
How are you so good in geo...

#geoishard
by Kayak, Sep 16, 2018, 4:45 PM
Nice work Genius :O
by TheDarkPrince, Sep 13, 2018, 1:08 PM
Thanks a lot!
by math_pi_rate, Aug 20, 2018, 12:17 PM
First shout! Lovely tangency theorem
by silverivy, Aug 19, 2018, 4:08 PM