Volume 9, Number 4 October 2004 – December 2004
Homothety
Kin Y. Li
Olympiad Corner
The Czech-Slovak-Polish Match this year took place in Bilovec on June 21-22, 2004.Here are the problems.
Problem 1. Show that real numbers p, q,
r satisfy the condition
p4(q – r)2 + 2p2(q + r) + 1 = p4
if and only if the quadratic equations x2 + px + q = 0 and y2 – py + r = 0
have real roots (not necessarily distinct) which can be labeled by x1, x2 and y1, y2,
respectively, in such way that the equality x1y1 – x2y2 = 1 holds.
Problem 2. Show that for each natural
number k there exist at most finitely many triples of mutually distinct primes p, q, r for which the number qr – k is a multiple of p, the number pr – k is a multiple of q, and the number pq – k is a multiple of r.
Problem 3. In the interior of a cyclic
quadrilateral ABCD, a point P is given such that |∠BPC|=|∠BAP|+|∠PDC|. Denote by E, F and G the feet of the perpendiculars from the point P to the lines AB, AD and DC, respectively. Show that the triangles FEG and PBC are similar.
(continued on page 4)
Editors: Ի ஶ(CHEUNG Pak-Hong), Munsang College, HK
ଽ υ ࣻ (KO Tsz-Mei)
గ ႀ ᄸ (LEUNG Tat-Wing)
፱ (LI Kin-Yin), Dept. of Math., HKUST
֔ ᜢ ݰ (NG Keng-Po Roger), ITC, HKPU
Artist: ྆ ؾ ़ (YEUNG Sau-Ying Camille), MFA, CU
Acknowledgment: Thanks to Elina Chiu, Math. Dept.,
HKUST for general assistance.
On-line:
http://www.math.ust.hk/mathematical_excalibur/ The editors welcome contributions from all teachers and students. With your submission, please include your name, address, school, email, telephone and fax numbers (if available). Electronic submissions, especially in MS Word, are encouraged. The deadline for receiving material for the next issue is January 20, 2005.
For individual subscription for the next five issues for the 03-04 academic year, send us five stamped self-addressed envelopes. Send all correspondence to:
Dr. Kin-Yin LI Department of Mathematics
The Hong Kong University of Science and Technology Clear Water Bay, Kowloon, Hong Kong
Fax: (852) 2358 1643 Email: [email protected]
A geometric transformation of the plane is a function that sends every point on the plane to a point in the same plane. Here we will like to discuss one type of geometric transformations, called homothety, which can be used to solve quite a few geometry problems in some international math competitions. A homothety with center O and ratio k is a function that sends every point X on the plane to the point X’ such that OXuuuur'=k OXuuur.
So if |k| > 1, then the homothety is a magnification with center O. If |k| < 1, it is a reduction with center O. A homothety sends a figure to a similar figure. For instance, let D, E, F be the midpoints of sides BC, CA, AB respectively of ∆ABC. The homothety with center A and ratio 2 sends ∆AFE to ∆ABC. The homothety with center at the centroid G and ratio –1/2 sends ∆ABC to ∆DEF.
Example 1. (1978 IMO) In ABC, AB = AC. A circle is tangent internally to the circumcircle of ABC and also to the sides AB, AC at P, Q, respectively. Prove that the midpoint of segment PQ is the center of the incircle of ∆ABC.
D O A B' C' P Q B C I
Solution. Let O be the center of the circle. Let the circle be tangent to the circumcircle of ∆ABC at D. Let I be the midpoint of PQ. Then A, I, O, D are collinear by symmetry. Consider the homothety with center A that sends ∆ABC to AB’C’ such that D is on B’C’. As right triangles AIP, ADB’, ABD, APO are similar, we have
AI /AO = (AI / AP)(AP / AO)
= (AD /AB’)(AB /AD) = AB/AB’, Since O is the incenter of ∆AB’C’,this implies I is the incenter of ∆ABC.
Example 2. (1981 IMO) Three congruent circles have a common point O and lie inside a given triangle. Each circle touches a pair of sides of the triangle. Prove that the incenter and the circumcenter of the triangle and the point O are collinear.
A' B' C' O A B C
Solution. Consider the figure shown. Let A’, B’, C’ be the centers of the circles. Since the radii are the same, so A’B’ is parallel to AB, B’C’ is parallel to BC, C’A’ is parallel to CA. Since AA’, BB’ CC’ bisect ∠ A, ∠ B, ∠ C respectively, they concur at the incenter I of ∆ABC. Note O is the circumcenter of ∆A’B’C’ as it is equidistant from A’, B’, C’. Then the homothety with center I sending ∆A’B’C’ to ∆ABC will send O to the circumcenter P of ∆ABC. Therefore, I, O, P are collinear.
Example 3. (1982 IMO) A non-isosceles triangle A1A2A3 is given
with sides a1, a2, a3 (ai is the side
opposite Ai). For all i=1, 2, 3, Mi is the
midpoint of side ai, and Ti is the point
where the incircle touchs side ai. Denote
by Si the reflection of Ti in the interior
bisector of angle Ai.
Prove that the lines M1S1, M2S2 and M3S3
are concurrent. I T2 A2 A3 A 1 B2 B3 B1T1 S2 S3 S1 T3
Mathematical Excalibur, Vol. 9, No. 4, Oct. 04- Dec. 04 Page 2
Solution. Let I be the incenter of ∆A1A2A3. Let B1, B2, B3 be the points
where the internal angle bisectors of ∠A1, ∠ A2, ∠ A3 meet a1, a2, a3
respectively. We will show SiSj is
parallel to MiMj. With respect to A1B1,
the reflection of T1 is S1 and the
reflection of T2 is T3. So ∠T3IS1 = ∠
T2IT1. With respect to A2B2, the
reflection of T2 is S2 and the reflection
of T1 is S3. So ∠T3IS2 = ∠T1IT2. Then
∠ T3IS1 = ∠ T3IS2. Since IT3 is
perpendicular to A1A2, we get S2S1 is
parallel to A1A2. Since A1A2 is parallel
to M2M1, we get S2S1 is parallel to
M2M1. Similarly, S3S2 is parallel to
M3M2 and S1S3 is parallel to M1M3.
Now the circumcircle of ∆S1S2S3 is the
incircle of ∆A1A2A3 and the
circumcircle of ∆M1M2M3 is the nine
point circle of ∆A1A2A3. Since ∆A1A2A3
is not equilateral, these circles have different radii. Hence ∆S1S2S3 is not
congruent to ∆M1M2M3 and there is a
homothety sending ∆S1S2S3 to
∆M1M2M3. Then M1S1, M2S2 and M3S3
concur at the center of the homothety.
Example 4. (1983 IMO) Let A be one of the two distinct points of intersection of two unequal coplanar circles C1 and C2 with centers O1 and
O2 respectively. One of the common
tangents to the circles touches C1 at P1
and C2 at P2, while the other touches C1
at Q1 and C2 at Q2. Let M1 be the
midpoint of P1Q1 and M2 be the
midpoint of P2Q2. Prove that ∠O1AO2
= ∠M1AM2. O1 A O2 O P2 Q2 P1 Q1 M2 M1 B
Solution. By symmetry, lines O2O1,
P2P1, Q2Q1 concur at a point O.
Consider the homothety with center O which sends C1 to C2. Let OA meet C1
at B, then A is the image of B under the homothety. Since ∆BM1O1 is sent to
∆AM2O2, so ∠M1BO1 = ∠M2AO2.
Now ∆OP1O1 similar to ∆OM1P1
implies OO1/OP1 = OP1/OM1. Then
OO1 ·OM1 = OP12 = OA · OB,
which implies points A, B, M1, O are
concyclic. Then ∠ M1BO1 = ∠ M1AO1.
Hence ∠ M1AO1 = ∠ M2AO2. Adding
∠O1AM2 to both sides, we have ∠O1AO2
= ∠M1AM2.
Example 5. (1992 IMO) In the plane let C be a circle, L a line tangent to the circle C, and M a point on L. Find the locus of all points P with the following property: there exist two points Q, R on L such that M is the midpoint of QR and C is the inscribed circle of ∆PQR. C L C' P Q R S U T,T' M V
Solution. Let L be the tangent to C at S. Let T be the reflection of S with respect to M. Let U be the point on C diametrically opposite S. Take a point P on the locus. The homothety with center P that sends C to the excircle C’ will send U to T’, the point where QR touches C’. Let line PR touch C’ at V. Let s be the semiperimeter of ∆PQR, then
TR = QS = s – PR = PV – PR =VR = T’R so that P, U, T are collinear. Then the locus is on the part of line UT, opposite the ray U Tuuur.
Conversely, any point P on the part of line UT, opposite the ray U Tuuur, will result in T
= T’ and QS = TR by reversing the steps above. Then
QM = QS – MS =TR – MT = RM.
For the next example, the solution involves the concepts of power of a point with respect to a circle and the radical axis. We will refer the reader to the article
“Power of Points Respect to Circles,” in Math Excalibur, vol. 4, no. 3, pp. 2, 4.
Example 6. (1999 IMO) Two circles Γ1
and Γ2 are inside the circle Γ, and are
tangent to Γ at the distinct points M and N, respectively. Γ1 passes through the
center of Γ2. The line passing through
the two points of intersection of Γ1 and
Γ2 meets Γ at A and B. The lines MA
and MB meet Γ1 at C and D,
respectively. Prove that CD is tangent to Γ2. Γ L O2 N F E M C,C' A,A' B O1 D
Solution. (Official Solution) Let EF be the chord of Γ which is the common tangent to Γ1 and Γ2 on the same side of
line O1O2 as A. Let EF touch Γ1 at C’.
The homothety with center M that sends Γ1 to Γ will send C’ to some point
A’ and line EF to the tangent line L of Γ at A’. Since lines EF and L are parallel, A’ must be the midpoint of arc FA’E. Then ∠A’EC’ = ∠A’FC’ = ∠A’ME. So ∆A’EC is similar to ∆A’ME. Then the power of A’ with respect to Γ1 is
A’C’ ·A’M = A’E2. Similar, the power
of A’ with respect to Γ2 is A’F2. Since
A’E = A’F, A’ has the same power with respect to Γ1 and Γ2. So A’ is on the
radical axis AB. Hence, A’ = A. Then C’ = C and C is on EF.
Similarly, the other common tangent to Γ1 and Γ2 passes through D. Let Oi be
the center of Γi. By symmetry with
respect to O1O2, we see that O2 is the
midpoint of arc CO2D. Then
∠DCO2 = ∠CDO2 = ∠FCO2.
This implies O2 is on the angle bisector
of ∠FCD. Since CF is tangent to Γ2,
therefore CD is tangent to Γ2.
Problem Corner
We welcome readers to submit their solutions to the problems posed below for publication consideration. The solutions should be preceded by the solver’s name, home (or email) address and school affiliation. Please send submissions to Dr. Kin Y. Li, Department of Mathematics, The Hong Kong University of Science & Technology, Clear Water Bay, Kowloon, Hong Kong. The deadline for submitting solutions is January 20, 2005.
Problem 211. For every a, b, c, d in
[1,2], prove that
a b c d 4 a c.
b c d a b d
+ + + ≤ +
+ + +
(Source: 32nd Ukranian Math
Olympiad)
Problem 212. Find the largest positive
integer N such that if S is any set of 21 points on a circle C, then there exist N arcs of C whose endpoints lie in S and each of the arcs has measure not exceeding 120°.
Problem 213. Prove that the set of all
positive integers can be partitioned into 100 nonempty subsets such that if three positive integers a, b, c satisfy a + 99 b = c, then at least two of them belong to the same subset.
Problem 214. Let the inscribed circle
of triangle ABC be tangent to sides AB, BC at E and F respectively. Let the angle bisector of ∠ CAB intersect segment EF at K. Prove that ∠CKA is a right angle.
Problem 215. Given a 8×8 board.
Determine all squares such that if each one is removed, then the remaining 63 squares can be covered by 21 3×1 rectangles.
*****************
Solutions
****************Problem 206. (Due to Zdravko F.
Starc, Vršac, Serbia and Montenegro) Prove that if a, b are the legs and c is the hypotenuse of a right triangle, then
(a b+ ) a +(a b− ) b< 2 2c c.
lution. Cheng HAO (The Second igh School Attached to Beijing
eorem,
So H
Normal University), HUI Jack (Queen’s College, Form 5), D. Kipp JOHNSON (Valley Catholic School, Teacher, Beaverton, Oregon, USA), POON Ming
Fung(STFA Leung Kau Kui College,
Form 7), Achilleas P. PORFYRIADIS (American College of Thessaloniki “Anatolia”, Thessaloniki, Greece), Problem Group
Discussion Euler-Teorema(Fortaleza, Brazil), Anna Ying PUN (STFA Leung Kau Kui
College, Form 6), TO Ping Leung (St. Peter’s Secondary School) and YIM Wing Yin (South Tuen Mun Government Secondary School, Form 4). y Pythagoras’ th B 2 (a b) (a + ≤ + + −b)2 = 2 .c
quality if and only if a = b. By the
a b E Cauchy-Schwarz inequality, (a +b) a +(a−b) b ≤ (a +b)2 + (a− b)2 a+ b ≤ 2c 2c.
equality to hol hout, we need
For d throug
: : 1 :1
a b a b+ − = a b = , which is not possible for legs of a triangle. So we must have strict inequality.
Other commended solvers: HUDREA
cond High al
s of A with 1, 2 or 3
m
undation Secondary School, Form
D be a point on the minor arc BC of
Greece),
Mihail (High School “Tiberiu Popoviciu”
Cluj-Napoca Romania) and TONG Yiu
Wai (Queen Elizabeth School, Form 7). Problem 207. Let A = { 0, 1, 2, …, 9} and
B1, B2, …, Bk be nonempty subsets of A
such that Bi and Bj have at most 2 common
elements whenever i ≠ j. Find the maximum possible value of k.
Solution. Cheng HAO (The Se
chool Attached to Beijing Norm S
University), HUI Jack (Queen’s College, Form 5), POON Ming Fung(STFA Leung Kau Kui College, Form 7) and
Achilleas P. PORFYRIADIS (American
College of Thessaloniki “Anatolia”, Thessaloniki, Greece).
we take all subset If
elements, then these 10 + 45 + 120 = 175 subsets satisfy the condition. So k ≥ 175. Let B1, B2, …, Bk satisfying the condition
with k maxi um. If there exists a Bi with
at least 4 elements, then every 3 element subset of Bi cannot be one of the Bj, j ≠ i,
since Bi and Bj can have at most 2 common
elements. So adding these 3 element subsets to B1, B2, …, Bk will still satisfy the
conditions. Since Bi has at least four 3
element subsets, this will increase k, which contradicts maximality of k. Then every Bi has at most 3 elements. Hence, k
≤ 175. Therefore, the maximum k is 175.
Other commended solvers: CHAN
Wai Hung (Carmel Divine Grace
o F
6), LI Sai Ki (Carmel Divine Grace Foundation Secondary School, Form 6), LING Shu Dung,
Anna Ying PUN
(STFA Leung Kau Kui College, Form 6) and YIM Wing Yin (South Tuen Mun Government Secondary School, Form 4).
Problem 208. In ∆ABC, AB > AC > BC.
Let
the circumcircle of ∆ABC. Let O be the circumcenter of ∆ABC. Let E, F be the intersection points of line AD with the perpendiculars from O to AB, AC, respectively. Let P be the intersection of lines BE and CF. If PB = PC + PO, then find ∠BAC with proof.
Solution. Achilleas P. PORFYRIADIS (American College of Thessaloniki
Anatolia”, Thessaloniki, “
Problem Group Discussion Euler - Teorema ( Fortaleza, Brazil) and Anna Ying PUN (STFA Leung Kau Kui College,
Form 6). O A B C D E F P
Since E is on the perpendicular bisector of chord AB and F is on t ∠ ∠ ∠ ∠ = PO·BC. Since d so he perpendicular bisector of chord AC, AE = BE and AF = CF. Applying exterior angle theorem,
∠BPC =∠AEP + ∠CFD = 2 ( BAD+ CAD)
= 2 BAC = BOC. Hence, B, C, P, O are concyclic. By Ptolemy’s theorem,
PB·OC = PC·OB + PO·BC. hen (PB – PC)·OC
T
PB – PC = PO, we get OC = BC an ∆OBC is equilateral. Then
∠BAC=1
2∠BOC = 30°
Other commended solvers: Cheng HAO
ig hed to
Un I Jack
ueen’s College, Form 5), POON
ing Fung(STFA Leung Kau Kui
(The Second H h School Attac Beijing Normal iversity), HU (Q
M
Mathematical Excalibur, Vol. 9, No. 4, Oct. 04- Dec. 04 Page 4
sible by n – 1.
Kau Kui College,
m Group Discussion
2 + 2 is another such number. ince N – 1 = 2n + 1= (n – 1)k is odd, so k
1) is N
o is
(Queen Elizabeth School, Form 7) and
YIM Wing Yin (South Tuen Mun
Government Secondary School, Form 4).
Problem 209. Prove that there are
infinitely many positive integers n such that 2n + 2 is divisible by n and 2n + 1 is
ivi d
Solution. D. Kipp JOHNSON (Valley Catholic School, Teacher, Beaverton, Oregon, USA), POON Ming
Fung(STFA Leung Proble
Form 7) and
Euler-Teorema(Fortaleza, Brazil).
As 22 + 2 = 6 is divisible by 2 and 22 + 1
= 5 is divisible by 1, n = 2 is one such number.
Next, suppose 2n + 2 is divisible by n
and 2n + 1 is divisible by n – 1. We will
prove N = n
S
is odd and n is even. Since N = 2n + 2 =
2(2n–1 + 1) = nm and n is even, so m must
be odd. Recall the factorization xi + 1 = (x + 1)(xi–1 – xi–3 + … + 1)
for odd positive integer i. Since k is odd, 2N + 2 = 2(2N–1 + 1) = 2(2(n–1)k +
n
divisible by 2(2n–1 + 1) = 2 + 2 =
sing the factorization ab ve. Since m u
odd, 2N + 1 = 2nm + 1 is divisible by
2n + 1 = N – 1. Hence, N is also such a
number. As N > n, there will be infinitely many such numbers.
Problem 210. Let a1 = 1 and
1 1 2 n n a a a + = + n r every
for n = 1, 2, 3, … . Prove that fo integer n > 1, 2 2 2 n a − is an integer.
.R.A. 20 roblem Group
(Roma, Italy), HUDREA Mihail (High eriu Popoviciu” Cluj-
Problem Group iscussion Euler – Teorema
Solution. G P
School “Tib
Napoca Romania),
D (Fortaleza,
Brazil), TO Ping Leung (St. Peter’s Secondary School) and YIM Wing Yin (South Tuen Mun Government Secondary School, Form 4).
Note an = pn / qn, where p1 = q1 = 1, pn+1 = pn2 + 2qn2 , qn+1 = 2pnqn for n = 1, 2, 3, …. Then 2 2 2 2 2 . 2 2 n n n n q a − = p − q
It suffices to show by mathematical
n that pn2 – 2qn2 = 1 for n > 1. We
have 2 – 2q22 = 32 – 2·22 = 1. Assuming
case n is true, we get
vers: Ellen CHAN ru h Normal ollege, inductio p2 pn+12 – 2qn+12 = (pn2 + 2qn2)2 –2(2pnqn) = (pn2 – 2qn2)2 = 1.
Other commended sol
On Ting (T e Light Girls’ College, Form
5), Cheng HAO (The Second Hig School Attached to Beijing
niversity), HUI Jack (Queen’s C U
Form 5), D. Kipp JOHNSON (Valley Catholic School, Teacher, Beaverton, Oregon, USA), LAW Yau Pui (Carmel Divine Grace Foundation Secondary School, Form 6), Asger OLESEN (Toender Gymnasium (grammar school), Denmark), POON Ming Fung(STFA Leung Kau Kui College, Form 7),
Achilleas P. PORFYRIADIS (American
College of Thessaloniki “Anatolia”, Thessaloniki, Greece), Anna Ying PUN (STFA Leung Kau Kui College, Form 6),
Steve ROFFE, TONG Yiu Wai (Queen
Elizabeth School, Form 7) and YEUNG
Wai Kit (STFA Leung Kau Kui College,
Form 4).
lymp
O
iad Corner
(continued from page 1 roblem 4. Solve the system of equations
) P 1 1, x = + xy z 1 1, y yz = x + 1 1 z zx = y +
Problem 5. In the interiors of the sides AB,
BC and CA of a given triangle ABC, points iven such in the domain of real numbers.
K, L and M, respectively, are g hat t | | | | | |. | | | | | | A K B L C M K B = L C = M A
Show that the triangles ABC and KLM have a common orthocenter if and only if
gle ABC is equilateral.
the heaps n the table, merge them into a single new the trian
Problem 6. On the table there are k heaps
of 1, 2, …, k stones, where k ≥3. In the first step, we choose any three of
o
heap, and remove 1 stone (throw it away from the table) from this new heap. In the second step, we again merge some three of the heaps together into a single new heap, and then remove 2 stones from this new heap. In general, in the i-th step we choose any three of the heaps, which contain more than i stones when combined, we merge them into a single new heap, then remove i stones from this new heap. Assume that after a number of steps, there
is a single heap left on the table, containing p stones. Show that the number p is a perfect square if and only if the numbers 2k+2 and 3k+1 are perfect squares. Further, find the least number k for which p is a perfect square.
Homothety
(continued from page 2)
Example 7. (20 a t
00APMO) Let ABC be h
isector, respectively at A meet the side
riangle. Let M and N be the points in ich the median and the angle w
b
BC. Let Q and P be the points in which the perpendicular at N to NA meets MA and BA respectively and O the point in which the perpendicular at P to BA meets AN produced.
Prove that QO is perpendicular to BC. A B C D M N P O C' B' K,Q
Solution (due to Bobby Poon). The case AB = AC is clear.
t loss of generality, we m
Withou ay
assume AB > AC. Let AN intersect the circumcircle of ∆ABC at D. Then
∠DBC = ∠DAC = 1
2∠BAC
= ∠DAB =∠DCB. So DB = DC and MD is perpendicular to BC
Consider the homothety with center A en B’ = OC’ and BC is parallel to B’C’. ∠OB’K = ∠DBC = ∠DAB and K = C’K. Since BC is parallel to B’C’, = Q. B’C’, we get QO is perpendicular to .
that sends ∆DBC to ∆OB’C’. Th O
Let B’C’ intersect PN at K. Then
= 90° – ∠AOP = ∠OPK. So points P, B’, O, K are concyclic. Hence ∠B’KO =∠B’PO = 90° B’
this implies K is on AM. Hence, K Since ∠B’KO = 90° and BC is parallel to
w First part Assume that the equations (2) have real roots satisfying xryr - ssyz = 1. By the familiar formula, the roots of the quadratic equations are given by
-pfK md PfL
x1,2 z ~
2 Yl,i = -1
2 . (3)
where the real numbers K, L satisfy Ka = p2 - 4q and L2 = P2 - 4r (we choose the signs of K, L in accordance with the labelling of the roots). Thus
l~zlyl_z2~22 (-p+K)b+L)-(-P-KWL) =P(K-.L)
4 2 ’
whence p # 0 and K - L = 2/p. Substituting this into the equality
(K+L)(K-L)=K’-L2=(p2-4q)-@-4r)=4(r-q)
yieldsKi-L=2p(r-q). FkomthesevaluesofKfLandK-LweobtainK=
l/p-p(q-r), so upon squaring, K2 = l/p2-2(q-r)+p2(q-r)2. Comparing this with the equality K2 = p2 - 4q, an easy manipulation leads to the desired equation (1).
Second part. Assume that (1) holds. Then clearly p # 0. The equation (1) can be rewritten in either of the following two forms,
p4(q - r)’ + 2p2(q - r) + 1 = p” - 4p2r and p”(r 1 q)” t 2p”(r - q) + 1 = p” - 4p’q.
Upon dividing by p2 we ikd that the discrimmants of the equations (2) are equal to
p2 - 4q = pyr-q)+l 2
>
and p2-4r= P
hence, they are nonnegative and the (real) roots of (2) have the form (3), where
K = P"(--
q) +
1 mdL_p2krr)+1
P P
.
The signs of the numbers K and L have been chosen so that (see First Part)
xlyl _ x2y2 =
P(K - L) =
21 .2 2 (
p’(r - q) + 1 + P2(P - T) + 1 = I
P P > *
’ (Mutually distmct) primes p, q, r satisfy the desired conditions if and, only if the number ~q +pr + qr - k is divisible by each of the primes p, q, r; that is, by the product ~qr. The equality PQ f pr -!- qr - k = n . pqr, for a suitable integer n, can be rewritten as k = ~q + pr + qr - n. pqr. If n < 0, then the last equality implies th@
max{pq, pr, qr} < k; however, then each of the primes P, q, r is less than or equal to k/2 (and there is only a finite number af, such triples). If n 2 1, then we get the estimate k < PQ + pr + qr - pqr. Let us show that the last expression is negative (contradicting the fact that k > 0) unless the triple in question is (p, q,r} = {2,3,5}. Wecanassumethat2<p<q<randr>7. ThenPq>/2~3=6audtheinequality (p - 2)(q - 2) > 0 implies that p + q < $pq + 2, hence
W+pr+qr-pqr~~++)r+W-Wrg(~pq+2)r+W-Wr
= 2r -pq(!jr - 1) < 2r - S(ir - 1) = 6 -r < 0. ___--
cumcircles of the triangles PAB and PCD, respectively. In the interior of the angle
BPC, consider the half-line PT such that lLBPT/ = JLBAPI. Then the hypothesis on P implies that (Fig. 1)
lLTPC/ = /LBPCl - lLBPTl= ILBPCl - ILBAP( = ILPDCI.
Thus PT is the common interior tangent of the circles kl and k2.
Fig. 1
Let us fkst sssume that the sides AB and CD of the given cyclic quadrangle are not parallel. Since the segments AB and CD are comnion chords of the Arcles ICI,
k and k2, k, respectively, there exists a unique point Q having the same power with respect to all three circles k, kr and k2. The point Q is the intersection of the, three limes AB, DC and PT. Without loss of generality, we can assume that the point Q is located nn the half-line B-4 beyond the point A (Fig.‘l). Then we have
..
(LQPAI = ILPBAI.
Since ILAEPI = liAFP/ = QO”, the quadrangle AEPF is cyclic, and
0)
ILFEPI = ILFAPI = ILDAPJ. ’
(2)
Similarly we see that the quadrangle DGPF is cyclic. It follows that]LBPCl = /LBAPl + ILPDCI = ]LEFPj t ILPFGI = ILEFGI.
(3)
Since further (/PiQI = ILPGQl = 90”;the quadrangle QEPG is also cyclic and
!LGEPI = (LGQPJ = (LDQPI. (4)
F’rom therelations (2), (4) andtheequality ILDAPI+ILQPAI = ILQDAI+ILDQPJ,
we further have
ILFEGI = ILFEPI - !LGEPI = [LDAPI - ILDQPI = ILQDAI - !LQPAI. (5)
Since the quadrangle /&CD is cyclic, ILQDAj = ILQBCI. From the relations (1)
and (5) we thus get
ILFEGI = ILQBCI - lLPl?Al = ILPBCI.
Using finaily the relations (3) and (6) we see that the triangles FEG and PBC are
similar (aa they have two congruent angles). I
An analogous argument can be used when the point Q is located on the half-line AB beyond the point B. If the lines AB and CD ace parallel, then ABED is an equilateral trapezoid, with bases AB and CD. Since the points E, P, G are collinear
and the common interior tangent of the circles kr and ks is parallel to both lir.~ AI3 and CD, the triangles APD and BPC are congruent. The simkity of the triangles
EFG and APD thus implies also the similarity of the triangles EFG and BPC. This completes the proof, .
k%J!?&- From the form of the equations it is immediate that syz # 0. Two of the numbers x, v, t have to be of the same sign; then the right-hand side of the equation where the ratio of these two numbers occurs is positive, hence so must be the corresponding left-hand side, which implies that the third of the numbers 5, y, z must also have the same sign as the f&t and the second. Thus either t.y.t > 0; or z, g, z < 0. Let us consider only the former case (the latter can be reduced to it by passing from the soiution (z, y,z) to the solution (-2,--y, -z)). Multiply the first two equations of the system by the expression xyt and then subtract them; this gives, upon a smaIi manipuiation, z - 5 = y(z2 - yz). If a triple (2, y, 2) is a solution, then so are also the triples (y, z, z
x = m8x{?,y, z}. Then z - z < 0 and z a
and (z, x, y); thus we may assume that - yz 2 0 (remember that z, y, I > 0), so the equality z - z = &z2 - yz), together with the condition y > 0, implies that z - I = x2 - yz = 0, which means that x = y = z. The system then reduces to the single equation’l/s’ = 1+ 1, which has a (unique) positive root z = d/2.
Conclusion The system has exactly two solutions, z = g I= z = &a/2.
t+ A point V of the plane containing a triangle ABC is its orthocenter if end only if, at the same time, AV I BC and BV I AC; that is, AV . BC = 6
and BV * AC = 0. Substituting BC = BV - CV and AC = AV - CV, an easy
manipulation leads to the equivalent condition in the form of the equality of the scalar products
AV.BV=AV.CV=BV.CV. (1)
Our goal is thus to find out when the system (1) is satisfied together with the analogous system
KV.LV=KV.MV=LV-MV, (2)
expressing the fact that the point V is the orthocenter of the triangle KLM. We now express the vectors from (2) as linear combinations of the vectors from (1). By hy- pothesis, there exists a number p, 0 < p < 1, for which
AK=pAB, BL=pBC, CM=pCA.
gubstituting into the first equality AK = AV - KV and AB = AV - BV, we g&t after a small manipulation the first of the following three equalities
KV=(l-P)AV+PBV, LV=(l-p)BV+pCv,
---m____ MVzz(l-pjC~~+pA~~;
the other two can be derived similarly. Taking products, we get
KV . LV = (1 -- p)'AV . BV + ~(1 - p)AV . CV + p(1 - p)BV2 = = (1 -p)s+p(l -p)BV’,
where s denotes the common value of the products from (1). Simiiarly, KVMV=(l-p)s+p(l-p)AV’ and LV&I.V=(~-~)~+P(~-~)BV? We see that the system (2) is equivalent to the system of equalities
p(l - p)AVa’= ~(1 - p)BV’ = ?‘(I - P)cv’,
which, in view of the condition p(1 - p) # 0, is fuifiiied if and only if iAvl = JBVI =
jCV(. The last condition meass that the orthocenter V of the triangle ABC coincides
with its circumcenter. This happens if and only if the triangle ABC is equiI&raL ___ . ftib’*!?--After i steps, there will be k - 2i heaps left on the table; thus if a single
heap is to remain in the end, the number k must be odd and the t&d number of steps has to be ‘(k - 1). Let us distinguish two cases, according as the remainder of
k upon division a y 4 is 1 or 3.
Thccaseofk-4cfi. Inthebeginningtherearel+~~~+k=~k(k+l)=
(4c + 1)(2c + 1) stones on the table, from which in course of the 2c steps we wiii remove 1 + . . . $2~ = c(2c + 1) stones; thus the number of stones in the l8st heap wiii be
p = (4c + 1)(2c + 1) - c(2c + 1) = (zc + 1)(3c f 1).
Since the numbers 2c + 1 and 3c + 1 are coprime, p is a perfect square if and only if both 2c + 1 and 3c + 1 are perfect squares; that is, if and only if their qrmdruples 4(2c + 1) = 2k + 2 and 4(3c+ 1) = 3k + 1 8re perfect squares.
‘i’hecaseofk=4c+3. Inthebeginningtherearel+~~~+k=~k(k+l)=
2(c + 1) (4~ + 3) stones on the table, from which we remove in course of a!lthe2c+l ste&r a total of l+.:. ‘. + (zc.$-i) I (0 f i j(&+ q+&& :.&* &_._lg+t IBM.&+“‘, ,wti contain ,, ;’ i-, :< :. _ . .._. j’ ,, ,’ :. . . .
