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Italian WinterCamps test07 Problem4
mattilgale   88
N 34 minutes ago by OronSH
Source: ISL 2006, G3, VAIMO 2007/5
Let $ ABCDE$ be a convex pentagon such that
\[ \angle BAC = \angle CAD = \angle DAE\qquad \text{and}\qquad \angle ABC = \angle ACD = \angle ADE.
\]The diagonals $BD$ and $CE$ meet at $P$. Prove that the line $AP$ bisects the side $CD$.

Proposed by Zuming Feng, USA
88 replies
mattilgale
Jan 29, 2007
OronSH
34 minutes ago
Hard to approach it !
BogG   129
N 38 minutes ago by OronSH
Source: Swiss Imo Selection 2006
Let $\triangle ABC$ be an acute-angled triangle with $AB \not= AC$. Let $H$ be the orthocenter of triangle $ABC$, and let $M$ be the midpoint of the side $BC$. Let $D$ be a point on the side $AB$ and $E$ a point on the side $AC$ such that $AE=AD$ and the points $D$, $H$, $E$ are on the same line. Prove that the line $HM$ is perpendicular to the common chord of the circumscribed circles of triangle $\triangle ABC$ and triangle $\triangle ADE$.
129 replies
BogG
May 25, 2006
OronSH
38 minutes ago
Simple triangle geometry [a fixed point]
darij grinberg   48
N 42 minutes ago by OronSH
Source: German TST 2004, IMO ShortList 2003, geometry problem 2
Three distinct points $A$, $B$, and $C$ are fixed on a line in this order. Let $\Gamma$ be a circle passing through $A$ and $C$ whose center does not lie on the line $AC$. Denote by $P$ the intersection of the tangents to $\Gamma$ at $A$ and $C$. Suppose $\Gamma$ meets the segment $PB$ at $Q$. Prove that the intersection of the bisector of $\angle AQC$ and the line $AC$ does not depend on the choice of $\Gamma$.
48 replies
darij grinberg
May 18, 2004
OronSH
42 minutes ago
IMO ShortList 1998, geometry problem 4
orl   14
N 44 minutes ago by OronSH
Source: IMO ShortList 1998, geometry problem 4
Let $ M$ and $ N$ be two points inside triangle $ ABC$ such that
\[ \angle MAB = \angle NAC\quad \mbox{and}\quad \angle MBA = \angle NBC.
\]
Prove that
\[ \frac {AM \cdot AN}{AB \cdot AC} + \frac {BM \cdot BN}{BA \cdot BC} + \frac {CM \cdot CN}{CA \cdot CB} = 1.
\]
14 replies
orl
Oct 22, 2004
OronSH
44 minutes ago
Functional equation
Nima Ahmadi Pour   98
N an hour ago by ezpotd
Source: ISl 2005, A2, Iran prepration exam
We denote by $\mathbb{R}^+$ the set of all positive real numbers.

Find all functions $f: \mathbb R^ + \rightarrow\mathbb R^ +$ which have the property:
\[f(x)f(y)=2f(x+yf(x))\]
for all positive real numbers $x$ and $y$.

Proposed by Nikolai Nikolov, Bulgaria
98 replies
Nima Ahmadi Pour
Apr 24, 2006
ezpotd
an hour ago
Geometry
noneofyou34   0
an hour ago
Please can someone help me prove that orthocenter of a triangle exists by using Menelau's Theorem!
0 replies
noneofyou34
an hour ago
0 replies
Hard combi
EeEApO   0
an hour ago
In a quiz competition, there are a total of $100 $questions, each with $4$ answer choices. A participant who answers all questions correctly will receive a gift. To ensure that at least one member of my family answers all questions correctly, how many family members need to take the quiz?

Now, suppose my spouse and I move into a new home. Every year, we have twins. Starting at the age of $16$, each of our twin children also begins to have twins every year. If this pattern continues, how many years will it take for my family to grow large enough to have the required number of members to guarantee winning the quiz gift?
0 replies
EeEApO
an hour ago
0 replies
Inequality with mathematical means
StefanSebez   12
N an hour ago by Sh309had
Source: Serbia JBMO TST 2022 P1
Prove that for all positive real numbers $a$, $b$ the following inequality holds:
\begin{align*}
\sqrt{\frac{a^2+b^2}{2}}+\frac{2ab}{a+b}\ge \frac{a+b}{2}+ \sqrt{ab}
\end{align*}When does equality hold?
12 replies
StefanSebez
Jun 1, 2022
Sh309had
an hour ago
Really fun geometry problem
Sadigly   4
N 2 hours ago by Double07
Source: Azerbaijan Senior MO 2025 P6
In the acute triangle $ABC$ with $AB<AC$, the foot of altitudes from $A,B,C$ to the sides $BC,CA,AB$ are $D,E,F$, respectively. $H$ is the orthocenter. $M$ is the midpoint of segment $BC$. Lines $MH$ and $EF$ intersect at $K$. Let the tangents drawn to circumcircle $(ABC)$ from $B$ and $C$ intersect at $T$. Prove that $T;D;K$ are colinear
4 replies
Sadigly
3 hours ago
Double07
2 hours ago
Orthocenter
jayme   8
N 2 hours ago by cj13609517288
Dear Mathlinkers,

1. ABC an acuatangle triangle
2. H the orthcenter of ABC
3. DEF the orthic triangle of ABC
4. A* the midpoint of AH
5. X the point of intersection of AH and EF.

Prove : X is the orthocenter of A*BC.

Sincerely
Jean-Louis
8 replies
jayme
Mar 25, 2015
cj13609517288
2 hours ago
Constructing graphs satisfying conditions on degrees
jlammy   19
N 2 hours ago by de-Kirschbaum
Source: EGMO 2017 P4
Let $n\geq1$ be an integer and let $t_1<t_2<\dots<t_n$ be positive integers. In a group of $t_n+1$ people, some games of chess are played. Two people can play each other at most once. Prove that it is possible for the following two conditions to hold at the same time:

(i) The number of games played by each person is one of $t_1,t_2,\dots,t_n$.

(ii) For every $i$ with $1\leq i\leq n$, there is someone who has played exactly $t_i$ games of chess.
19 replies
jlammy
Apr 9, 2017
de-Kirschbaum
2 hours ago
Equal angles with midpoint of $AH$
Stuttgarden   2
N Apr 24, 2025 by HormigaCebolla
Source: Spain MO 2025 P4
Let $ABC$ be an acute triangle with circumcenter $O$ and orthocenter $H$, satisfying $AB<AC$. The tangent line at $A$ to the circumcicle of $ABC$ intersects $BC$ in $T$. Let $X$ be the midpoint of $AH$. Prove that $\angle ATX=\angle OTB$.
2 replies
Stuttgarden
Mar 31, 2025
HormigaCebolla
Apr 24, 2025
Equal angles with midpoint of $AH$
G H J
G H BBookmark kLocked kLocked NReply
Source: Spain MO 2025 P4
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Stuttgarden
34 posts
#1 • 1 Y
Y by Rounak_iitr
Let $ABC$ be an acute triangle with circumcenter $O$ and orthocenter $H$, satisfying $AB<AC$. The tangent line at $A$ to the circumcicle of $ABC$ intersects $BC$ in $T$. Let $X$ be the midpoint of $AH$. Prove that $\angle ATX=\angle OTB$.
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WLOGQED1729
46 posts
#2 • 1 Y
Y by ehuseyinyigit
Nice property! Here is my solution.

Let $M$ be midpoint of side $BC$. It is well-known that $AXMO$ is parallelogram.
Claim $X$ is orthocenter of triangle $ATM$
Proof We have $AX \perp TM$. Since $OA \perp AT$ and $XM \parallel AO$, we deduce that $MX \perp AT$. $\square$

The rest is just angle chasing.
Note that $\angle ATX = \angle AMX = \angle OAM =\angle OTM = \angle OTB$ $\blacksquare$
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HormigaCebolla
3 posts
#3 • 1 Y
Y by Steve12345
My solution during the exam: parallelogram isogonality lemma applied to parallelogram $AXMO$ and triangle $TAM$, where $M$ is the midpoint of side $BC$.
This post has been edited 1 time. Last edited by HormigaCebolla, Apr 25, 2025, 1:46 PM
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