Plan ahead for the next school year. Schedule your class today!

Contests & Programs AMC and other contests, summer programs, etc.
AMC and other contests, summer programs, etc.
3 M G
BBookmark  VNew Topic kLocked
Contests & Programs AMC and other contests, summer programs, etc.
AMC and other contests, summer programs, etc.
3 M G
BBookmark  VNew Topic kLocked
G
Topic
First Poster
Last Poster
k a July Highlights and 2025 AoPS Online Class Information
jwelsh   0
Jul 1, 2025
We are halfway through summer, so be sure to carve out some time to keep your skills sharp and explore challenging topics at AoPS Online and our AoPS Academies (including the Virtual Campus)!

[list][*]Over 60 summer classes are starting at the Virtual Campus on July 7th - check out the math and language arts options for middle through high school levels.
[*]At AoPS Online, we have accelerated sections where you can complete a course in half the time by meeting twice/week instead of once/week, starting on July 8th:
[list][*]MATHCOUNTS/AMC 8 Basics
[*]MATHCOUNTS/AMC 8 Advanced
[*]AMC Problem Series[/list]
[*]Plus, AoPS Online has a special seminar July 14 - 17 that is outside the standard fare: Paradoxes and Infinity
[*]We are expanding our in-person AoPS Academy locations - are you looking for a strong community of problem solvers, exemplary instruction, and math and language arts options? Look to see if we have a location near you and enroll in summer camps or academic year classes today! New locations include campuses in California, Georgia, New York, Illinois, and Oregon and more coming soon![/list]

MOP (Math Olympiad Summer Program) just ended and the IMO (International Mathematical Olympiad) is right around the corner! This year’s IMO will be held in Australia, July 10th - 20th. Congratulations to all the MOP students for reaching this incredible level and best of luck to all selected to represent their countries at this year’s IMO! Did you know that, in the last 10 years, 59 USA International Math Olympiad team members have medaled and have taken over 360 AoPS Online courses. Take advantage of our Worldwide Online Olympiad Training (WOOT) courses
and train with the best! Please note that early bird pricing ends August 19th!
Are you tired of the heat and thinking about Fall? You can plan your Fall schedule now with classes at either AoPS Online, AoPS Academy Virtual Campus, or one of our AoPS Academies around the US.

Our full course list for upcoming classes is below:
All classes start 7:30pm ET/4:30pm PT unless otherwise noted.

Introductory: Grades 5-10

Prealgebra 1 Self-Paced

Prealgebra 1
Wednesday, Jul 16 - Oct 29
Sunday, Aug 17 - Dec 14
Tuesday, Aug 26 - Dec 16
Friday, Sep 5 - Jan 16
Monday, Sep 8 - Jan 12
Tuesday, Sep 16 - Jan 20 (4:30 - 5:45 pm ET/1:30 - 2:45 pm PT)
Sunday, Sep 21 - Jan 25
Thursday, Sep 25 - Jan 29
Wednesday, Oct 22 - Feb 25
Tuesday, Nov 4 - Mar 10
Friday, Dec 12 - Apr 10

Prealgebra 2 Self-Paced

Prealgebra 2
Friday, Jul 25 - Nov 21
Sunday, Aug 17 - Dec 14
Tuesday, Sep 9 - Jan 13
Thursday, Sep 25 - Jan 29
Sunday, Oct 19 - Feb 22
Monday, Oct 27 - Mar 2
Wednesday, Nov 12 - Mar 18

Introduction to Algebra A Self-Paced

Introduction to Algebra A
Tuesday, Jul 15 - Oct 28
Sunday, Aug 17 - Dec 14
Wednesday, Aug 27 - Dec 17
Friday, Sep 5 - Jan 16
Thursday, Sep 11 - Jan 15
Sunday, Sep 28 - Feb 1
Monday, Oct 6 - Feb 9
Tuesday, Oct 21 - Feb 24
Sunday, Nov 9 - Mar 15
Friday, Dec 5 - Apr 3

Introduction to Counting & Probability Self-Paced

Introduction to Counting & Probability
Wednesday, Jul 2 - Sep 17
Sunday, Jul 27 - Oct 19
Monday, Aug 11 - Nov 3
Wednesday, Sep 3 - Nov 19
Sunday, Sep 21 - Dec 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Friday, Oct 3 - Jan 16
Sunday, Oct 19 - Jan 25
Tuesday, Nov 4 - Feb 10
Sunday, Dec 7 - Mar 8

Introduction to Number Theory
Tuesday, Jul 15 - Sep 30
Wednesday, Aug 13 - Oct 29
Friday, Sep 12 - Dec 12
Sunday, Oct 26 - Feb 1
Monday, Dec 1 - Mar 2

Introduction to Algebra B Self-Paced

Introduction to Algebra B
Friday, Jul 18 - Nov 14
Thursday, Aug 7 - Nov 20
Monday, Aug 18 - Dec 15
Sunday, Sep 7 - Jan 11
Thursday, Sep 11 - Jan 15
Wednesday, Sep 24 - Jan 28
Sunday, Oct 26 - Mar 1
Tuesday, Nov 4 - Mar 10
Monday, Dec 1 - Mar 30

Introduction to Geometry
Monday, Jul 14 - Jan 19
Wednesday, Aug 13 - Feb 11
Tuesday, Aug 26 - Feb 24
Sunday, Sep 7 - Mar 8
Thursday, Sep 11 - Mar 12
Wednesday, Sep 24 - Mar 25
Sunday, Oct 26 - Apr 26
Monday, Nov 3 - May 4
Friday, Dec 5 - May 29

Paradoxes and Infinity
Mon, Tue, Wed, & Thurs, Jul 14 - Jul 16 (meets every day of the week!)
Sat & Sun, Sep 13 - Sep 14 (1:00 - 4:00 PM PT/4:00 - 7:00 PM ET)

Intermediate: Grades 8-12

Intermediate Algebra
Sunday, Jul 13 - Jan 18
Thursday, Jul 24 - Jan 22
Friday, Aug 8 - Feb 20
Tuesday, Aug 26 - Feb 24
Sunday, Sep 28 - Mar 29
Wednesday, Oct 8 - Mar 8
Sunday, Nov 16 - May 17
Thursday, Dec 11 - Jun 4

Intermediate Counting & Probability
Sunday, Sep 28 - Feb 15
Tuesday, Nov 4 - Mar 24

Intermediate Number Theory
Wednesday, Sep 24 - Dec 17

Precalculus
Wednesday, Aug 6 - Jan 21
Tuesday, Sep 9 - Feb 24
Sunday, Sep 21 - Mar 8
Monday, Oct 20 - Apr 6
Sunday, Dec 14 - May 31

Advanced: Grades 9-12

Calculus
Sunday, Sep 7 - Mar 15
Wednesday, Sep 24 - Apr 1
Friday, Nov 14 - May 22

Contest Preparation: Grades 6-12

MATHCOUNTS/AMC 8 Basics
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)
Sunday, Aug 17 - Nov 9
Wednesday, Sep 3 - Nov 19
Tuesday, Sep 16 - Dec 9
Sunday, Sep 21 - Dec 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Monday, Oct 6 - Jan 12
Thursday, Oct 16 - Jan 22
Tues, Thurs & Sun, Dec 9 - Jan 18 (meets three times a week!)

MATHCOUNTS/AMC 8 Advanced
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)
Sunday, Aug 17 - Nov 9
Tuesday, Aug 26 - Nov 11
Thursday, Sep 4 - Nov 20
Friday, Sep 12 - Dec 12
Monday, Sep 15 - Dec 8
Sunday, Oct 5 - Jan 11
Tues, Thurs & Sun, Dec 2 - Jan 11 (meets three times a week!)
Mon, Wed & Fri, Dec 8 - Jan 16 (meets three times a week!)

AMC 10 Problem Series
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)
Sunday, Aug 10 - Nov 2
Thursday, Aug 14 - Oct 30
Tuesday, Aug 19 - Nov 4
Mon & Wed, Sep 15 - Oct 22 (meets twice a week!)
Mon, Wed & Fri, Oct 6 - Nov 3 (meets three times a week!)
Tue, Thurs & Sun, Oct 7 - Nov 2 (meets three times a week!)

AMC 10 Final Fives
Friday, Aug 15 - Sep 12
Sunday, Sep 7 - Sep 28
Tuesday, Sep 9 - Sep 30
Monday, Sep 22 - Oct 13
Sunday, Sep 28 - Oct 19 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Wednesday, Oct 8 - Oct 29
Thursday, Oct 9 - Oct 30

AMC 12 Problem Series
Wednesday, Aug 6 - Oct 22
Sunday, Aug 10 - Nov 2
Monday, Aug 18 - Nov 10
Mon & Wed, Sep 15 - Oct 22 (meets twice a week!)
Tues, Thurs & Sun, Oct 7 - Nov 2 (meets three times a week!)

AMC 12 Final Fives
Thursday, Sep 4 - Sep 25
Sunday, Sep 28 - Oct 19
Tuesday, Oct 7 - Oct 28

AIME Problem Series A
Thursday, Oct 23 - Jan 29

AIME Problem Series B
Tuesday, Sep 2 - Nov 18

F=ma Problem Series
Tuesday, Sep 16 - Dec 9
Friday, Oct 17 - Jan 30

WOOT Programs
Visit the pages linked for full schedule details for each of these programs!


MathWOOT Level 1
MathWOOT Level 2
ChemWOOT
CodeWOOT
PhysicsWOOT


Programming

Introduction to Programming with Python
Thursday, Aug 14 - Oct 30
Sunday, Sep 7 - Nov 23
Tuesday, Dec 2 - Mar 3

Intermediate Programming with Python
Friday, Oct 3 - Jan 16

USACO Bronze Problem Series
Wednesday, Sep 3 - Dec 3
Thursday, Oct 30 - Feb 5
Tuesday, Dec 2 - Mar 3

Physics

Introduction to Physics
Tuesday, Sep 2 - Nov 18
Sunday, Oct 5 - Jan 11
Wednesday, Dec 10 - Mar 11

Physics 1: Mechanics
Sunday, Sep 21 - Mar 22
Sunday, Oct 26 - Apr 26
0 replies
1 viewing
jwelsh
Jul 1, 2025
0 replies
AMC training
BAM10   2
N 8 minutes ago by Soupboy0
At what point do past AMC 8/10/12 tests become unrealistic in terms of difficulty to the actual competition. Is there a certain year that tests just got a lot harder? Are there some tests that are way too easy then ones after that are still realistic? etc. thx :D :surf:
2 replies
BAM10
Yesterday at 11:53 PM
Soupboy0
8 minutes ago
OpenAI won gold on 2025 IMO
centslordm   139
N 35 minutes ago by IncredibleMongoose15
it got 35/42
3 years ago it got 2; aura imo

[its solutions]
139 replies
1 viewing
centslordm
Jul 20, 2025
IncredibleMongoose15
35 minutes ago
Stressed spelled backwards
centslordm   28
N 39 minutes ago by NamelyOrange
Source: AIME 2025 #3
The 9 members of a baseball team went to an ice-cream parlor after their game. Each player had a singlescoop cone of chocolate, vanilla, or strawberry ice cream. At least one player chose each flavor, and the number of players who chose chocolate was greater than the number of players who chose vanilla, which was greater than the number of players who chose strawberry. Let $N$ be the number of different assignments of flavors to players that meet these conditions. Find the remainder when $N$ is divided by $1000.$
28 replies
centslordm
Feb 7, 2025
NamelyOrange
39 minutes ago
USAJMO #5 - points on a circle
hrithikguy   234
N 4 hours ago by TPColor
Points $A,B,C,D,E$ lie on a circle $\omega$ and point $P$ lies outside the circle. The given points are such that (i) lines $PB$ and $PD$ are tangent to $\omega$, (ii) $P, A, C$ are collinear, and (iii) $DE \parallel AC$. Prove that $BE$ bisects $AC$.
234 replies
hrithikguy
Apr 28, 2011
TPColor
4 hours ago
No more topics!
apparently circles have two intersections :'(
itised   78
N Jul 11, 2025 by eg4334
Source: 2020 USOJMO Problem 2
Let $\omega$ be the incircle of a fixed equilateral triangle $ABC$. Let $\ell$ be a variable line that is tangent to $\omega$ and meets the interior of segments $BC$ and $CA$ at points $P$ and $Q$, respectively. A point $R$ is chosen such that $PR = PA$ and $QR = QB$. Find all possible locations of the point $R$, over all choices of $\ell$.

Proposed by Titu Andreescu and Waldemar Pompe
78 replies
itised
Jun 21, 2020
eg4334
Jul 11, 2025
apparently circles have two intersections :'(
G H J
G H BBookmark kLocked kLocked NReply
Source: 2020 USOJMO Problem 2
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Tafi_ak
310 posts
#67
Y by
Let $I$ be the center of the equilateral triangle $ABC$. Let the circles with radius $AP$ and $BQ$ intersect each other at $R_1$ and $R_2$. The locus of the all points of such $R$ is circles actually, the circles with radius $IR_1$ (this is the circumcircle of $ABC$) and $IR_2$.

The point $D$ lies on $R_1R_2$, which follows by the PoP.

[asy]
 /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(32.04009188686031cm); 
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ 
pen dotstyle = black; /* point style */ 
real xmin = -13.55596900292541, xmax = 18.4841228839349, ymin = -18.814372246868626, ymax = 6.0214386240112;  /* image dimensions */
pen ffxfqq = rgb(1.,0.4980392156862745,0.); 
 /* draw figures */
draw(circle((1.3581702394607207,-4.918497679616305), 2.116571695140565), linewidth(0.8)); 
draw((xmin, 2.9836120376494693*xmin-15.63104086425135)--(xmax, 2.9836120376494693*xmax-15.63104086425135), linewidth(0.8)); /* line */
draw(circle((2.8849340119995452,-7.023517018225128), 6.526953344280609), linewidth(0.8) + red); 
draw(circle((3.6713070312575016,-4.677285011884334), 6.422444339501898), linewidth(0.8) + green); 
draw(circle((1.358407455868242,-4.918479292321957), 4.233082233658332), linewidth(0.8) + ffxfqq); 
draw((-2.6555321573673805,-3.5732482483295986)--(1.3259710422622277,-0.6854767519122333), linewidth(0.8)); 
draw((1.3259710422622277,-0.6854767519122333)--(1.3581702394607207,-4.918497679616305), linewidth(0.8)); 
draw((1.3581702394607207,-4.918497679616305)--(5.040173496202865,-7.007122820712979), linewidth(0.8)); 
draw((1.3581702394607207,-4.918497679616305)--(1.3742698380599672,-7.035008143468338), linewidth(0.8)); 
draw((-0.48283138891035104,-3.8741851090679673)--(1.3581702394607207,-4.918497679616305), linewidth(0.8)); 
draw((1.3259710422622277,-0.6854767519122333)--(-2.291633820082928,-7.062893466223699), linewidth(0.8)); 
draw((-2.291633820082928,-7.062893466223699)--(5.040173496202865,-7.007122820712979), linewidth(0.8)); 
draw((5.040173496202865,-7.007122820712979)--(1.3259710422622277,-0.6854767519122333), linewidth(0.8)); 
draw((1.3259710422622277,-0.6854767519122333)--(2.8849340119995452,-7.023517018225128), linewidth(0.8) + red); 
draw((2.8849340119995452,-7.023517018225128)--(-2.6555321573673805,-3.5732482483295986), linewidth(0.8) + red); 
draw((-2.6555321573673805,-3.5732482483295986)--(9.385575033116927,-7.608996542189704), linewidth(0.8)); 
draw(circle((1.3574812381604935,-4.918232986367356), 8.466908766081035), linewidth(0.8)); 
draw((2.8849340119995452,-7.023517018225128)--(9.385575033116927,-7.608996542189704), linewidth(0.8) + red); 
 /* dots and labels */
dot((1.3259710422622277,-0.6854767519122333),dotstyle); 
label("$A$", (1.0105816478069403,0.4602392432543942), NE * labelscalefactor); 
dot((-2.291633820082928,-7.062893466223699),dotstyle); 
label("$B$", (-3.2867087827778745,-7.8183643803722145), NE * labelscalefactor); 
dot((5.040173496202865,-7.007122820712979),dotstyle); 
label("$C$", (5.813435658460557,-7.344398524057714), NE * labelscalefactor); 
dot((1.3581702394607207,-4.918497679616305),linewidth(4.pt) + dotstyle); 
label("$I$", (1.9269156366816433,-4.721787452450811), NE * labelscalefactor); 
dot((1.3742698380599672,-7.035008143468338),linewidth(4.pt) + dotstyle); 
label("$E$", (0.7577998577725393,-8.071146170406616), NE * labelscalefactor); 
dot((3.3650214378747707,-5.591122395259655),dotstyle); 
label("$D$", (3.5699972719052493,-6.459662258937313), NE * labelscalefactor); 
dot((3.6713070312575016,-4.677285011884334),linewidth(4.pt) + dotstyle); 
label("$Q$", (4.328342642008452,-4.721787452450811), NE * labelscalefactor); 
dot((2.8849340119995452,-7.023517018225128),linewidth(4.pt) + dotstyle); 
label("$P$", (3.001238244327847,-8.229134789178115), NE * labelscalefactor); 
dot((-2.6555321573673805,-3.5732482483295986),linewidth(4.pt) + dotstyle); 
label("$R_1$", (-3.760674639092376,-3.0787058172272097), NE * labelscalefactor); 
dot((9.385575033116927,-7.608996542189704),linewidth(4.pt) + dotstyle); 
label("$R_2$", (9.76315112774807,-8.450318855458216), NE * labelscalefactor); 
dot((-0.48283138891035104,-3.8741851090679673),linewidth(4.pt) + dotstyle); 
label("$M$", (-0.8220863299424661,-3.45787850227881), NE * labelscalefactor); 
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); 
 /* end of picture */
[/asy]

Claim: $IA=IR_1$. And similarly $IB=IR_1$.

Proof. The triangles $\triangle AEP$ and $\triangle AR_1D$ are congruent. Because $EP=DP, R_1P=AP$ and there have a right angle. Therefore $IE=R_1D$ and hence done. $\square$

Since we have $IA=IB=IR_1$, we can conclude that the point $R_1$ moves on the circumcircle of $ABC$. (Basically on minor arc $AB$).

Now notice that the triangles $\triangle PR_1D$ and $\triangle PR_2D$ are congruent. Since the length $IR_1$ doesn't change therefore $DR_2$ also does not change. Since the line $R_1R_2$ are moving with respect to the fixed point $I$ with the same length, so the locus of the point of $R_2$ is a circle.
This post has been edited 1 time. Last edited by Tafi_ak, Nov 27, 2021, 7:55 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
blackbluecar
307 posts
#68
Y by
Let $E$, $F$, and $T$ be the tangency point of $AC$, $AB$, and $\ell$ with $\omega$ and let $TI$ intersect $(ABC)$ at $L$. We claim $BLET$ is cyclic. Indeed, since $BI=LI$ and $TI=EI$ it follows that $BLET$ is an isosceles trapezoid. Now, consider the inversion around $(ABC)$. $L \to L$, $B \to B$, and $E \to E^*$ where $E$ is the midpoint of $E^*B$. We claim $Q$ is the circumcenter of $LBE^*$. Indeed, $QB=QE^*$ by the midpoint property so let us show $QB=QL$. $LB=IL$ so by SAS it is sufficient to show $\angle QIL = \angle QIB$. Clearly, $\angle EIL = \angle TIB$ and $QE=QT$. Thus, by SSS, $\angle EIQ = \angle TIQ$. So, \[ \angle QIL = \angle EIL + \angle EIQ = \angle TIB + \angle TIQ = \angle QIB \]Proving the claim. Using the fact that $T$ lies on $(BLE)$ we have $T^*Q=BQ$. By symmetry, $T^*P=AP$ so $T^*=R$. Notice that a point $X$ is a valid choice of $T$ iff $X$ is on minor arc $EF$. Thus, all valid choices of $T^*$ is the arc $E^*F^*$ on $\omega^*$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
AwesomeYRY
579 posts
#69
Y by
I did not notice that there was a second intersection point until reading the first few posts on the thread lol. Anyways, being able to directly construct $R$ is pretty cool.
The two circles intersect at exactly two points. Let $PQ$ be tangent to the incircle at $T$. I claim that the two intersections are the points $R_1, R_2$ such that $RT\perp PQ$ and $RT = \frac{\sqrt3}{2} s$ where $s$ is the sidelength of $\triangle ABC$. It suffices to show that this definition of $R$ satisfies $QB=QR$ and $PA = PR$.

Let $X,Y$ be the tangency points on $CB,CA$ respectively. Then
\[PA^2 = PX^2 + AX^2 = PT^2 + (\frac{\sqrt3}{2} s)^2 = PT^2 + TR^2 = PR^2.\]A similar argument shows that $QB = QR$. Thus, we have described all points $R$ that lie on the intersection of the two circles.

Now, note that we always have $IT\perp PQ$. Since $r= \frac{s\sqrt3}{6}$, the locus of points $R$ is the points $s(\frac{\sqrt3}{2} - \frac{\sqrt3}{6})= s\frac{\sqrt{3}}{3} = 2r$ in the direction opposite $T$ from the incenter, and $s(\frac{\sqrt3}{2} + \frac{\sqrt3}{6}) = \frac{s\cdot 2\sqrt3}{3} = 2R=4r$ in the direction of $T$ from $I$.

$T$ must lie on the $120^{\circ}$ arc $XY$. Note that a homothety with factor $-2$ and the homothety with factor 4 take $T$ to $R_1$ and $R_2$. Thus, the locus of points $R$ is the arc $XY$ under homothety with factor $-2$ or $4$ centered at $I$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Math4Life2020
2967 posts
#70
Y by
How many points would missing the other case get?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Mogmog8
1080 posts
#71 • 1 Y
Y by centslordm
Let $I$ be the incenter, $D,E,F$ be the contact points, $s$ a side length, $r$ the circumradius, $T=\omega\cap\ell,$ and $R_1,R_2=\odot(P,PA)\cap\odot(Q,QB).$

Claim: $\overline{R_1T}\perp\overline{PQ}$ so $I,T,R_1,$ and $R_2$ are collinear.
Proof. Notice \begin{align*}(R_1P^2-TP^2)-(R_1Q^2-TQ^2)&=(AP^2-PD^2)-(BQ^2-QE^2)\\&=AD^2-BE^2\\&=0\end{align*}and the second part follows as $\overline{TI}\perp\overline{PQ}.$ $\blacksquare$

We also see $\triangle R_1TP\cong\triangle ADP$ so $$IR_1=IT+AD=\frac{s\sqrt{3}}{6}+\frac{s\sqrt{3}}{2}=2r.$$Also, $$IR_2=BE-IT=\frac{s\sqrt{3}}{6}-\frac{s\sqrt{3}}{2}=\frac{r}{2}.$$
Hence, all $R$ are on the circle with radius $\tfrac{1}{2}r$ or $2r.$ Notice $T$ must lie on minor arc $DE$ on $\omega,$ and so by homothety, $R$ must lie on the $120$ degree arcs bounded by $\overline{AD}$ and $\overline{BE}.$ That is, $R$ is on minor arc $\overline{AB}$ of $\odot(I,\tfrac{1}{2}r)$ or minor arc $\overline{A_1B_1}$ of $\odot(I,2r),$ where $A_1=\overline{AD}\cap\odot(I,2r)$ and $B_1=\overline{BE}\cap\odot(I,2r).$ $\square$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Pleaseletmewin
1574 posts
#72
Y by
Very fun problem :) I would appreciate it if anyone could look over my writeup and give me some feedback.
Let $I$ be the incenter of $\triangle ABC$, $\omega_1$ be the circle with center $P$ and radius $PA$, and $\omega_2$ be the circle with center $Q$ and radius $QB$. Denote $D, E, F$ as the tangency points of $\omega$ with $BC, CA$, and $AB$ respectively. Clearly, the intersections of $\omega_1$ and $\omega_2$ will be possible locations of the point $R$. The key claim is the following:
Claim: $TI$ is the radical axis of $\omega_1$ and $\omega_2$.
Proof. Since $PQ$ is the line connecting the centers of $\omega_1$ and $\omega_2$ and $TI$ is clearly perpendicular to $PQ$, it suffices to show that $T$ has equal power to $\omega_1$ and $\omega_2$. Let $PD=PT=x, QT=QE=y$, and $DB=BF=FA=EA=d$. By the Law of Cosines, we find
\begin{align*}
PA^2&=BP^2+BA^2-2(BP)(BA)\cos 60^\circ=(x+d)^2+(2d)^2-(x+d)(2d)=x^2+3d^2 \\
QB^2&=AQ^2+AB^2-2(AQ)(AB)\cos 60^\circ=(y+d)^2+(2d)^2-(y+d)(2d)=y^2+3d^2.
\end{align*}Therefore,
\begin{align*}
\text{Pow}(T,\omega_1)=TP^2-PA^2=x^2-(x^2+3d^2)=-3d^2=y^2-(y^2+3d^2)=\text{Pow}(T,\omega_2),
\end{align*}as desired.
Let $R_1$ denote the intersection of $\omega_1$ and $\omega_2$ that lies on the opposite side of $C$ with respect to $AB$. We have $T, I, R_1$ collinear. By the Pythagorean Theorem, we have
\begin{align*}
R_1T=\sqrt{PR_1^2-PT^2}=\sqrt{PA^2-PT^2}=\sqrt{(x^2+3d^2)-x^2}=\sqrt{3}d.
\end{align*}But as $TI=\tfrac{d}{\sqrt{3}}$ (inradius), we have $IR_1=\tfrac{2d}{\sqrt{3}}$. This implies that there is a negative homothety centered at $I$ with ratio $-2$ that sends $T$ to $R_1$. Letting $R_2$ denote the intersection of $\omega_1$ and $\omega_2$ that lies on the same side of $C$ with respect to $AB$, we find with similar methods that $IR_2=\tfrac{4d}{\sqrt{3}}$, so there is a positive homothety centered at $I$ with ratio $4$ that sends $T$ to $R_2$. Let $D'$ and $E'$ denote the images of $D$ and $E$, respectively, after a homothety of ratio $4$ centered at $I$. Then, as $T$ can be any point on minor arc $\widehat{DE}$, the desired locus is the union of two $120^\circ$ circular arcs $\widehat{D'E'}$ and $\widehat{AB}$ centered at $I$.
This post has been edited 1 time. Last edited by Pleaseletmewin, Mar 19, 2022, 6:12 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
OronSH
1842 posts
#73
Y by
The answer is minor arc $AB$ on the circumcircle together with its image under a homothety at the center with scale factor $-2.$

Let $X$ be the tangency point of $PQ$ to the incircle and let $DEF$ be the contact triangle, with $D$ opposite $A,$ $E$ opposite $B$ and $F$ opposite $C.$ It is clear that $X$ is on minor arc $DE$ of the incircle. Let $I$ be the incenter. We claim that $X,I,R$ are collinear.

To show this, let $AE=AF=BF=BD=c,$ $QE=QX=a$ and $PD=PX=b.$ Notice that $R$ is defined as the intersection of circles centered at $P,Q.$ Thus $R$ is on their radical axis, which is perpendicular to $PQ.$ Since $IX\perp PQ$ it suffices to show that $X$ lies on the radical axis. This is equivalent to showing that $QB^2-QX^2=PA^2-PX^2.$ By Law of Cosines we have $QB^2=a^2+3c^2$ so $QB^2-QX^2=3c^2.$ Similarly $PA^2-PX^2=3c^2$ so we finish.

Now we show that $RX=3IX.$ Letting $r$ be the inradius, we have $RX=\sqrt{QR^2-QX^2}=\sqrt{QB^2-QX^2}=c\sqrt3=r$ so we finish.

Now if $R,I$ are on the same side of $PQ$ we have that $R$ is the image of $X$ under a homothety at $I$ with scale factor $-2$ which is minor arc $AB.$ If $R,I$ are on opposite sides of $PQ$ we see that $R$ is the image of $X$ under a homothety at $I$ with scale factor $4,$ which is the aforementioned image of $AB.$ Thus we are done.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
HamstPan38825
8904 posts
#74
Y by
This is fake geometry :(

The answer is the minor arc $\widehat{AB}$ along with its image under a homothety at $O$ with ratio $-2$.

To see the first locus, let $E$ be the tangency point of $\ell$. I claim that $R = \ell \cap (ABC)$ satisfies the given conditions. Set $N = \overline{OP} \cap \overline{AR}$ and $A'$ the incircle touchpoint on $\overline{BC}$. As $OA = OR$ and $PE = PA'$, by symmetry, $N$ is the midpoint of $\overline{AR}$. In particular, we must have $PA=PR$. Similarly, $QB=QR$, hence $\widehat{AB}$ describes one set of the locus of $R$.

To see the other locus, notice that $(P)$ and $(Q)$ have radical axis $\overline{OR}$. Their other intersection $R'$ is a point with $ER' = ER$ and hence $OR'=2OR$ fixed.
This post has been edited 1 time. Last edited by HamstPan38825, Mar 10, 2024, 3:30 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
joshualiu315
2535 posts
#75
Y by
yea this is not such a beatiful problem :|


Let $T$ be the intersection of $\ell$ and $\omega$. The loci are minor arc $AB$ in $(ABC)$ and its reflection over $T$.

Let $R = (ABC) \cup \overline{IT}$ be the point on minor arc $AB$ in $(ABC)$. We claim that $R$ is a desired point $R$ and over all configurations of $\ell$, $R$ makes up the first locus.

Suppose that $PD=PT=a$, $QD=QT=b$, and the inradius of $\omega$ is $r$. Notice that

\begin{align*}
AP^2 &= AB^2+BP^2-2 (AB) (BP) \cos 60^\circ \\
&=(2r\sqrt{3})^2+(a+r\sqrt{3})^2 - (2r\sqrt{3})(a+r\sqrt{3}) \\
&=12r^2+(a^2+2ar\sqrt{3}+3r^2) - (2ar\sqrt{3}+6r^2) \\
&= a^2+9r^2 = a^2+(3r)^2 \\
&= TP^2+TR^2 = RP^2.
\end{align*}
Hence, $AP=RP$, and analogously, $BQ=RQ$.

Now that we know that $R$ is one of our solutions, we can reflect it about $T$; this preserves the length of $RP$ and $RQ$, so this produces the latter locus.

The complete solution set has been proved, so we are done.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
shendrew7
818 posts
#76
Y by
Define $X = \ell \cap \omega$, and $R_1$, $R_2$ as the images of $I$ under homotheties centered at $X$ with scale -3 and 3, respectively. We claim $R_1$, $R_2$ are the two desired points of $R$ for given $\ell$. Note we have at most two choices of $R$ for each $\ell$, so it suffices to show both satisfy the length condition. But this simply holds from
\[\triangle PMA \cong \triangle PXR_1 \cong \triangle PXR_2, \quad \triangle QNB \cong \triangle QXR_1 \cong \triangle QXR_2,\]
where $M$, $N$ are the midpoints of $BC$, $CA$. Hence our locus is the union of the images of minor arc $MN$ on $\omega$ under homotheties of scale -2 and 4. $\blacksquare$

[asy]
size(300); defaultpen(linewidth(0.7)+fontsize(11));

pair I, A, B, C, M, N, X, P, Q, R1, R2;
I = (0,0);
A = dir(210);
B = dir(330);
C = dir(90);
M = .5*B + .5*C;
N = .5*C + .5*A;
X = .5*dir(95);
P = extension(B,C,X,rotate(90,X)*I);
Q = extension(C,A,X,rotate(90,X)*I);
R1 = 4*X;
R2 = -2*X;

draw(arc(I, 2, 30, 150)^^arc(I, 1, 210, 330), green+dashed+linewidth(1));
draw(A--B--C--cycle^^circle(I, .5), gray+linewidth(0.2));
draw(P--A--M--cycle^^Q--B--N--cycle^^P--Q^^R1--R2);
draw(P--A^^P--R1^^P--R2^^Q--B^^Q--R1^^Q--R2);
draw(P--A^^P--R1^^P--R2, red);
draw(Q--B^^Q--R1^^Q--R2, blue);

dot("$A$", A, dir(225));
dot("$B$", B, dir(315));
dot("$C$", C, dir(90));
dot("$M$", M, dir(30));
dot("$N$", N, dir(150));
dot("$I$", I, dir(250));
dot("$X$", X, dir(50));
dot("$P$", P, dir(25));
dot("$Q$", Q, dir(155));
dot("$R_1$", R1, dir(90));
dot("$R_2$", R2, dir(270));
[/asy]
This post has been edited 1 time. Last edited by shendrew7, Jun 9, 2024, 4:41 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
N3bula
316 posts
#77
Y by
Let $I$ be the incentre, $R_1$ and $R_2$ be the two choices of $R$ for a specific line and $D$, $E$, $F$ the intouch points opposite $A$, $B$, $C$ respectively, reflect $A$ over $D$ to get $A'$ and reflect $B$ over $E$ to get $B'$,
we clearly get that the circle centred at $P$ through $A$ also goes through $A'$, and the circle centred at $Q$ through $B$ also passes through $B'$,
thus from the fact that its equilateral and equal lengths and radax we get that $R_1$, $I$ and $R_2$ are collinear, let $G$ be the point where $l$ meets $\omega$,
form the fact that $R_1$ and $R_2$ are perpendicular to $l$ we get that $G$ lies on $R_1R_2$. Thus because $P$ is the centre of the circle through $R_1R_2AA'$ and from
equal tangents length we get that the shape is infact an isoscelles trapezium which implies that $R_1$ lies on the circle centred at $I$ through $A$ and that $R_2$
lies on the circle centred at $I$ and through $A'$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Ihatecombin
77 posts
#78
Y by
flashsonic wrote:
I can't be the only one who misread PR=RA, QR=RB for the entirety of the test...
Bruh, I definitely didn't just misread it in the exact same way :whistling: :whistling:
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Ilikeminecraft
734 posts
#79
Y by
The answer is the minor arc $AB$ in circumcircle $(ABC),$ and its $-2\times$ homothety centered at $C.$
Let $T$ be the tangency point of $\ell$ and $\omega.$
Let $D$ be the tangency point of $BC$ and $\omega.$
Let $R’$ be the intersection of $TI$ and $(ABC)$ along minor arc $AB.$
We claim $R’ = R.$
We have that $TQ = TD, AD = AI + ID = IR’ + IT = TR’.$
Using pythagorean theorem, we get $R$s desired traits.
For the other point, do the same thing.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
BS2012
1072 posts
#80
Y by
Too easy for j2?

Define points $D,E,$ and $F$ as in the attached diagram, let $R_1$ be the intersection of $DF$ with $(ABC),$ and let $R_2$ be the image of $R_1$ under a homothety of ratio $-2$ centered at $D.$ Consider $f$ as the transformation which is the reflection over the line $PD.$ we have that $f(E)=F$ and $f(D)=D$, so $f(DE)=DF.$ Additionally, we have $f((ABC))=(ABC),$ so $f(DE\cap (ABC))=DF\cap (ABC),$ which means that $f(A)=R_1.$ This means that $PA=PR_1.$ Additionally, we have that $R_2F=R_2D-DF=2R_1D-DF=R_1F$ since the inradius of an equilateral triangle is half its circumradius, and since $PF\perp DF$ we have $PF\perp R_1R_2$ so $PA=PR_1=PR_2.$ Additionally, by a similar argument we have $QA=QR_1=QR_2.$ Since distinct circles intersect at most twice, we have that $R_1$ and $R_2$ are the only possible $R$ for a specific line. We have that $R_1$ is in arc $AB$ of $(ABC),$ and it is clear that $R_1$ can be any point on arc $AB$ that is not $A$ or $B.$ Thus, the locus is arc $AB$ (not including $A$ or $B$) of $(ABC),$ and the image of that under a homothety of ratio $2$ centered at $D.$
Attachments:
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
eg4334
739 posts
#81
Y by
Let $T$ be the tangency point between $\ell$ and $\omega$. Let $D$ be the point on $TI$ such that $ID = 2r$ where $r$ is the inradius of $\triangle ABC$ and $DIT$ are collinear in that order. Then we need to prove that $DP = PA$ (by symmetry the other side is the same). Let $AI \cap BC = X$. Then we just need $\triangle AXP = \triangle DTP$. This is true because $PT = PX$ by tangency and $DT = 3r$ and $AX = 3r$. The other side, let $E$ be the point with $TE = 3r$ where $ETI$ are collinear in that order. The other side is the same, with proving $\triangle ETP = \triangle APX$. Thus the answer is the arcs that everyone else has described.
This post has been edited 1 time. Last edited by eg4334, Jul 11, 2025, 3:51 PM
Z K Y
N Quick Reply
G
H
=
a