[Signups Now!] - Inaugural Academy Math Tournament

by elements2015, May 12, 2025, 8:13 PM

Hello!

Pace Academy, from Atlanta, Georgia, is thrilled to host our Inaugural Academy Math Tournament online through Saturday, May 31.

AOPS students are welcome to participate online, as teams or as individuals (results will be reported separately for AOPS and Georgia competitors). The difficulty of the competition ranges from early AMC to mid-late AIME, and is 2 hours long with multiple sections. The format is explained in more detail below. If you just want to sign up, here's the link:

https://forms.gle/ih548axqQ9qLz3pk7

If participating as a team, each competitor must sign up individually and coordinate team names!

Detailed information below:

Divisions & Teams

Junior Varsity: Students in 10th grade or below who are enrolled in Algebra 2 or below.
Varsity: All other students.
Teams of up to four students compete together in the same division.
- (If you have two JV‑eligible and two Varsity‑eligible students, you may enter either two teams of two or one four‑student team in Varsity.)
- You may enter multiple teams from your school in either division.
- Teams need not compete at the same time. Each individual will complete the test alone, and team scores will be the sum of individual scores.

Competition Format
Both sections—Sprint and Challenge—will be administered consecutively in a single, individually completed 120-minute test. Students may allocate time between the sections however they wish to.

Sprint Section
- 25 multiple‑choice questions (five choices each)
- recommended 2 minutes per question
- 6 points per correct answer; no penalty for guessing
Challenge Section
- 18 open‑ended questions
- answers are integers between 1 and 10,000
- recommended 3 or 4 minutes per question
- 8 points each

You may use blank scratch/graph paper, rulers, compasses, protractors, and erasers. No calculators are allowed on this examination.

Awards & Scoring

There are no cash prizes.
Team Awards: Based on the sum of individual scores (four‑student teams have the advantage). Top 8 teams in each division will be recognized.
Individual Awards: Top 8 individuals in each division, determined by combined Sprint + Challenge scores, will receive recognition.

How to Sign Up
Please have EACH STUDENT INDIVIDUALLY reserve a 120-minute window for your team's online test in THIS GOOGLE FORM:
https://forms.gle/ih548axqQ9qLz3pk7
EACH STUDENT MUST REPLY INDIVIDUALLY TO THE GOOGLE FORM.
You may select any slot from now through May 31, weekdays or weekends. You will receive an email with the questions and a form for answers at the time you receive the competition. There will be a 15-minute grace period for entering answers after the competition.

math

Math Competitions

Tournament

Math Tournaments

JMO<200?

by DreamineYT, May 10, 2025, 5:37 PM

Loading poll details...

Just wanted to ask

JMO qual

poll

Jane street swag package? USA(J)MO

by arfekete, May 7, 2025, 4:34 PM

Hey! People are starting to get their swag packages from Jane Street for qualifying for USA(J)MO, and after some initial discussion on what we got, people are getting different things. Out of curiosity, I was wondering how they decide who gets what.
Please enter the following info:

- USAMO or USAJMO
- Grade
- Score
- Award/Medal/HM
- MOP (yes or no, if yes then color)
- List of items you got in your package

I will reply with my info as an example.

ranttttt

by alcumusftwgrind, Apr 30, 2025, 11:04 PM

rant

summer program

Ross Mathematics Program

PROMYS

Mathcamp

MathILy

Circle in a Parallelogram

by djmathman, Feb 9, 2022, 7:04 PM

Let $ABCD$ be a parallelogram with $\angle BAD < 90^{\circ}$ . A circle tangent to sides $\overline{DA}$ , $\overline{AB}$ , and $\overline{BC}$ intersects diagonal $\overline{AC}$ at points $P$ and $Q$ with $AP < AQ$ , as shown. Suppose that $AP = 3$ , $PQ = 9$ , and $QC = 16$ . Then the area of $ABCD$ can be expressed in the form $m\sqrt n$ , where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n$ .

$[asy] defaultpen(linewidth(0.6)+fontsize(11)); size(8cm); pair A,B,C,D,P,Q; A=(0,0); label("$A$", A, SW); B=(6,15); label("$B$", B, NW); C=(30,15); label("$C$", C, NE); D=(24,0); label("$D$", D, SE); P=(5.2,2.6); label("$P$", (5.8,2.6), N); Q=(18.3,9.1); label("$Q$", (18.1,9.7), W); draw(A--B--C--D--cycle); draw(C--A); draw(Circle((10.95,7.45), 7.45)); dot(A^^B^^C^^D^^P^^Q); [/asy]$

Evan's mean blackboard game

by hwl0304, Apr 18, 2019, 10:58 PM

Two rational numbers $\tfrac{m}{n}$ and $\tfrac{n}{m}$ are written on a blackboard, where $m$ and $n$ are relatively prime positive integers. At any point, Evan may pick two of the numbers $x$ and $y$ written on the board and write either their arithmetic mean $\tfrac{x+y}{2}$ or their harmonic mean $\tfrac{2xy}{x+y}$ on the board as well. Find all pairs $(m,n)$ such that Evan can write 1 on the board in finitely many steps.

Proposed by Yannick Yao

This post has been edited 1 time. Last edited by djmathman, Apr 19, 2019, 2:31 PM

Lots of Cyclic Quads

by Vfire, Apr 19, 2018, 11:00 PM

In convex cyclic quadrilateral $ABCD$ , we know that lines $AC$ and $BD$ intersect at $E$ , lines $AB$ and $CD$ intersect at $F$ , and lines $BC$ and $DA$ intersect at $G$ . Suppose that the circumcircle of $\triangle ABE$ intersects line $CB$ at $B$ and $P$ , and the circumcircle of $\triangle ADE$ intersects line $CD$ at $D$ and $Q$ , where $C,B,P,G$ and $C,Q,D,F$ are collinear in that order. Prove that if lines $FP$ and $GQ$ intersect at $M$ , then $\angle MAC = 90^\circ$ .

Proposed by Kada Williams

This post has been edited 2 times. Last edited by djmathman, Jun 22, 2020, 5:49 AM

Ptolemy triangle and QA-Orthopole-circle

by TelvCohl, Feb 28, 2018, 11:08 AM

Notation

For two lines $\tau, \eta$ and a points $P,$ let $\mathbb{T}_{\tau,\eta}(P)$ be the midpoint between the reflection of $P$ in $\tau$ and the reflection of $P$ in $\eta.$

Preliminaries

Lemma 1 : Given two lines $\ell_1, \ell_2$ and two points $P,Q.$ Let $X$ be the intersection of $\ell_1, \ell_2.$ Then $\triangle XPQ \stackrel{-}{\sim} \triangle X\mathbb{T}_{\ell_1,\ell_2} (P)\mathbb{T}_{\ell_1,\ell_2} (Q).$

Proof : Let $P_i, Q_i$ be the projection of $P,Q$ on $\ell_i,$ respectively where $i \in \{1,2 \}$ and let $Y \equiv \mathbb{T}_{\ell_1,\ell_2} (P), Z \equiv \mathbb{T}_{\ell_1,\ell_2} (Q),$ then $Y, Z$ is the orthocenter of $\triangle XP_1P_2, \triangle XQ_1Q_2,$ respectively, so $\frac{XP}{XQ} = \frac{XY}{XZ}.$ Since $XY, XZ$ is the isogonal conjugate of $XP, XQ$ WRT $\angle (\ell_1, \ell_2),$ respectively, so we conclude that $\triangle XPQ \stackrel{-}{\sim} \triangle XYZ.$ $\qquad \blacksquare$

Corollary 1.1 : Given two lines $\ell_1, \ell_2$ and two points $P,Q.$ Then $PQ$ and $\mathbb{T}_{\ell_1,\ell_2} (P)\mathbb{T}_{\ell_1,\ell_2} (Q)$ are antiparallel WRT $\angle (\ell_1,\ell_2).$

Corollary 1.2 : Given two lines $\ell_1, \ell_2$ and three points $P,Q, R.$ Then $\triangle PQR \stackrel{-}{\sim} \triangle \mathbb{T}_{\ell_1,\ell_2} (P)\mathbb{T}_{\ell_1,\ell_2} (Q)\mathbb{T}_{\ell_1,\ell_2} (R).$

Corollary 1.3 : Given a $\triangle ABC$ and a point $P \in BC.$ Let $E, F$ be the projection of $B, C$ on $CA, AB,$ respectively. Then $\mathbb{T}_{AB,AC} (P)$ lies on $EF.$

Lemma 2 : Given a quadrangle $P_1P_2P_3P_4$ with Euler-Poncelet point $E$ and Isogonal center $I.$ Then $\mathbb{T}_{P_1P_2,P_3P_4} (I) = \mathbb{T}_{P_1P_3,P_2P_4} (I) = \mathbb{T}_{P_1P_4,P_2P_3} (I) = E.$

Proof : It suffices to prove $\mathbb{T}_{P_1P_4, P_2P_3} (I) = E.$ Let $O_4$ be the circumcenter of $\triangle P_1P_2P_3$ and let the perpendicular from $E$ to $P_2P_3$ cuts $P_1P_4, IO_4$ at $U, V,$ respectively. By Miquel triangle and Isogonal center (Property 11), $I$ is the image of the isogonal conjugate $Q_4$ of $P_4$ WRT $\triangle P_1P_2P_3$ under the inversion WRT $\odot (O_4),$ so $\measuredangle VIP_1 = \measuredangle Q_4P_1O_4 = \measuredangle (\perp P_2P_3, P_1P_4) = \measuredangle VUP_1$ $\Longrightarrow$ $P_1, I, U, V$ are concyclic. Notice that $E$ is the orthopole of $O_4Q_4$ WRT $\triangle P_1P_2P_3$ we get $P_1V \perp O_4Q_4,$ so $U$ is the projection of $I$ on $P_1P_4.$ Similarly, the intersection of the perpendicular from $I,E$ to $P_2P_3, P_1P_4,$ respectively lies on $P_2P_3,$ so $\mathbb{T}_{P_1P_4, P_2P_3} (I) = E.$ $\qquad \blacksquare$

Lemma 3 : Given a quadrilateral $P_1P_2P_3P_4$ and a point $T.$ Then the pedal circle of $T$ WRT $\triangle P_2P_3P_4, \triangle P_1P_3P_4, \triangle P_1P_2P_4, \triangle P_1P_2P_3$ are concurrent.

Proof : Let $H_{ij}$ be the projection of $T$ on $P_iP_j$ where $i<j$ and $i, j \in \{ 1,2,3,4 \}.$ Consider an inversion with center $T$ and denote the image of $\Box$ as ${\Box}^*,$ then $H^*_{ij}, H^*_{ik}, H^*_{il}$ are collinear where $\{ i, j, k, l \} = \{ 1,2,3,4 \},$ so $\odot (H^*_{23}H^*_{24}H^*_{34}), \odot (H^*_{13}H^*_{14}H^*_{34}), \odot (H^*_{12}H^*_{14}H^*_{24}), \odot (H^*_{12}H^*_{13}H^*_{23})$ are concurrent $\Longrightarrow$ $\odot (H_{23}H_{24}H_{34}), \odot (H_{13}H_{14}H_{34}), \odot (H_{12}H_{14}H_{24}), \odot (H_{12}H_{13}H_{23})$ are concurrent. $\qquad \blacksquare$

Lemma 4 : Given a $\triangle ABC$ with orthocenter $H$ and a point $P.$ Let $Y, Z$ be the reflection of $P$ in $CA, AB,$ respectively. Then the anti-Steiner point $T$ of $HP$ WRT $\triangle ABC$ lies on $\odot (AYZ).$

Proof : Let $E, F$ be the reflection of $H$ in $CA, AB,$ respectively, then $T$ lies on $EY, FZ$ and $E, F$ lie on $\odot (ABC),$ so $\measuredangle YAZ = \measuredangle EAF = \measuredangle ETF = \measuredangle YTZ$ $\Longrightarrow$ $T \in \odot (AYZ).$ $\qquad \blacksquare$

Main result

In this section, I'll give a brief introduction to Ptolemy triangle, QA-Orthopole-circle and related transformation WRT a quadrangle. All symbols in this section bear the same meaning.

Notation

Given a quadrangle $P_1P_2P_3P_4$ with Euler-Poncelet point $E,$ Isogonal center $I$ and a point $T.$ Let $T_X \equiv \mathbb{T}_{P_1P_4, P_2P_3} (T), T_Y \equiv \mathbb{T}_{P_1P_3, P_2P_4} (T), T_Z \equiv \mathbb{T}_{P_1P_2, P_3P_4} (T).$

Property 1 : $\triangle T_XT_YT_Z$ is the image of the pedal triangle of $P_i$ WRT $\triangle P_jP_kP_l$ under the spiral similarity with center $E,$ ratio $\frac{IT}{IP_i}$ and angle $\measuredangle TIP_i$ where $\{ i,j,k,l \} = \{ 1,2,3,4 \}.$

Proof : It suffices to prove the case when $i=4.$ Let $\triangle A_4B_4C_4$ be the pedal triangle of $P_4$ WRT $\triangle P_1P_2P_3.$ From Lemma 2 $\Longrightarrow$ $\mathbb{T}_{P_1P_4, P_2P_3} (I) = E,$ so combining Corollary 1.2 for lines $P_1P_4, P_2P_3$ and points $P_4, I, T$ we get $\triangle ITP_4 \stackrel{-}{\sim} \triangle ET_XA_4.$ Similarly, we can prove $\triangle ITP_4 \stackrel{-}{\sim} \triangle ET_YB_4,$ $\triangle ITP_4 \stackrel{-}{\sim} \triangle ET_ZC_4,$ so $\triangle T_XT_YT_Z$ is the image of $\triangle A_4B_4C_4$ under the spiral similarity with center $E,$ ratio $\frac{IT}{IP_4}$ and angle $\measuredangle TIP_4.$ $\qquad \blacksquare$

Corollary 2 : $E$ lies on $\odot (T_XT_YT_Z).$

Corollary 3 : Let $O_T$ be the circumcenter of $\triangle T_XT_YT_Z.$ Then the mapping $T \mapsto O_T$ is linear.

Proof : Let $M_4$ be the circumcenter $\triangle A_4B_4C_4,$ then from Property 1 we get $\triangle ITP_4 \stackrel{-}{\sim} \triangle EO_TM_4,$ so the mapping $T \mapsto O_T$ is linear. $\qquad \blacksquare$

Remark : $\triangle T_XT_YT_Z$ is called the Ptolemy triangle of $T$ and the mapping $T \mapsto O_T$ is called the QA-Orthopole transformation WRT the quadrangle $P_1P_2P_3P_4.$

Property 4 : Let $\varrho$ be a line passing through $T,$ $R_{ij}$ be the intersection of $\varrho$ with $P_iP_j$ and $H_{kl}$ be the projection of $R_{ij}$ on $P_kP_l$ where $\{ i,j,k,l \} = \{ 1,2,3,4 \}.$ Then $H_{12}H_{34}, H_{13}H_{24}, H_{14}H_{23}$ are concurrent on $\odot (O_T).$

Proof : From Corollary 1.3 $\Longrightarrow$ $T_Y \in H_{13}H_{24},$ $T_Z \in H_{12}H_{34},$ so note that $\{ \varrho, H_{13}H_{24} \}, \{ \varrho, H_{12}H_{34}\}$ are antiparallel WRT $\angle (P_1P_3, P_2P_4), \angle (P_1P_2, P_3P_4),$ respectively we get $\measuredangle (H_{13}H_{24}, H_{14}H_{23}) = \measuredangle (H_{13}H_{24}, P_2P_4) + \measuredangle (P_2P_4, P_3P_4) + \measuredangle (P_3P_4, H_{12}H_{34}) = \measuredangle (P_1P_3, \varrho) + \measuredangle (P_2P_4, P_3P_4) + \measuredangle (\varrho, P_1P_2) = \measuredangle (P_1P_3, P_1P_2) + \measuredangle (P_2P_4, P_3P_4) = \measuredangle T_YT_XT_Z,$ hence the intersection of $H_{13}H_{24}, H_{12}H_{34}$ lies on $\odot (O_T).$ Similarly, the intersection of $\{ H_{12}H_{34}, H_{14}H_{23} \}, \{ H_{14}H_{23}, H_{13}H_{24}\}$ lies on $\odot (O_T),$ so $H_{12}H_{34}, H_{13}H_{24}, H_{14}H_{23}, \odot (O_T)$ are concurrent at $T_{\varrho}.$ $\qquad \blacksquare$

Remark : The mapping $\varrho \mapsto T_{\varrho}$ is called the Quang Duong’s transformation WRT the quadrangle $P_1P_2P_3P_4.$

Property 5 : Let $O_i$ be the circumcenter of $\triangle P_jP_kP_l$ and $V_i$ be the orthopole of $O_iT$ WRT $\triangle P_jP_kP_l$ where $\{ i,j,k,l \} = \{ 1,2,3,4 \}.$ Then $V_i$ lies on $\odot (O_T)$ for any $i \in \{ 1,2,3,4 \}.$

Proof : Let $U$ be the midpoint of $P_2P_3,$ $Q$ be the projection of $T$ on $P_2P_3$ and $E_1, E_4$ be the Poncelet point of $P_2P_3P_4T, P_1P_2P_3T,$ respectively, then $E, U, E_1, V_1$ lie on the 9-point circle of $\triangle P_2P_3P_4$ and $E, U, E_4, V_4$ lie on the 9-point circle of $\triangle P_1P_2P_3.$ From Lemma 3 $\Longrightarrow$ the pedal circle of $T$ WRT $\triangle P_2P_3P_4, \triangle P_1P_3P_4, \triangle P_1P_2P_4, \triangle P_1P_2P_3$ are concurrent at $R,$ so note that $Q, U, E_1, E_4$ lie on the 9-point circle of $\triangle TP_2P_3$ we get $\measuredangle EV_1R + \measuredangle RV_4E = \measuredangle EV_1E_1 + \measuredangle E_1V_1R + \measuredangle RV_4E_4 + \measuredangle E_4V_4E = \measuredangle EUE_1 + \measuredangle E_1QR + \measuredangle RQE_4 + \measuredangle E_4UE = \measuredangle E_1QE_4 + \measuredangle E_4UE_1 = 0,$ hence $E, U, V_1, V_4$ are concyclic. Similarly, we can prove $V_2, V_3$ lie on $\odot (EUV_4),$ so $E, U, V_1, V_2, V_3, V_4$ lie on a circle $\odot (O_T^*).$

Let $N_i$ be the 9-point center of $\triangle P_jP_kP_l$ where $\{ i,j,k,l \} = \{ 1,2,3,4 \}.$ Since $O_1O_4, O_1T$ is the Steiner line of $U, V_1$ WRT the medial triangle of $\triangle P_2P_3P_4,$ respectively, so $\measuredangle O_T^*N_1N_4 = \measuredangle V_1EU = \measuredangle O_4O_1T.$ Similarly, we can prove $\measuredangle O_T^*N_4N_1 = \measuredangle O_1O_4T,$ so $\triangle O_T^*N_1N_4 \stackrel{-}{\sim} \triangle TO_1O_4.$ Analogously, we can prove $\triangle O_T^*N_2N_4 \stackrel{-}{\sim} \triangle TO_2O_4,$ $\triangle O_T^*N_3N_4 \stackrel{-}{\sim} \triangle TO_3O_4,$ so $\triangle N_1N_2N_3 \cup O_T^* \stackrel{-}{\sim} \triangle O_1O_2O_3 \cup T.$

On the other hand, from Interesting Properties related to Four 9-point Centers (Property 1 and Corollary 11) we get $\triangle N_1N_2N_3 \cup M_4 \cup E \stackrel{-}{\sim} \triangle O_1O_2O_3 \cup P_4 \cup I$ and $E, I$ is the image of $M_4, P_4$ under the inversion WRT $\odot (N_1N_2N_3), \odot (O_1O_2O_3),$ so combining $\triangle EO_TM_4 \stackrel{-}{\sim} \triangle ITP_4,$ by Property 1, we conclude that $\triangle N_1N_2N_3 \cup O_T \stackrel{-}{\sim} \triangle O_1O_2O_3 \cup T$ $\Longrightarrow$ $O_T \equiv O_T^*$ and $\odot (O_T) \equiv \odot (O_T^*).$ $\qquad \blacksquare$

Remark : $\odot (O_T)$ is called the QA-Orthopole-circle of $T$ WRT the quadrangle $P_1P_2P_3P_4.$

Property 6 : The image of $\varrho$ under the QA-Orthopole transformation WRT $P_1P_2P_3P_4$ is the perpendicular bisector of $ET_{\varrho}.$

Proof : Let $\gamma$ be the perpendicular bisector of $ET_{\varrho},$ then note that $ET_X$ and $IT$ are antiparallel WRT $\angle (P_1P_4, P_2P_3),$ by Corollary 1.1 and Lemma 2, we get $\measuredangle (EO_T, \gamma) = \measuredangle (ET_X, T_{\varrho}T_X) = \measuredangle (\varrho, IT),$ so $\gamma$ is the image of $\varrho$ under the QA-Orthopole transformation WRT $P_1P_2P_3P_4.$ $\qquad \blacksquare$

Let $\triangle XYZ$ be the diagonal triangle of the quadrangle $P_1P_2P_3P_4$ where $X, Y, Z$ is the intersection of $\{ P_1P_4, P_2P_3 \}, \{ P_1P_3, P_2P_4 \}, \{ P_1P_2, P_3P_4 \},$ respectively, then from New proof of the property of Poncelet point and Corollary 2 we know $E$ is one of the intersection of $\odot (O_T)$ and $\odot (XYZ).$ The following property gives the character of another intersection of these two circles.

Property 7 : Let $H^{\mathbf{D}}$ be the orthocenter of $\triangle XYZ.$ Then the anti-Steiner point $S$ of $H^{\mathbf{D}}T$ WRT $\triangle XYZ$ lies on $\odot (O_T).$

Proof : Let $X^T, Y^T. Z^T$ be the reflection of $T$ in $YZ, ZX, XY,$ respectively, then $T_X, T_Y, T_Z$ lie on $\odot (XY^TZ^T), \odot (YZ^TX^T), \odot (ZX^TY^T),$ respectively (well-known). On the other hand, from Lemma 4 we know $S$ lies on $\odot (XY^TZ^T), \odot (YZ^TX^T), \odot (ZX^TY^T),$ so $\measuredangle T_YST_Z = \measuredangle T_YSX^T + \measuredangle X^TST_Z = \measuredangle T_YYX^T + \measuredangle X^TZT_Z.$ Note that $\{ YT, YT_Y \}, \{ ZT, ZT_Z \}$ are isogonal conjugate WRT $\angle (T_1T_3, T_2T_4), \angle (T_1T_2, T_3T_4),$ respectively, so we get $\measuredangle T_YST_Z = \measuredangle T_YYP_1 + \measuredangle P_1YX^T + \measuredangle X^TZP_1 + \measuredangle P_1ZT_Z = \measuredangle P_4YT + \measuredangle P_1YX^T + \measuredangle X^TZP_1 + \measuredangle TZP_4 = \measuredangle P_4YZ + \measuredangle P_1YZ + \measuredangle YZP_1 + \measuredangle YZP_4 = \measuredangle P_2P_4P_3 + \measuredangle P_3P_1P_2,$ hence by Property 1 we conclude that $\measuredangle T_YST_Z = \measuredangle T_YT_XT_Z$ $\Longrightarrow$ $S \in \odot (O_T).$ $\qquad \blacksquare$

isogonal center

Ptolemy triangle

QA-Orthopole-circle

Circle Incident

by MSTang, Mar 4, 2016, 3:27 PM

Circles $\omega_1$ and $\omega_2$ intersect at points $X$ and $Y$ . Line $\ell$ is tangent to $\omega_1$ and $\omega_2$ at $A$ and $B$ , respectively, with line $AB$ closer to point $X$ than to $Y$ . Circle $\omega$ passes through $A$ and $B$ intersecting $\omega_1$ again at $D \neq A$ and intersecting $\omega_2$ again at $C \neq B$ . The three points $C$ , $Y$ , $D$ are collinear, $XC = 67$ , $XY = 47$ , and $XD = 37$ . Find $AB^2$ .

2016 AIME I

circles

AIME

Points Collinear iff Sum is Constant

by djmathman, Apr 29, 2014, 10:41 PM

Prove that there exists an infinite set of points $\dots, \; P_{-3}, \; P_{-2},\; P_{-1},\; P_0,\; P_1,\; P_2,\; P_3,\; \dots$ in the plane with the following property: For any three distinct integers $a,b,$ and $c$ , points $P_a$ , $P_b$ , and $P_c$ are collinear if and only if $a+b+c=2014$ .

Cyclic Quad

by worthawholebean, May 1, 2008, 5:01 PM

Let $ABC$ be an acute, scalene triangle, and let $M$ , $N$ , and $P$ be the midpoints of $\overline{BC}$ , $\overline{CA}$ , and $\overline{AB}$ , respectively. Let the perpendicular bisectors of $\overline{AB}$ and $\overline{AC}$ intersect ray $AM$ in points $D$ and $E$ respectively, and let lines $BD$ and $CE$ intersect in point $F$ , inside of triangle $ABC$ . Prove that points $A$ , $N$ , $F$ , and $P$ all lie on one circle.

geometry

circumcircle

geometric transformation