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AD=BE implies ABC right
v_Enhance   115
N 18 minutes ago by Adywastaken
Source: European Girl's MO 2013, Problem 1
The side $BC$ of the triangle $ABC$ is extended beyond $C$ to $D$ so that $CD = BC$. The side $CA$ is extended beyond $A$ to $E$ so that $AE = 2CA$. Prove that, if $AD=BE$, then the triangle $ABC$ is right-angled.
115 replies
v_Enhance
Apr 10, 2013
Adywastaken
18 minutes ago
Geometry
gggzul   6
N 22 minutes ago by Captainscrubz
In trapezoid $ABCD$ segments $AB$ and $CD$ are parallel. Angle bisectors of $\angle A$ and $\angle C$ meet at $P$. Angle bisectors of $\angle B$ and $\angle D$ meet at $Q$. Prove that $ABPQ$ is cyclic
6 replies
gggzul
Yesterday at 8:22 AM
Captainscrubz
22 minutes ago
Need help on this simple looking problem
TheGreatEuler   0
27 minutes ago
Show that 1+2+3+4....n divides 1^k+2^k+3^k....n^k when k is odd. Is this possible to prove without using congruence modulo or binomial coefficients?
0 replies
TheGreatEuler
27 minutes ago
0 replies
Geometry
Lukariman   5
N an hour ago by Lukariman
Given circle (O) and point P outside (O). From P draw tangents PA and PB to (O) with contact points A, B. On the opposite ray of ray BP, take point M. The circle circumscribing triangle APM intersects (O) at the second point D. Let H be the projection of B on AM. Prove that $\angle HDM$ = 2∠AMP.
5 replies
Lukariman
Yesterday at 12:43 PM
Lukariman
an hour ago
inq , not two of them =0
win14   0
an hour ago
Let a,b,c be non negative real numbers such that no two of them are simultaneously equal to 0
$$\frac{1}{a + b} + \frac{1}{b + c} + \frac{1}{c + a} \ge \frac{5}{2\sqrt{ab + bc + ca}}.$$
0 replies
1 viewing
win14
an hour ago
0 replies
IMO Genre Predictions
ohiorizzler1434   62
N an hour ago by ehuseyinyigit
Everybody, with IMO upcoming, what are you predictions for the problem genres?


Personally I predict: predict
62 replies
ohiorizzler1434
May 3, 2025
ehuseyinyigit
an hour ago
Number theory
MathsII-enjoy   5
N 2 hours ago by MathsII-enjoy
Prove that when $x^p+y^p$ | $(p^2-1)^n$ with $x,y$ are positive integers and $p$ is prime ($p>3$), we get: $x=y$
5 replies
MathsII-enjoy
Monday at 3:22 PM
MathsII-enjoy
2 hours ago
Number theory
Foxellar   0
2 hours ago
It is known that for all positive integers $k$,
\[
1^2 + 2^2 + 3^2 + \ldots + k^2 = \frac{k(k + 1)(2k + 1)}{6}
\]Find the smallest positive integer $k$ such that $1^2 + 2^2 + 3^2 + \ldots + k^2$ is divisible by 200.
0 replies
Foxellar
2 hours ago
0 replies
Combinatorics
P162008   4
N 2 hours ago by cazanova19921
Let $m,n \in \mathbb{N}.$ Let $[n]$ denote the set of natural numbers less than or equal to $n.$

Let $f(m,n) = \sum_{(x_1,x_2,x_3, \cdots, x_m) \in [n]^{m}} \frac{x_1}{x_1 + x_2 + x_3 + \cdots + x_m} \binom{n}{x_1} \binom{n}{x_2} \binom{n}{x_3} \cdots \binom{n}{x_m} 2^{\left(\sum_{i=1}^{m} x_i\right)}$

Compute the sum of the digits of $f(4,4).$
4 replies
P162008
Today at 5:38 AM
cazanova19921
2 hours ago
Aime type Geo
ehuseyinyigit   4
N 2 hours ago by ehuseyinyigit
Source: Turkish First Round 2024
In a scalene triangle $ABC$, let $M$ be the midpoint of side $BC$. Let the line perpendicular to $AC$ at point $C$ intersect $AM$ at $N$. If $(BMN)$ is tangent to $AB$ at $B$, find $AB/MA$.
4 replies
ehuseyinyigit
Monday at 9:04 PM
ehuseyinyigit
2 hours ago
n variables with n-gon sides
mihaig   1
N 2 hours ago by mihaig
Source: Own
Let $n\geq3$ and let $a_1,a_2,\ldots, a_n\geq0$ be reals such that $\sum_{i=1}^{n}{\frac{1}{2a_i+n-2}}=1.$
Prove
$$\frac{24}{(n-1)(n-2)}\cdot\sum_{1\leq i<j<k\leq n}{a_ia_ja_k}\geq3\sum_{i=1}^{n}{a_i}+n.$$
1 reply
mihaig
Apr 25, 2025
mihaig
2 hours ago
trolling geometry problem
iStud   4
N Apr 22, 2025 by GreenTea2593
Source: Monthly Contest KTOM April P3 Essay
Given a cyclic quadrilateral $ABCD$ with $BC<AD$ and $CD<AB$. Lines $BC$ and $AD$ intersect at $X$, and lines $CD$ and $AB$ intersect at $Y$. Let $E,F,G,H$ be the midpoints of sides $AB,BC,CD,DA$, respectively. Let $S$ and $T$ be points on segment $EG$ and $FH$, respectively, so that $XS$ is the angle bisector of $\angle{DXA}$ and $YT$ is the angle bisector of $\angle{DYA}$. Prove that $TS$ is parallel to $BD$ if and only if $AC$ divides $ABCD$ into two triangles with equal area.
4 replies
iStud
Apr 21, 2025
GreenTea2593
Apr 22, 2025
trolling geometry problem
G H J
Source: Monthly Contest KTOM April P3 Essay
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iStud
268 posts
#1
Y by
Given a cyclic quadrilateral $ABCD$ with $BC<AD$ and $CD<AB$. Lines $BC$ and $AD$ intersect at $X$, and lines $CD$ and $AB$ intersect at $Y$. Let $E,F,G,H$ be the midpoints of sides $AB,BC,CD,DA$, respectively. Let $S$ and $T$ be points on segment $EG$ and $FH$, respectively, so that $XS$ is the angle bisector of $\angle{DXA}$ and $YT$ is the angle bisector of $\angle{DYA}$. Prove that $TS$ is parallel to $BD$ if and only if $AC$ divides $ABCD$ into two triangles with equal area.
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iStud
268 posts
#2
Y by
i can prove the $\Longleftarrow$ direction but not pretty sure with the $\Longrightarrow$ direction, can anyone help me?
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mathuz
1524 posts
#3
Y by
Note that $EFGH$ is a parallelogram and $FG\parallel EH \parallel BD$. Therefore, $TS\parallel BD$ iff $ES:SG = HT:TF$.

Assume now $TS\parallel BD$. Then $ES:SG = HT:TF$.
We have $\triangle XCD \sim \triangle XAB$ and $\triangle YBC\sim \triangle YDA$. This implies $XS$ is the angle bisector of $\angle EXG$ and $YT$ is the angle bisector of $\angle HYF$. Therefore, we have \[ \frac{EX}{GX} = \frac{ES}{SG} = \frac{HT}{TF} = \frac{HY}{FY}. \]Again, by the similarity this means \[ \frac{AB}{CD} = \frac{DA}{BC} \quad \Rightarrow \quad AB\cdot BC = CD\cdot DA, \]which essentially means the areas of $ABC$ and $ADC$ are equal.

This gives $\Longrightarrow $ direction.
This post has been edited 1 time. Last edited by mathuz, Apr 22, 2025, 1:35 AM
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iStud
268 posts
#4
Y by
mathuz wrote:
Note that $EFGH$ is a parallelogram and $FG\parallel EH \parallel BD$. Therefore, $TS\parallel BD$ iff $ES:SG = HT:TF$.

Assume now $TS\parallel BD$. Then $ES:SG = HT:TF$.
We have $\triangle XCD \sim \triangle XAB$ and $\triangle YBC\sim \triangle YDA$. This implies $XS$ is the angle bisector of $\angle EXG$ and $YT$ is the angle bisector of $\angle HYF$. Therefore, we have \[ \frac{EX}{GX} = \frac{ES}{SG} = \frac{HT}{TF} = \frac{HY}{FY}. \]Again, by the similarity this means \[ \frac{AB}{CD} = \frac{DA}{BC} \quad \Rightarrow \quad AB\cdot BC = CD\cdot DA, \]which essentially means the areas of $ABC$ and $ADC$ are equal.

This gives $\Longrightarrow $ direction.

demn, all of my Menelause bash worth nothing in front of this dude's solution
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GreenTea2593
236 posts
#5
Y by
Oops, I took this from 2016 Japan MO Finals Problem 2
This post has been edited 1 time. Last edited by GreenTea2593, Apr 22, 2025, 3:12 AM
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