V olum e 9, N um ber 2 M ay 2004 – July 2004
Olympiad Corner Inversion
The XVI Asian Pacific Mathematical
Olympiad took place on March 2004. Kin Y. Li
Here are the problems. Time allowed:
In algebra, the method of logarithm Now for the method of inversion, let O
transforms tough problems involving be a point on the plane and r be a positive
Problem 1. Determine all finite nonempty multiplications and divisions into number. The inversion with center O and
sets S of positive integers satisfying simpler problems involving additions radius r is the function on the extended
i + j is an element of S for all i, j in S, and subtractions. For every positive plane that sends a point X ≠ O to the
(i, j ) image point X′ on the ray OX such that
number x, there is a unique real number
where (i, j) is the greatest common divisor log x in base 10. This is a one-to-one OX·OX′ = r2.
of i and j.
correspondence between the positive
When X = O, X′ is taken to be the point at
Problem 2. Let O be the circumcenter numbers and the real numbers.
infinity. When X is infinity, X′ is taken to
and H the orthocenter of an acute triangle In geometry, there are also be O. The circle with center O and
ABC. Prove that the area of one of the radius r is called the circle of inversion.
transformation methods for solving
triangles AOH, BOH, COH is equal to the
problems. In this article, we will discuss
sum of the areas of the other two. The method of inversion is based on
one such method called inversion. To the following facts.
Problem 3. Let a set S of 2004 points in present this, we will introduce the
extended plane, which is the plane (1) The function sending X to X′
the plane be given, no three of which are
together with a point that we would like described above is a one-to-one
collinear. Let ℒ denote the set of all lines
correspondence between the extended
(extended indefinitely in both directions) to think of as infinity. Also, we would
plane with itself. (This follows from
determined by pairs of points from the like to think of all lines on the plane will
checking (X′ )′ = X. )
set. Show that it is possible to color the go through this point at infinity! To
points of S with at most two colors, such understand this, we will introduce the (2) If X is on the circle of inversion, then
that for any points p, q of S, the number stereographic projection, which can be X′ = X. If X is outside the circle of
described as follow. inversion, then X′ is the midpoint of the
(continued on page 4)
chord formed by the tangent points T1, T2
Consider a sphere sitting on a point O of the tangent lines from X to the circle
of a plane. If we remove the north pole N of inversion. (This follows from
Editors: 張 百 康 (CHEUNG Pak-Hong), Munsang College, HK
of the sphere, we get a punctured sphere. OX·OX′ = (r sec ∠T1OX )(r cos ∠T1OX)
高 子 眉 (KO Tsz-Mei)
梁 達 榮 (LEUNG Tat-Wing) For every point P on the plane, the line
李 健 賢 (LI Kin-Yin), Dept. of Math., HKUST NP will intersect the punctured sphere at = r2. )
吳 鏡 波 (NG Keng-Po Roger), ITC, HKPU
a unique point SP. So this gives a (3) A circle not passing through O is sent
Artist: 楊 秀 英 (YEUNG Sau-Ying Camille), MFA, CU
one-to-one correspondence between the to a circle not passing through O. In this
Acknowledgment: Thanks to Elina Chiu, Math. Dept., HKUST
for general assistance. plane and the punctured sphere. If we case, the images of concyclic points are
On-line: http://www.math.ust.hk/mathematical_excalibur/ consider the points P on a circle in the concyclic. The point O, the centers of
The editors welcome contributions from all teachers and plane, then the SP points will form a the circle and the image circle are
students. With your submission, please include your name,
circle on the punctured sphere. collinear. However, the center of the
address, school, email, telephone and fax numbers (if available).
Electronic submissions, especially in MS Word, are encouraged. However, if we consider the points P on circle is not sent to the center of the
The deadline for receiving material for the next issue is August image circle!
9, 2004. any line in the plane, then the SP points
For individual subscription for the next five issues for the 03-04 will form a punctured circle on the (4) A circle passing through O is sent to
academic year, send us five stamped self-addressed envelopes.
Send all correspondence to: sphere with N as the point removed from a line which is not passing through O and
Dr. Kin-Yin LI the circle. If we move a point P on any is parallel to the tangent line to the circle
Department of Mathematics
The Hong Kong University of Science and Technology
line on the plane toward infinity, then SP at O. Conversely, a line not passing
Clear Water Bay, Kowloon, Hong Kong will go toward the same point N! Thus, through O is sent to a circle passing
Fax: (852) 2358 1643 in this model, all lines can be thought of through O with the tangent line at O
as going to the same infinity. parallel to the line.
M athematical Excalibur Vol. 9, N o. 2, M ay 04- July 04 Page 2
(5) A line passing through O is sent to Example 2. (1993 USAMO) Let ABCD similar and ∆APC, ∆AC′P′ are similar.
itself. be a convex quadrilateral such that Now
(6) If two curves intersect at a certain diagonals AC and BD intersect at right ∠B′C′P′ =∠AC′P′ – ∠AC′B′
angle at a point P ≠ O, then the image angles, and let O be their intersection =∠APC – ∠ABC
curves will also intersect at the same point. Prove that the reflections of O = ∠APB – ∠ACB
angle at P′. If the angle is a right angle, across AB, BC, CD, DA are concyclic. = ∠AB′P – ∠AB′C′
the curves are said to be orthogonal. So =∠C′B′P′.
Solution. Let P, Q, R, S be the feet of
in particular, orthogonal curves at P are
perpendiculars from O to AB, BC, CD, So ∆B′C′P′ is isosceles and P′B′ = P′C′.
sent to orthogonal curves at P’. A circle
DA, respectively. The problem is From ∆APB, ∆AB′P′ similar and ∆APC,
orthogonal to the circle of inversion is
sent to itself. Tangent curves at P are sent equivalent to showing P, Q, R, S are ∆AC′P′ similar, we get
to tangent curves at P’. concyclic (since they are the midpoints of BA P ′A P ′A CA
= = = .
O to its reflections). Note OSAP, OPBQ, BP P ′B ′ P ′C ′ CP
(7) If points A, B are different from O and
OQCR, ORDS are cyclic quadrilaterals.
points O, A, B are not collinear, then the Therefore, X = Y.
Let their circumcircles be called CA, CB,
equation OA·OA′ = r = OB·OB′ implies
CC, CD, respectively.
OA/OB=OB′/OA′. Along with ∠AOB = Example 4. (1995 Israeli Math
∠B′OA′, they imply ∆OAB, ∆OB′A′ are Olympiad) Let PQ be the diameter of
Consider the inversion with center O semicircle H. Circle O is internally
and any radius r. By fact (5), lines AC and tangent to H and tangent to PQ at C. Let
A ′B ′ OA′ r2 BD are sent to themselves. By fact (4), A be a point on H and B a point on PQ
AB OB OA ⋅ OB circle CA is sent to a line LA parallel to BD, such that AB ⊥ PQ and is tangent to O.
so that circle CB is sent to a line LB parallel to AC, Prove that AC bisects ∠PAB.
circle CC is sent to a line LC parallel to BD,
A ′B ′ = AB . circle CD is sent to a line LD parallel to AC. Solution. Consider the inversion with
OA ⋅ OB
center C and any radius r. By fact (7),
The following are some examples that Next CA intersects CB at O and P. This ∆CAP, ∆CP′A′ similar and ∆CAB, ∆CB′A′
similar. So AC bisects PAB if and only if
illustrate the powerful method of implies LA intersects LB at P′. Similarly, LB
inversion. In each example, when we do ∠CAP =∠CAB if and only if ∠CP′A′ =
intersects LC at Q′, LC intersects LD at R′
inversion, it is often that we take the point ∠CB′A′.
and LD intersects LA at S′.
that plays the most significant role and By fact (5), line PQ is sent to itself.
where many circles and lines intersect. Since circle O passes through C, circle O
Since AC ⊥ BD, P′Q′R′S′ is a rectangle, is sent to a line O′ parallel to PQ. By fact
hence cyclic. Therefore, by fact (3), P, Q, (6), since H is tangent to circle O and is
Example 1. (Ptolemy’s Theorem) For
R, S are concyclic. orthogonal to line PQ, H is sent to the
coplanar points A, B, C, D, if they are
semicircle H′ tangent to line O′ and has
Example 3. (1996 IMO) Let P be a point diameter P′Q′. Since segment AB is
AB·CD + AD·BC = AC·BD. inside triangle ABC such that tangent to circle O and is orthogonal to
PQ, segment AB is sent to arc A′B′ on the
Solution. Consider the inversion with ∠APB – ∠ACB =∠APC – ∠ABC.
semicircle tangent to line O′ and has
center D and any radius r. By fact (4), the
Let D, E be the incenters of triangles APB, diameter CB’. Now observe that arc A′Q′
circumcircle of ∆ABC is sent to the line
APC, respectively. Show that AP, BD, CE and arc A′C are symmetrical with respect
through A′, B′, C′. Since A′B′ + B′C′ =
meet at a point. to the perpendicular bisector of CQ′ so we
A′C′, we have by fact (7) that
get ∠CP′A′ = ∠CB′A′.
Solution. Let lines AP, BD intersect at X,
r2 r2 r2
AB + BC = AC . lines AP, CE intersect at Y. We have to
AD ⋅ BD BD ⋅ CD AD ⋅ CD In the solutions of the next two
show X = Y. By the angle bisector
examples, we will consider the
Multiplying by (AD·BD·CD)/r2, we get theorem, BA/BP = XA/XP. Similarly,
nine-point circle and the Euler line of a
the desired equation. CA/CP = YA/YP. As X, Y are on AP, we
triangle. Please consult Vol. 3, No. 1 of
Remarks. The steps can be reversed to get X = Y if and only if BA/BP = CA/CP.
Mathematical Excalibur for discussion if
get the converse statement that if necessary.
Consider the inversion with center A
AB·CD + AD·BC = AC·BD, (continued on page 4)
and any radius r. By fact (7), ∆ABC,
then A,B,C,D are concyclic. ∆AC′B′ are similar, ∆APB, ∆AB′P′ are
M athematical Excalibur Vol. 9, N o. 2, M ay 04- July 04 Page 3
Problem Corner Athens, Greece) Let x1 , x2 ,..., xn be 2n─3−1. The result follows.
positive real numbers with sum equal to 1. Other commended solvers: NGOO Hung
We welcome readers to submit their Prove that for every positive integer m, Wing (Valtorta College).
solutions to the problems posed below
Problem 198. In a triangle ABC, AC =
for publication consideration. The m m m
n ≤ n m ( x1 + x 2 + ... + x n ). BC. Given is a point P on side AB such
solutions should be preceded by the
solver’s name, home (or email) address Solution. CHENG Tsz Chung (La Salle that ∠ACP = 30○. In addition, point Q
and school affiliation. Please send College, Form 5), Johann Peter Gustav outside the triangle satisfies ∠CPQ =
submissions to Dr. Kin Y. Li, Lejeune DIRICHLET (Universidade de ∠CPA + ∠APQ = 78○. Given that all
Sao Paulo – Campus Sao Carlos), KWOK angles of triangles ABC and QPB,
Department of Mathematics, The Hong Tik Chun (STFA Leung Kau Kui College,
Kong University of Science & Form 6), POON Ming Fung (STFA Leung measured in degrees, are integers,
Technology, Clear Water Bay, Kowloon, Kau Kui College, Form 6), Achilleas P. determine the angles of these two
Hong Kong. The deadline for PORFYRIADIS (American College of triangles. (Source: KöMaL C. 524)
Thessaloniki “Anatolia”, Thessaloniki, Greece),
submitting solutions is August 9, 2004. SIU Ho Chung (Queen’s College, Form 5) Solution. CHAN On Ting Ellen (True
and YU Hok Kan (STFA Leung Kau Kui Light Girls’ College, Form 4), CHENG
Problem 201. (Due to Abderrahim College, Form 6). Tsz Chung (La Salle College, Form 5),
OUARDINI, Talence, France) Find POON Ming Fung (STFA Leung Kau
which nonright triangles ABC satisfy Applying Jensen’s inequality to f (x) = Kui College, Form 6), TONG Yiu Wai
xm on [0, 1] or the power mean inequality, (Queen Elizabeth School, Form 6),
tan A tan B tan C we have YEUNG Yuen Chuen (La Salle College,
> [tan A] + [tan B] + [tan C], Form 4) and YU Hok Kan (STFA Leung
x1 + L + x n m m
x m + L + xn Kau Kui College, Form 6).
where [t] denotes the greatest integer ( ) ≤ 1 .
n n As ∠ACB >∠ACP = 30○, we get
less than or equal to t. Give a proof.
Using x1 + L + xn = 1 and multiplying ∠CAB = ∠CBA < (180○− 30○) / 2 = 75○.
Problem 202. (Due to LUK Mee Lin, both sides by nm+1, we get the desired
La Salle College) For triangle ABC, let Hence ∠CAB ≤ 74○. Then
D, E, F be the midpoints of sides AB,
BC, CA, respectively. Determine Other commended solvers: TONG Yiu ∠CPB = ∠CAB + ∠ACP
which triangles ABC have the property Wai (Queen Elizabeth School, Form 6), ≤ 74○+ 30○ = 104○.
YEUNG Wai Kit (STFA Leung Kau Kui Now
that triangles ADF, BED, CFE can be College, Form 3) and YEUNG Yuen Chuen (La
folded above the plane of triangle DEF Salle College, Form 4). ∠QPB = 360○ – ∠QPC − ∠CPB
to form a tetrahedron with AD
≥ 360○ – 78○ – 104○ = 178○.
coincides with BD; BE coincides with Problem 197. In a rectangular box, the
CE; CF coincides with AF. lengths of the three edges starting at the Since the angles of triangle QPB are
same vertex are prime numbers. It is also positive integers, we must have
Problem 203. (Due to José Luis
given that the surface area of the box is a ∠QPB = 178○, ∠PBQ = 1○ =∠PQB
DÍAZ-BARRERO, Universitat Politec-
power of a prime. Prove that exactly one
nica de Catalunya, Barcelona, Spain) and all less-than-or-equal signs must be
of the edge lengths is a prime number of
Let a, b and c be real numbers such that equalities so that
the form 2k −1. (Source: KöMaL Gy.3281)
a + b + c ≠ 0. Prove that the equation
Solution. CHAN Ka Lok (STFA Leung ∠CAB = ∠CBA = 74○ and ∠ACB = 32○.
(a+b+c)x + 2(ab+bc+ca)x + 3abc = 0 Kau Kui College, Form 4), KWOK Tik Other commended solvers: CHAN Ka Lok
Chun (STFA Leung Kau Kui College, Form (STFA Leung Kau Kui College, Form 4),
has only real roots. 6), John PANAGEAS (Kaisari High KWOK Tik Chun (STFA Leung Kau Kui
School, Athens, Greece), POON Ming College, Form 6), Achilleas P.
Problem 204. Let n be an integer with Fung (STFA Leung Kau Kui College, Form PORFYRIADIS (American College of
n > 4. Prove that for every n distinct 6), Achilleas P. PORFYRIADIS (American Thessaloniki “Anatolia”, Thessaloniki,
integers taken from 1, 2, …, 2n, there College of Thessaloniki “Anatolia”, Greece), SIU Ho Chung (Queen’s College,
Thessaloniki, Greece), SIU Ho Chung Form 5), YEUNG Wai Kit (STFA Leung Kau
always exist two numbers whose least (Queen’s College, Form 5), TO Ping Kui College, Form 3), Richard YEUNG Wing
common multiple is at most 3n + 6. Leung (St. Peter’s Secondary School), Fung (STFA Leung Kau Kui College, Form 6)
YEUNG Wai Kit (STFA Leung Kau Kui and YIP Kai Shing (STFA Leung Kau Kui
Problem 205. (Due to HA Duy Hung, College, Form 3), YEUNG Yuen Chuen (La College, Form 4).
Hanoi University of Education, Salle College, Form 4) and YU Hok Kan
(STFA Leung Kau Kui College, Form 6). Problem 199. Let R+ denote the
Vietnam) Let a, n be integers, both
greater than 1, such that an – 1 is Let the prime numbers x, y, z be the positive real numbers. Suppose
divisible by n. Prove that the greatest lengths of the three edges starting at the f : R + → R + is a strictly decreasing
common divisor (or highest common same vertex. Then 2(xy + yz + zx) = pn for function such that for all x, y ∈ R + ,
factor) of a – 1 and n is greater than 1. some prime p and positive integer n. Since f (x + y) + f (f (x) + f (y))
the left side is even, we get p = 2. So xy + = f (f (x + f (y)) + f (y + f (x))).
***************** yz + zx = 2n─1. Since x, y, z are at least 2, Prove that f (f (x)) = x for every x > 0.
Solutions the left side is at least 12, so n is at least 5. (Source: 1997 Iranian Math Olympiad)
**************** If none or exactly one of x, y, z is even, Solution. Johann Peter Gustav Lejeune
then xy + yz + zx would be odd, a DIRICHLET (Universidade de Sao
Problem 196. (Due to John contradiction. So at least two of x, y, z are Paulo – Campus Sao Carlos) and Achilleas
PANAGEAS, High School “Kaisari”, even and prime, say x = y = 2. Then z = P. PORFYRIADIS (American College of
M athematical Excalibur Vol. 9, N o. 2, M ay 04- July 04 Page 4
Thessaloniki “Anatolia”, Thessaloniki, moving west. Combining the movement
swept out by A and ∀, we get two
(continued from page 2)
Setting y = x gives continuous paths on the equator. At the
same moment, each point in one path will
f (2x) + f (2f (x)) = f (2f ( x + f (x))). have its opposite point in the other path. Example 5. (1995 Russian Math
Olympiad) Given a semicircle with
Setting both x and y to f(x) in the given Let N be the initial point of A in his travel
equation gives and let P(N) denote the path beginning diameter AB and center O and a line,
with N. Let W be the westernmost point which intersects the semicircle at C and D
f (2f (x)) + f (2f (f (x))) on P(N). Let N’ and W’ be the opposite and line AB at M (MB < MA, MD < MC).
= f (2f (f (x) + f (f (x)))). points of N and W respectively. By the
Let K be the second point of intersection
westward travel condition on A, W cannot
Subtracting this equation from the one of the circumcircles of triangles AOC and
be as far as N’.
DOB. Prove that ∠MKO = 90○.
Assume the conclusion of the problem is
f (2f (f (x))) – f (2x)=f (2f ( f (x) + f (f (x)))) false. Then the easternmost point reached
– f ( 2f (x + f (x))). Solution. Consider the inversion with
by P(N) cannot be as far as N’. So P(N)
center O and radius r = OA. By fact (2),
Assume f (f (x)) > x. Then 2f (f (x)) > 2x. will not cover the inside of minor arc WN’
and the other path will not cover the inside A, B, C, D are sent to themselves. By fact
Since f is strictly decreasing , we have
f(2f (f (x))) < f (2x). This implies the left of minor arc W’N. Since A have walked (4), the circle through A, O, C is sent to
side of the last displayed equation is over all points of the equator (and hence A line AC and the circle through D, O, B is
negative. Hence, and∀ together walked every point at least
sent to line DB. Hence, the point K is sent
twice), P(N) must have covered every
f (2f ( f (x) + f ( f (x)))) < f ( 2f ( x + f (x))). point of the minor arc W’N at least twice. to the intersection K′ of lines AC with DB
Since P(N) cannot cover the entire equator, and the point M is sent to the intersection
Again using f strictly decreasing, this every point of minor arc W’N must be
inequality implies M′ of line AB with the circumcircle of
traveled westward at least once by A or ∀.
∆OCD. Then the line MK is sent to the
2f ( f (x) + f ( f (x))) > 2f ( x + f (x)), Then A travelled westward at least a
distance equal to the sum of lengths of circumcircle of OM′K′.
which further implies minor arcs W’N and NW, i.e. half of the To solve the problem, note by fact (7),
equator. We got a contradiction. ∠MKO=90○ if and only if ∠K′M′O= 90○.
f (x) + f (f (x)) < x + f (x).
Other commended solvers: POON Ming Since BC⊥AK′, AD⊥BK′ and O is the
Canceling f (x) from both sides leads to Fung (STFA Leung Kau Kui College, Form
6). midpoint of AB, so the circumcircle of
the contradiction that f (f (x)) < x.
∆OCD is the nine-point circle of ∆ABK′,
Similarly, f (f (x)) < x would also lead to a which intersects side AB again at the foot
contradiction as can be seen by reversing of perpendicular from K′ to AB. This
all inequality signs above. Therefore, we Olympiad Corner
point is M′. So ∠K′M′O = 90○ and we are
must have f (f (x)) = x. (continued from page 1)
Problem 200. Aladdin walked all over of lines in ℒ which separate p from q is
the equator in such a way that each odd if and only if p and q have the same Example 6. (1995 Iranian Math
moment he either was moving to the color. Olympiad) Let M, N and P be points of
west or was moving to the east or Note: A line ℓ separates two points p intersection of the incircle of triangle
applied some magic trick to get to the and q if p and q lie on opposite sides of ℓ ABC with sides AB, BC and CA
opposite point of the Earth. We know with neither point on ℓ. respectively. Prove that the orthocenter
that he travelled a total distance less
of ∆MNP, the incenter of ∆ABC and the
than half of the length of the equator Problem 4. For a real number x, let x
altogether during his westward moves. circumcenter of ∆ABC are collinear.
Prove that there was a moment when stand for the largest integer that is less
Solution. Note the incircle of ∆ABC is
the difference between the distances he than or equal to x. Prove that the circumcircle of ∆MNP. So the first
had covered moving to the east and
moving to the west was at least half of ( n − 1)! two points are on the Euler line of ∆MNP.
n ( n + 1)
the length of the equator. (Source: Consider inversion with respect to the
KöMaL F. 3214) incircle of ∆ABC with center I. By fact
is even for every positive integer n.
Solution. (2), A, B, C are sent to the midpoints A′,
Let us abbreviate Aladdin by A. At every Problem 5. Prove that B′, C′ of PM, MN, NP, respectively. The
moment let us consider a twin, say ∀, of circumcenter of ∆A′B′C′ is the center of
(a2 +2) (b2 +2) (c2 +2) ≥ 9 (ab+bc+ca)
A located at the opposite point of the the nine point circle of ∆MNP, which is
position of A. Now draw the equator for all real numbers a, b, c > 0. on the Euler line of ∆MNP. By fact (3),
circle. Observe that at every moment
the circumcircle of ∆ABC is also on the
either both are moving east or both are
Euler line of ∆MNP.