GEO Max(Shafiee Al Ehtesam)
K.M. Tamim
March 2025
1 USA TSTST 2011/4 Solution
Reflect H over M,N and name the intersections at w as M ′ ,N ′ . 2HM =
HM ′ , 2HN = HN ′ .Using power of point we get,2HM.HP = 2HN.HP .Hence,M N P Q(w1 )
is concyclic. Using homothety at A, we deduce that(AM N ) or (w2 ) is inter-
nally tangent to w. w∩w1 −−−→QP , w1 ∩w2 −−−→MN ,BP∩MN− → R which is
the radical center for all three circles w,w1 andw2 . Since w and w2 are inter-
nally tangent to A,the radical axis will also be tangent to A. RA is tangent to
w and w2 . Finally, we got RA ⊥OA
2 APMO 1989 P3 Solution
Using Menelaus’ theorem and the fact GA : GD = GB : GE = GC : GF = 2 : 1,
we get GP:GA = GQ:GB = GR:GC = 2:5
and we also get GN:FG = GS:GD = GM:GE = 4:7
Applying the homothety of the scale factor 52 to G and 47 , we deduce that
△P QR is homothetic to △ABC and △N SM is homothetic to the medial tri-
angle, respectively.
so, the area ratio of △P QR and △N SM is
( 27 )2 : ( 47 )2 = 25
49
3 APMO 1991 P1
Let GC, GB meet AB, AC at O, N . Using the ceva lemma on △OBC and
△N BC, we get that O, Q, M and B, P, Y are collinear.
Then, we get XY ∥ P Q ∥ BC ,QM ∥ AC, P M ∥ AB.
So, in general, it implies that △QP M is similar to △ABC.
4 APMO 1998
A is the center of the spiral similarity taking BC to EF .
So, we deduce △ABC ∼ △AEF .
1
Since M, N are the midpoints of BC and EF , respectively, we deduce
∠AM D = ∠AN D.
So, AN M D is concyclic and it implies AN ⊥ N M .
5 APMO 1999
We noticed that △QAP ∼ △QRB. This implies that ∠QAR = ∠QBR and
QABR is a cyclic quadrilateral. It generally states that
∠AQB = ∠ARB = ∠P QR. Since AB is tangent to both L1 and L2 , we
deduce that ∠P QB = ∠P BA and ∠RAB = ∠P AB = ∠BQR = ∠AQP . It
implies ∠BP R = ∠P QR = ∠BRP or BP and BR are tangents to P QR.
6 Zanimax 1
Applying Menelaus’ theorem at △M BC implies
MP BZ CA
BP × CZ × AM = 1
BZ 1 BP AM 1
Or, CZ = 2 × M P [ CA = 2 and AM = CM ]
BZ BP
Or, CZ = 2M P
We also noticed △ABP ∼ △AP M [BM ∩ AZ]
Applying the thale theorem implies that
AB BP AP
CM = AP = P M
From this we deduce that,
AB 2
( CM ) = PBP M
Or,( CM ) × 12 = PBP
AB 2 1
M × 2
AB AB
Or, CM × 2CM = PBP 1
M × 2
AB BP 1 BZ
Or, CM = P M × 2 = CZ [AB = AC]
And, it implies that △ABZ ∼ △M ZC or ∠AZB = ∠M ZC
7 Ashrafulmax 1
F is the center of (AEC).
Since E is the radical center of (AEB), (BEC), (AEC), F O2 ⊥ AE, F O1 ⊥
EC, BD ⊥ O1 O2 . Let us name the intersection points as P, Q, R.
It implies ∠O2 F O1 = 90,F P EQ is a rectangle, and O1 O2 ∥ AC So, we
deduce ∠AF D = ∠F DQ = ∠DO2 O1 = ∠F O2 O1 .
Join F, L and F L ∩ DQ = O, which is the circumcenter of (F DEQ).
Since F, O, E and F, O, L are collinear, this implies that F, O, E, R lies on
the same line as desired.
8 APMO 2025 P1
We construct △ABC as extending B1 A1 intersects (ABC) at O and ∠ACO =
90. Assume, P lies on OB1 . It generally states that AO is the diameter of
2
(ABC). Then, let’s drop an altitude from O to A1 C1 and DC1 ∩the altitude
from O= Q, and we got a rectangle C1 QOC.
It implies that OQ = CC1 . So, [△A1 OC1 ] = [△A1 CC1 ].
So, [AB1 OC1 ] = [AB1 A1 ] + [AA1 C1 ] + [A1 CC1 ].
Since P cannot lie inside △ABC and has to be on OB1 , to fulfill the condi-
tion, P has lie outside.
So, the answer is No. P can reside inside (ABC) and outside (ABC).
9 Zanimax 2
Extend CD to the circle and name the intersection point as S. Join S, T.
Since ∠EDT = 90 = ∠ECT , DECT is a cyclic quadrilateral. It implies that
∠P EC = ∠P CE = ∠ST A = ∠DT E [Since ACT S, DECT are cyclic]
∠P EC + ∠P CE = ∠ST A + ∠DT E = ∠ST E = ∠SP E
it implies that SEP T is a cyclic quadrilateral, and
∠T SP = ∠T SC = ∠T EP = ∠T LC = ∠T BC = ∠T AC = 90 − B
Let M K ∩ ET = N, EP ∩ T C = R, DC ∩ BE = Q. BE ∥ T C ∥ M K and
EP = P C = CQ, we deduce that EQRC is a rectangle, and EP = P R. It also
implies that EN = N T . Since N E 2 = N P.N K = N T 2 , ∠N T P = ∠ET P =
∠P KT = ∠M KT = ∠KT C = ∠LT C.
The whole text says △ET P = △LT C
10 Zanimax 3
∠M AD = ∠M AE = ∠M BE = ∠F BE = 60 =⇒ M ABE is cyclic.
It implies ∠EAB = ∠EM C = 90, ∠M BA = ∠M EA = ∠EBC = ∠AEB =
∠AM B = ∠BF C = 15, ∠DF E = ∠DEF = 45
If EF = x,
ED = DF = sin(45) × EF = sin(45) × x = √x2 .
q √
2
EM = (x)2 − ( x2 )2 = 3x4 = 23x [Since M the midpoint of BF ,BF = x2 ]
p
Let’s drop an altitude
√ √
from M to AE and name its feet as N .
N E = cos(15)
√
3x
= 6+3 2
8 x
2 √ √ √ √
N D = N A = N E − DE = 6+3 2
x − x2 = 6− 2
√ √ √8 8
3x 3 2− 6
M N = sin(15) × 2 = 8 x
√
3x2
[△AM E] = [△M N E]+[△M N A] = 21 ×N E ×M N + 12 ×AN ×M N = 16
2
[△EDF ] = 12 × ED × F D = x4
√ √ √
(1+ 2)x
S△AM E = M E+M2 A+AE = 6+ 4
3
x S△EDF = 2
[△EDF ]
So, the inradius of △EDF = S△EDF = 2(1+x√2)
√
[△AM E]
and, the inradius of △AM E = S△AM √ 3x√
E = 4( 3+ 6)
so, the ratio of the inradii is 1 : 2 and the ratio of the areas of the incircles
are 1 : 4
3
11 ZANIMAX 4
Since this construction is based on isogonal conjugate duos (P, Q), the angle
produced by this construction should be true for all A, B, C. Let us draw for C.
Extend CQ to the circle and name the point of intersection as R. Join R, D.
RD ∩ AB = X. Join X, Q. As the problem, we have to prove ∠P DC = ∠AXQ.
Since D, P, A are collinear, ∠P DC = ∠ADC and ∠DRC = ∠DAC = ∠BAQ =
∠XRQ. It implies that XRAQ is a cyclic quadrilateral.So,∠QRA = ∠AXQ =
∠ADC = ∠P DC.
Figure 1: After Construction
