Stay ahead of learning milestones! Enroll in a class over the summer!

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Summer math contest prep
Abby0618   6
N 2 hours ago by A7456321
School is almost out, so I have a lot of time in the summer. I want to be able to make DHR on AMC 8 in 7th grade

(current 6th grader) and hopefully get an average score in AMC 10. What should I do during the summer to achieve

these goals? For context, I have many books from AOPS, have already taken the Intro to Algebra A course, and took

AMC 8 for the first time as a 6th grader. If there are any challenging math problems you think would benefit learning,

please post them here. Thank you! :-D
6 replies
Abby0618
Yesterday at 8:52 PM
A7456321
2 hours ago
The daily problem!
Leeoz   192
N 3 hours ago by valenbb
Every day, I will try to post a new problem for you all to solve! If you want to post a daily problem, you can! :)

Please hide solutions and answers, hints are fine though! :)

Problems usually get harder throughout the week, so Sunday is the easiest and Saturday is the hardest!

Past Problems!
192 replies
Leeoz
Mar 21, 2025
valenbb
3 hours ago
max number of candies
orangefronted   30
N 5 hours ago by Creeperboat
A store sells a strawberry flavoured candy for 1 dollar each. The store offers a promo where every 4 candy wrappers can be exchanged for one candy. If there is no limit to how many times you can exchange candy wrappers for candies, what is the maximum number of candies I can obtain with 100 dollars?
30 replies
orangefronted
Apr 3, 2025
Creeperboat
5 hours ago
Help to make it clear on basic Concept
Miranda2829   2
N 6 hours ago by UberPiggy
Where I use extraneous solution in equation?
Why square of both side we will have plus and minus answer?
2 replies
Miranda2829
Yesterday at 8:32 PM
UberPiggy
6 hours ago
If $x = (4096)^{7+4\sqrt{3}},$ then which of the following equals $64:$ (A) $\fr
Vulch   4
N 6 hours ago by pieMax2713
If $x = (4096)^{7+4\sqrt{3}},$ then which of the following equals $64:$
(A) $\frac{x^{\frac{7}{2}}}{x^{\frac{4}{\sqrt{3}}}}$
(B) $\frac{x^{7}}{x^{4\sqrt{3}}}$
(C) $\frac{x^{\frac72}}{x^{2\sqrt{3}}}$
(D) $\frac{x^{7}}{ x^{2\sqrt{3}}}$
4 replies
Vulch
Yesterday at 9:27 PM
pieMax2713
6 hours ago
Worst math problems
LXC007   6
N 6 hours ago by A7456321
What is the most egregiously bad problem or solution you have encountered in school?
6 replies
LXC007
Wednesday at 11:42 PM
A7456321
6 hours ago
Quick Question
b2025tyx   13
N Yesterday at 8:44 PM by b2025tyx
During my math final today at school, the question said stated "When every integer is raised to the power of zero, it is equal to 1". The answers were multiple choice and were : Always, sometimes, never, and I don't know.

I ended up putting the first one, and was informed that it was incorrect. My teacher told me that $0^0$ is not equal to one. I looked it up, and it said $0^0 = 1$. Can someone confirm and prove this. Thanks!
13 replies
b2025tyx
May 20, 2025
b2025tyx
Yesterday at 8:44 PM
AP calc?
Thayaden   22
N Yesterday at 7:53 PM by Inaaya
How are we all feeling on AP calc guys?
22 replies
Thayaden
May 20, 2025
Inaaya
Yesterday at 7:53 PM
What's the chance that two AoPS accounts generate with the same icon?
Math-lover1   19
N Yesterday at 7:30 PM by ZMB038
So I've been wondering how many possible "icons" can be generated when you first create an account. By "icon" I mean the stack of cubes as the first profile picture before changing it.

I don't know a lot about how AoPS icons generate, so I have a few questions:
- Do the colors on AoPS icons generate through a preset of colors or the RGB (red, green, blue in hexadecimal form) scale? If it generates through the RGB scale, then there may be greater than $256^3 = 16777216$ different icons.
- Do the arrangements of the stacks of blocks in the icon change with each account? If so, I think we can calculate this through considering each stack of blocks independently.
19 replies
Math-lover1
May 2, 2025
ZMB038
Yesterday at 7:30 PM
A Variety of Math Problems to solve
FJH07   41
N Yesterday at 7:22 PM by FJH07
Hi, so people can post different math problems that they think are hard, and I will post some (I think middle school math level) problems so that the community can help solve them. :)
41 replies
FJH07
Yesterday at 1:36 AM
FJH07
Yesterday at 7:22 PM
Parallelity and equal angles given, wanted an angle equality
BarisKoyuncu   5
N Apr 24, 2025 by SleepyGirraffe
Source: 2022 Turkey JBMO TST P4
Given a convex quadrilateral $ABCD$ such that $m(\widehat{ABC})=m(\widehat{BCD})$. The lines $AD$ and $BC$ intersect at a point $P$ and the line passing through $P$ which is parallel to $AB$, intersects $BD$ at $T$. Prove that
$$m(\widehat{ACB})=m(\widehat{PCT})$$
5 replies
BarisKoyuncu
Mar 15, 2022
SleepyGirraffe
Apr 24, 2025
Parallelity and equal angles given, wanted an angle equality
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Source: 2022 Turkey JBMO TST P4
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BarisKoyuncu
577 posts
#1 • 4 Y
Y by teomihai, HWenslawski, son7, oralayhan
Given a convex quadrilateral $ABCD$ such that $m(\widehat{ABC})=m(\widehat{BCD})$. The lines $AD$ and $BC$ intersect at a point $P$ and the line passing through $P$ which is parallel to $AB$, intersects $BD$ at $T$. Prove that
$$m(\widehat{ACB})=m(\widehat{PCT})$$
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BarisKoyuncu
577 posts
#2 • 3 Y
Y by teomihai, HWenslawski, oralayhan
Let the line passing through $D$ and parallel to $AB$ intersects $BP$ and $AT$ at $K$ and $L$, respectively. We have
$$\frac{|DK|}{|AB|}=\frac{|PD|}{|PA|}=\frac{|PD|}{|PA|}\cdot\frac{|PT|}{|DL|}\cdot\frac{|DL|}{|PT|}=\frac{|PD|}{|PA|}\cdot\frac{|AP|}{|AD|}\cdot\frac{|DL|}{|PT|}=$$$$\frac{|PD|}{|AD|}\cdot\frac{|DL|}{|PT|}=\frac{|PT|}{|AB|}\cdot\frac{|DL|}{|PT|}=\frac{|DL|}{|AB|}\Rightarrow |DK|=|DL|$$Also, $\angle DCK=\angle ABC=\angle DKC\Rightarrow |DK|=|DC|$.
Therefore, $\angle LCK=90^{\circ}$.
Hence, it suffices to prove that the angle bisector of $\angle ACT$ is $CL$.
Let $TA\cap PB=M$. We have
$$\frac{|MA|}{|MT|}=\frac{|AB|}{|TP|}=\frac{|AD|}{|DP|}=\frac{|LA|}{|LT|}\Rightarrow (M,L;A,T)=-1$$Also, $\angle LCM=90^{\circ}$. Hence, we conclude that the angle bisector of $\angle ACT$ is $CL$.
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hakN
429 posts
#3 • 2 Y
Y by HWenslawski, oralayhan
Let $AB \cap CD = Q$ and $TC \cap AB = E$. Note that by the given angle condition, $\triangle QBC$ is isosceles with $QB = QC$. By angle chasing, it suffices to prove $\angle QCA = \angle CEQ$, or that $QC^2 = QA \cdot QE$.

By Menelaus and similarity of triangles $\triangle BCE \sim \triangle PCT$ and $\triangle ADB \sim \triangle PDT$, we have $$1 = \dfrac{QA}{QB} \cdot \dfrac{BC}{CP} \cdot \dfrac{PD}{DA} = \dfrac{QA}{QB} \cdot \dfrac{BE}{PT} \cdot \dfrac{PT}{AB} \implies QA \cdot BE = QB \cdot AB.$$
Using the fact that $BE = QE - QB$, this gives $QA \cdot QE = QB^2 = QC^2$, which implies the result. $\square$
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bin_sherlo
733 posts
#4
Y by
Let $TA\cap BC=G, TA\cap CF=K$, foot of the altitude from $A$ to $BC$ be $H$, the perpendicular at $C$ to $BC$ meet $BA$ at $F$, foot of the altitude from $T$ to $BC$ be $L$, $BA\cap CD=E, TC\cap AB=R$, $GD$ meet $BF, TB$ at $M,N$ respectively.
We want to show that $BC$ is the angle bisector of $\angle RCA \iff (R,A;B,F)=-1$
\[(R,A;B,F)=(TC,TA;TB,TF)=(C,G;B,TF\cap BC)=(FC,FG;FB,FT)=(K,G;A,T)=(C,G;H,L)\]\[\frac{GH}{GL}=\frac{AH}{TL}=\frac{AB}{TP}=\frac{GB}{GP}=\frac{GM}{GN}=\frac{DM}{DN}=\frac{CH}{CL}\]As desired.$\blacksquare$
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bin_sherlo
733 posts
#5
Y by
$AB\cap CD=Q$
We will use method of moving points.
Take $QBCD$ fixed. Animate $A$ over $QB$. Denote $R(XY,ZT)$ as reflection of $XY$ to $ZT$.
\[f:A\rightarrow AC\rightarrow R(AC,BC)\rightarrow R(AC,BC)\cap BQ\]\[g: A\rightarrow AD\rightarrow AD\cap BC=T\rightarrow T(QB)_{\infty}\cap BD=P\rightarrow PC\cap QB\]$f,g$ has degree $2$.
$i)A=B$
$f: B\rightarrow BC\rightarrow BC\rightarrow B$
$g: B\rightarrow BD\rightarrow B\rightarrow B\rightarrow B$
So they are same at $A=B$.

$ii)A=Q$
Let the parallel line from $C$ to $QB$ intersect $BD$ at $S$.
$f:Q\rightarrow QC\rightarrow Q'C\rightarrow QB_{\infty}$
$g: Q\rightarrow QD\rightarrow C\rightarrow C(QB)_{\infty}\cap BD=S\rightarrow SC\cap QB=QB_{\infty}$
So they are same at $A=Q$.

Let the parallel from $D$ to $BC$ intersect $QB$ at $E$ and the parallel from $C$ to $BD$ intersect $QB$ at $F$.
$iii)A=E$
$f: E\rightarrow EC\rightarrow FC\rightarrow F$
$g: E\rightarrow ED\rightarrow BC_{\infty}\rightarrow (BC)_{\infty}(QB)_{\infty}\cap BD=BD_{\infty}\rightarrow F$
Thus $f,g$ are same at $3$ points as desired.$\blacksquare$
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SleepyGirraffe
1 post
#6
Y by
Let (ABC) intersect (CPT) in point E.
Claim: T,A,E are colinear!
Proof:
We have that : angle ABC = 180- angle TPC = angle TEC (1)
Also, CAEB is cyclic so angle ABC = angle AEC (2)
From (1) and (2) we get that :
angle TEC = angle AEC wich concludes our proof
Now, using this claim the problem is basically solved with a simple angle-chasing:
I will leave it up to the reader to figure it out
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