Art of Problem Solving
AoPS Online AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy AoPS Academy
Small live classes for advanced math
and language arts learners in grades 1-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Stay ahead of learning milestones! Enroll in a class over the summer!
geometry
B
No tags match your search
M
G
Topic
First Poster
Last Poster
H
Problem 1
SpectralS   145
N 18 minutes ago by IndexLibrorumProhibitorum
Given triangle $ABC$ the point $J$ is the centre of the excircle opposite the vertex $A.$ This excircle is tangent to the side $BC$ at $M$, and to the lines $AB$ and $AC$ at $K$ and $L$, respectively. The lines $LM$ and $BJ$ meet at $F$, and the lines $KM$ and $CJ$ meet at $G.$ Let $S$ be the point of intersection of the lines $AF$ and $BC$, and let $T$ be the point of intersection of the lines $AG$ and $BC.$ Prove that $M$ is the midpoint of $ST.$

(The excircle of $ABC$ opposite the vertex $A$ is the circle that is tangent to the line segment $BC$, to the ray $AB$ beyond $B$, and to the ray $AC$ beyond $C$.)

Proposed by Evangelos Psychas, Greece
145 replies
SpectralS
Jul 10, 2012
IndexLibrorumProhibitorum
18 minutes ago
J
H
Binomial Sum
P162008   0
2 hours ago
Compute $\sum_{r=0}^{n} \sum_{k=0}^{r} (-1)^k (k + 1)(k + 2) \binom {n + 5}{r - k}$
0 replies
P162008
2 hours ago
0 replies
J
H
Triple Sum
P162008   0
3 hours ago
Find the value of

$\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} \sum_{n=0}^{\infty} \sum_{m=0}^{\infty} \frac{(-1)^m}{k.2^n + 2m + 1}$
0 replies
P162008
3 hours ago
0 replies
J
H
Binomial Sum
P162008   0
3 hours ago
The numbers $p$ and $q$ are defined in the following manner:

$p = 99^{98} - \frac{99}{1} 98^{98} + \frac{99.98}{1.2} 97^{98} - \frac{99.98.97}{1.2.3} 96^{98} + .... + 99$

$q = 99^{100} - \frac{99}{1} 98^{100} + \frac{99.98}{1.2} 97^{100} - \frac{99.98.97}{1.2.3} 96^{100} + .... + 99$

If $p + q = k(99!)$ then find the value of $\frac{k}{10}.$
0 replies
P162008
3 hours ago
0 replies
J
H
Polynomial Limit
P162008   0
3 hours ago
If $P_{n}(x) = \prod_{k=0}^{n} \left(x + \frac{1}{2^k}\right) = \sum_{k=0}^{n} a_{k} x^k$ then find the value of $\lim_{n \to \infty} \frac{a_{n - 2}}{a_{n - 4}}.$
0 replies
P162008
3 hours ago
0 replies
J
H
Telescopic Sum
P162008   0
3 hours ago
Compute the value of $\Omega = \sum_{r=1}^{\infty} \frac{14 - 9r - 90r^2 - 36r^3}{7^r r(r + 1)(r + 2)(4r^2 - 1)}$
0 replies
P162008
3 hours ago
0 replies
J
H
Theory of Equations
P162008   0
4 hours ago
Let $a,b,c,d$ and $e\in [-2,2]$ such that $\sum_{cyc} a = 0, \sum_{cyc} a^3 = 0, \sum_{cyc} a^5 = 10.$ Find the value of $\sum_{cyc} a^2.$
0 replies
P162008
4 hours ago
0 replies
J
H
CHINA TST 2017 P6 DAY1
lingaguliguli   0
6 hours ago
When i search the china TST 2017 problem 6 day I i crossed out this lemme, but don't know to prove it, anyone have suggestion? tks
Given a fixed number n, and a prime p. Let f(x)=(x+a_1)(x+a_2)...(x+a_n) in which a_1,a_2,...a_n are positive intergers. Show that there exist an interger M so that 0<v_p((f(M))< n + v_p(n!)
0 replies
lingaguliguli
6 hours ago
0 replies
J
H
Math and physics camp
Snezana242   0
Today at 8:53 AM
Discover IMPSC 2025: International Math & Physics Summer Camp!

Are you a high school student (grades 9–12) with a passion for Physics and Math?
Join the IMPSC 2025, an online summer camp led by top IIT professors, offering a college-level education in Physics and Math.

What Can You Expect?

Learn advanced topics from renowned IIT professors

Connect with students worldwide

Strengthen your college applications

Get a recommendation letter for top universities!

How to Apply & More Info
For all the details you need about the camp, dates, application process, and more, visit our official website:
https://www.imc-impea.org/IMC/index.php

Don't miss out on this opportunity to elevate your academic journey!
Apply now and take your education to the next level.
0 replies
Snezana242
Today at 8:53 AM
0 replies
J
H
Combinatoric
spiderman0   1
N Today at 6:44 AM by MathBot101101
Let $ S = \{1, 2, 3, \ldots, 2024\}.$ Find the maximum positive integer $n \geq 2$ such that for every subset $T \subset S$ with n elements, there always exist two elements a, b in T such that:

$|\sqrt{a} - \sqrt{b}| < \frac{1}{2} \sqrt{a - b}$
1 reply
spiderman0
Yesterday at 7:46 AM
MathBot101101
Today at 6:44 AM
J
H
Combinatorial proof
MathBot101101   10
N Today at 6:20 AM by MathBot101101
Is there a way to prove
\frac{1}{(1+1)!}+\frac{2}{(2+1)!}+...+\frac{n}{(n+1)!}=1-\frac{1}{{n+1)!}
without induction and using only combinatorial arguments?

Induction proof wasn't quite as pleasing for me.
10 replies
MathBot101101
Apr 20, 2025
MathBot101101
Today at 6:20 AM
J
H
J
H
Yet another way to reflect H to the circumcircle!
Tintarn   3
N Nov 24, 2024 by Alex-Five
Source: Baltic Way 2024, Problem 14
Let $ABC$ be an acute triangle with circumcircle $\omega$. The altitudes $AD$, $BE$ and $CF$ of the triangle $ABC$ intersect at point $H$. A point $K$ is chosen on the line $EF$ such that $KH\parallel BC$. Prove that the reflection of $H$ in $KD$ lies on $\omega$.
3 replies
Tintarn
Nov 16, 2024
Alex-Five
Nov 24, 2024
J
Yet another way to reflect H to the circumcircle!
G H J
Reveal topic tags
geometry
circumcircle
geometry proposed
geometric transformation
reflection
L
G H BBookmark kLocked kLocked NReply
Source: Baltic Way 2024, Problem 14
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Tintarn
9038 posts
Tintarn
#1 h  Nov 16, 2024, 6:23 PM
Y by
Let $ABC$ be an acute triangle with circumcircle $\omega$. The altitudes $AD$, $BE$ and $CF$ of the triangle $ABC$ intersect at point $H$. A point $K$ is chosen on the line $EF$ such that $KH\parallel BC$. Prove that the reflection of $H$ in $KD$ lies on $\omega$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
bin_sherlo
707 posts
bin_sherlo
#2 h  Nov 16, 2024, 6:57 PM
Y by
Solution
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Lil_flip38
53 posts
Lil_flip38
#3 h  Nov 17, 2024, 10:44 AM • 1 Y
Y by eddiekopp
Alternate super long solution:
Lemma: in triangle $ABC$, let $E,F$ be the feet from $B,C$. Let $A'$ be the reflection of $A$ across the perpendicular bisector of $BC$. Let $Q$ be the $A$ queue point. Then, $QAEF\sim QA'CB$.
Proof: Consider the spiral similarity that maps $(AEF)$ to $(ABC)$. $F$ is sent to $B$, $E$ to $C$ so we need that $A$ is mapped to $A'$. as $HAA'=90^\circ$, $AA'$ is tangent to $(AEF)$, so by properties of spiral simiarities, the lemma is proven.

Now, let $H'$ be the intersection of the circle with radius $HK$ with center $K$ with $(ABC)$. Let $A'$ be the point s.t. $AA'\parallel BC$, with $A'\in (ABC)$, and $Q$ be the $A$-queue point.

Claim 1:$QK=KH$
proof. $KH$ is tangent to $(AEF)$ as $KH\perp AD$.
Now, as $(Q,H;E,F)=-1$ and $K\in EF$ we have that $KQ$ has to be tangent as well, which implies the conclusion.

Now by claim $1$, we have that $K$ is the center of $(QKH)'$ and $(QKH')$ is tangent to $AD$.
Claim 2: $H,H',A'$ is collinear.
proof: By lemma, $\measuredangle QFA = \measuredangle QBA'$, so $\measuredangle QHH' = \measuredangle QHA = \measuredangle QFA = \measuredangle QBA' =\measuredangle QH'A'$ as desired.

to finish, by claim 2 we have $\angle A'H'L = 90^\circ$, where $L=(ABC)\cap AD$, but by the reflection of orthocenter lemma, $D$ is the midpoint of $HL$, so $DH'=DL=DH$ so as $KH'=KH$, $KD$ is the perpendicular bisector of $HH'$ as desired.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Alex-Five
4 posts
Alex-Five
#4 h  Nov 24, 2024, 7:26 PM
Y by
For storage
solution
Z K Y
N Quick Reply
G
H
=
a