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Volume 9, Number 1 January 2004 – April 2004
Olympiad Corner Geometry via Complex Numbers
The Sixth Hong Kong (China) Kin Y. Li
Mathematical Olympiad took place on
December 20, 2003. Here are the Complex numbers are wonderful. In this
problems. Time allowed: 3 hours Z +WW Z =W +W
article we will look at some applications 1 2 1 2
and Z −W W Z = C −W W C
Problem 1. Find the greatest real K such of complex numbers to solving geometry respectively. 1 2 1 2
that for every positive u, v and w with u2 > problems. If a problem involves points
4vw, the inequality and chords on a circle, often we can By moving W toward W along the unit
2 1
2 2 2 2 without loss of generality assume it is the
(u - 4vw) > K(2v - uw)(2w - uv) circle, in the limit, we will get the
holds. Justify your claim. unit circle. In the following discussion, equation of the tangent line at W to the
we will use the same letter for a point to 1
denote the same complex number in the unit circle. It is Z +W 2Z = 2W .
Problem 2. Let ABCDEF be a regular 1 1
hexagon of side length 1, and O be the complex plane. To begin, we will study Similarly, the equation of the line
center of the hexagon. In addition to the the equation of lines through points. through C perpendicular to this tangent
sides of the hexagon, line segments are Suppose Z is an arbitrary point on the
drawn from O to each vertex, making a line through W and W . Since the vector line is Z −W 2Z = C −W 2C .
1 2 1 1
total of twelve unit line segments. Find from W to Z is a multiple of the vector
1 For a given triangle A A A with the unit
the number of paths of length 2003 along from W to W , so in terms of complex 1 2 3
1 2 circle as its circumcircle, in terms of
these line segments that start at O and numbers, we get Z − W = t(W −W ) for
1 2 1 complex numbers, its circumcenter is the
terminate at O. some real t. Now t = t and so origin O, its centroid is G = (A + A +
Z −W Z −W 1 2
1 1 A )/3, its orthocenter is H = A + A + A
Problem 3. Let ABCD be a cyclic W −W =W −W 3 1 2 3
quadrilateral. K, L, M, N are the 2 1 2 1 (because OH = 3OG) and the center of its
nine point circle is N = (A + A + A )/2
midpoints of sides AB, BC, CD and DA Reversing the steps, we can see that 1 2 3
respectively. Prove that the orthocentres every Z satisfying the equation (because N is the midpoint of OH).
of triangles AKN, BKL, CLM, DMN are corresponds to a point on the line Let us proceed to some examples.
through W and W . So this is the
the vertices of a parallelogram. 1 2
(continued on page 4) equation of a line through two points in Example 1. (2000 St. Petersburg City
the complex variable Z. Math Olympiad, Problem Corner 188)
The line S is tangent to the circumcircle
Next consider the line passing through a
Editors: Ի ஶ (CHEUNG Pak-Hong), Munsang College, HK point C and perpendicular to the line of acute triangle ABC at B. Let K be the
ଽ υ ࣻ (KO Tsz-Mei) through W and W . Let Z be on this line. projection of the orthocenter of triangle
గ ႀ ᄸ (LEUNG Tat-Wing) 1 2
፱ (LI Kin-Yin), Dept. of Math., HKUST Then the vector from C to Z is ABC onto line S (i.e. K is the foot of
perpendicular to the vector from W to perpendicular from the orthocenter of
֔ ᜢ ݰ (NG Keng-Po Roger), ITC, HKPU 1
W. In terms of complex numbers, we get
Artist: ྆ ؾ ़ (YEUNG Sau-Ying Camille), MFA, CU 2 triangle ABC to S). Let L be the midpoint
Z − C = it(W − W ) for some real t. So
Acknowledgment: Thanks to Elina Chiu, Math. Dept., HKUST 2 1 of side AC. Show that triangle BKL is
for general assistance and to Lee Man Fui and Poon Ming Fung Z −C = Z −C . isosceles.
for typesetting. i(W −W ) i(W −W )
On-line: http://www.math.ust.hk/mathematical_excalibur/ 2 1 2 1 Solution. (Due to POON Ming Fung,
The editors welcome contributions from all teachers and Again reversing steps, we can conclude this STFA Leung Kau Kui College, Form 6)
students. With your submission, please include your name, is the equation of the line through C Without loss of generality, let the
address, school, email, telephone and fax numbers (if available).
Electronic submissions, especially in MS Word, are encouraged. perpendicular to the line through W and W .
The deadline for receiving material for the next issue is May 25, 1 2 circumcircle of triangle ABC be the unit
2004. circle on the plane. Let A = a + bi, B =
In case the points W and W are on the
For individual subscription for the next five issues for the 03-04 1 2 −i, C = c + di. Then the orthocenter is H
academic year, send us five stamped self-addressed envelopes. unit circle, we have .
WW =1=WW
Send all correspondence to: 1 1 2 2 = A + B + C and K = (a + c) − i, L = (a +
Dr. Kin-Yin LI Multiplying the numerators and c)/2 + (b + d)i/2. Since
Department of Mathematics denominators of the right sides of the 1
The Hong Kong University of Science and Technology two displayed equations above by W W , LB= (a + c)2 + (b + d + 2)2 = KL,
Clear Water Bay, Kowloon, Hong Kong 1 2 2
Fax: (852) 2358 1643 we can simplify them to triangle BKL is isosceles.
Email: makyli@ust.hk
Mathematical Excalibur, Vol. 9, No. 1, Jan 04- Apr 04 Page 2
Example 2. Consider triangle ABC and Example 4. Let A , A , A be the Now the right side is
1 2 3
midpoints of W W , W W , W W
its circumcircle S. Reflect the circle with 2 3 3 1 1 2 Z - Z - Z Z W + Z Z W
respectively. From A drop a 1 2 1 3 2 3 .
respect to AB, AC and BC to get three new i Z - Z - Z Z W + Z Z W
circles S , S and S (with the same perpendicular to the tangent line to the 1 3 1 2 2 3
AB AC BC circumcircle of triangle W W W at W.
radius as S). Show that these three new 1 2 3 i Multiplying the numerator and
circles intersect at a common point. Prove that these perpendicular lines are denominator by Z Z Z W and using
concurrent. Identify this point of 1 2 3
Identify this point. concurrency. ZiZi = 1= WW , we get
Solution. Without loss of generality, we Z Z W −Z Z W −Z +Z
Solution. Without loss of generality, let 2 3 1 3 2 1 .
may assume S is the unit circle. Let the the circumcircle of triangle W W W be Z Z W −Z Z W −Z +Z
center of S be O´, then O´ is the mirror 1 2 3 2 3 1 2 3 1
AB the unit circle. The line perpendicular to This equals the left side (P − R)/(Q − R)
image of O with respect to the segment the tangent at W through A = (W + W )/2
1 1 2 3 and we complete the checking.
AB. So O´ = A + B (because segments has equation
OO´ and AB bisect each other). Similarly, Example 6. (2003 IMO, Problem 4) Let
2 W +W 2 W +W
the centers of S and S are A + C and B Z −W Z = 2 3 −W 2 3 . ABCD be a cyclic quadrilateral. Let P, Q
AC BC 1 2 1 2 and R be the feet of the perpendiculars
+ C respectively. We need to show there from D to the lines BC, CA and AB
is a point Z such that Z is on all three new Using WW =1, we may see that the right
1 1 respectively. Show that PQ = QR if and
circles, i.e. side is the same as only if the bisectors of ∠ABC and
|Z − (A + B)| = |Z − (A + C)| W +W +W W +W +W ∠ADC meet on AC.
1 2 3 −W2 1 2 3 .
= |Z − (B + C)| = 1. 2 1 2 Solution. (Due to SIU Tsz Hang, 2003
We easily see that the orthocenter of From this we see that N = (W + W + Hong Kong IMO team member) Without
triangle ABC, namely Z = H = A + B + C, 1 2
W)/2 satisfies the equation of the line and loss of generality, assume A, B, C, D lies
satisfies these equations. Therefore, the 3
so N is on the line. Since the expression on the unit circle and the perpendicular
three new circles intersect at the for N is symmetric with respect to W , W ,
1 2 bisector of AC is the real axis. Let A =
orthocenter of triangle ABC. W, we can conclude that N will also lie on
3 cosθ + isin θ, then C = A = cosθ −isinθ
Example 3. A point A is taken inside a the other two lines. Therefore, the lines so that AC = 1 and A + C = 2cosθ. Since
circle. For every chord of the circle concur at N, the center of the nine point the bisectors of ∠ABC and ∠ADC pass
circle of triangle W W W .
passing through A, consider the 1 2 3 through the midpoints of the major and
intersection point of the two tangents at Example 5. (Simson Line Theorem) Let minor arc AC, we may assume the
W be on the circumcircle of triangle ∠ABC and ∠ADC pass
the endpoints of the chord. Find the locus Z Z Z and P, Q, R be the feet of the bisectors of
of these intersection points. 1 2 3 through 1 and −1 respectively. Let AC
perpendiculars from W to Z Z , Z Z , Z Z
3 1 1 2 2 3 intersect the bisector of ∠ABC at Z, then
Solution. Without loss of generality we respectively. Prove that P, Q, R are Z satisfies Z + ACZ = A+C , (which is
may assume the circle is the unit circle collinear. (This line is called the Simson Z + Z = 2cosθ ), and .
line of triangle Z Z Z from W.) Z + BZ = B+1
and A is on the real axis. Let WX be a 1 2 3 Solving for Z, we get
chord passing through A with W and X on Solution. Without loss of generality, we 2Bcosθ −B−1
the circle. The intersection point Z of the may assume the circumcircle of triangle Z = .
tangents at W and X satisfies Z Z Z is the unit circle. B − 1
2 2 1 2 3
Z +W Z =2W and Z + X Z =2X . Then Z = Z = Z = W =1. Now P is Similarly, the intersection point Z′ of AC
Solving these equations together for Z, we 1 2 3 with the bisector of ∠ADC is
on the line Z Z and the line through W
find . 3 1
Z = 2/(W + X) perpendicular to Z Z . So P satisfies the 2Dcosθ +D−1
3 1 Z'= .
Since A is on the chord WX, the real equations P+ Z Z P = Z +Z and P – D+1
1 3 1 3
number A satisfies the equation for line Z Z P =W −Z Z W . Solving these
1 3 1 3 Next, R is on the line AB and the line
WX, i.e. A + WXA = W + X. Using together for P, we get through D perpendicular to AB. So
WW = 1 = XX , we see that Z + Z +W - Z Z W R+ABR=A+B andR−ABR=D−ABD .
P = 1 3 1 3 . Solving for R, we find
ReZ = 1 + 1 =WX+1= 1 . 2
W +X W+X W+X A Similarly, R = A+ B+ D− ABD .
So the locus lies on the vertical line Z + Z +W - Z Z W 2
through 1/A. Q = 1 2 1 2 Similarly,
Conversely, for any point Z on this line, 2
draw the two tangents from Z to the unit and P = B +C + D−BCD
circle and let them touch the unit circle at Z + Z +W - Z Z W 2
the point W and X. Then the above R = 2 3 2 3 . and
equations are satisfied by reversing the 2
argument. In particular, A + WXA = W + To show P, Q, R are collinear, it suffices to Q = C + A+ D−CAD .
X and so A is on the chord WX. Therefore, check that 2
the locus is the line perpendicular to OA P-R = P-R . (continued on page 4)
at a distance 1/OA from O. Q-R Q-R
Mathematical Excalibur, Vol. 9, No. 1, Jan 04- Apr 04 Page 3
Problem Corner Prove that there was a moment when the Problem 192. Inside a triangle ABC,
difference between the distances he had there is a point P satisfies ∠PAB = ∠
We welcome readers to submit their covered moving to the east and moving to PBC = ∠PCA = φ. If the angles of the
solutions to the problems posed below the west was at least half of the length of triangle are denoted by α, β and γ,
for publication consideration. The the equator. prove that
solutions should be preceded by the 1 1 1 1
solver’s name, home (or email) address ***************** sin2ϕ = sin2α + sin2 β + sin2γ .
and school affiliation. Please send Solutions
submissions to Dr. Kin Y. Li, **************** Solution. LEE Tsun Man Clement (St.
Department of Mathematics, The Hong Paul’s College), POON Ming Fung (STFA
Kong University of Science & Due to an editorial mistake in the last Leung Kau Kui College, Form 6), SIU Ho
Technology, Clear Water Bay, Kowloon, issue, solutions to problems 186, 187, 188 Chung (Queen’s College, Form 5) and
by POON Ming Fung (STFA Leung Kau Yufei ZHAO (Don Mills Collegiate
Hong Kong. The deadline for Institute, Tornoto, Canada, Grade 10).
May 25, 2004. Kui College, Form 6) were overlooked
submitting solutions is and his name was not listed among the Let AP meet BC at X. Since ∠XBP = ∠
solvers. We express our apology to him BAX and ∠BXP = ∠AXB, triangles
Problem 196. (Due to John and point out that his clever solution to XPB and XBA are similar. Then XB/XP =
PANAGEAS, High School “Kaisari”, problem 188 is printed in example 1 of the XA/XB. Using the sine law and the last
Athens, Greece) Let x ,x ,...,x be
1 2 n Geometry via Complex Numbers”
positive real numbers with sum equal article “ equation, we have
to 1. Prove that for every positive in this issue. sin2ϕ sin2 ∠XAB XB2
= =
integer m, Problem 191. Solve the equation sin2 β sin2 ∠XBA XA2
n≤nm(xm +xm +...+ xm). 3 XP⋅XA XP
1 2 n x −3x = x+2. = =
XA2 XA
Solution. Helder Oliveira de CASTRO
Problem 197. In a rectangular box, the (ITA-Aeronautic Institue of Technology, Using [ ] to denote area, we have
length of the three edges starting at the Sao Paulo, Brazil) and Yufei ZHAO (Don XP [XBP] [XCP] [BPC]
same vertex are prime numbers. It is Mills Collegeiate Institute, Toronoto, XA = [XBA] = [XCA] = [ABC]
Canada, Grade 10).
also given that the surface area of the
box is a power of a prime. Prove that If x < -2, then the right side of the equation Combining the last two equations, we
exactly one of the edge lengths is a is not defined. If x > 2, then have sin2ϕ/sin2 β =[BPC]/[ABC]. By
k 3 similar arguments, we have
prime number of the form 2 - 1. 3 x +3x(x+2)(x−2)
x −3x= 4 sin2ϕ sin2ϕ sin2ϕ
Problem 198. In a triangle ABC, AC = + +
BC. Given is a point P on side AB such x3 sin2α sin2φ sin2γ
that ∠ACP = 30∘. In addition, point > 4 > x+2. [APB] [BPC] [CPA]
Q outside the triangle satisfies ∠CPQ =[ABC]+[ABC]+[ABC]
= ∠CPA + ∠APQ = 78∘. Given that So the solution(s), if any, must be in [-2, 2]. [ABC]
all angles of triangles ABC and QPB, Write x = 2 cos a, where 0 ≤ a ≤π . The = =1
measured in degrees, are integers, equation becomes [ABC]
determine the angles of these two 8cos3 a −6cosa = 2cosa+2. The result follows.
triangles. Using the triple angle formula on the left Other commended solvers: CHENG Tsz
side and the half angle formula on the right Chung (La Salle College, Form 5), LEE
+
Problem 199. Let R denote the side, we get Man Fui (CUHK, Year 1) and Achilleas
positive real numbers. Suppose a P. PORFYRIADIS (American College
f : R+ → R+ is a strictly decreasing 2cos3a = 2cos (≥ 0). of Thessaloniki “Anatolia”, Thessaloniki,
function such that for all x, y∈R+ , 2 Greece).
f (x + y) + f (f (x) + f (y)) Then 3a ± (a/2) = 2nπ for some integer n. Comments: Professor Murray
= f (f (x + f (y)) + f (y + f (x))). Since 3a ± (a/2) ∈ [-π/2, 7π /2], we get KLAMKIN (University of Alberta,
Prove that f (f (x)) = x for every x > 0. Edmonton, Canada) informed us that the
n = 0 or 1. We easily checked that a = 0, 2 2 2 2
(Source: 1997 Iranian Math result csc φ = csc α + csc β + csc γ in the
4π /5, 4π/7 yield the only solutions x = 2, problem is a known relation for the
Olympiad) 2cos(4π/5), 2cos(4π/7). Brocard angle φ of a triangle. Also
known is cot φ = cot α + cot β + cot γ. He
Problem 200. Aladdin walked all over Other commended solvers: CHUNG Ho mentioned these relations and others are
the equator in such a way that each Yin (STFA Leung Kau Kui College, Form
moment he either was moving to the 7), LEE Man Fui (CUHK, Year 1), given in R.A. Johnson, Advanced
LING Shu Dung, POON Ming Fung Euclidean Geometry, Dover, N.Y., 1960,
west or was moving to the east or (STFA Leung Kau Kui College, Form 6), pp. 266-267. (For the convenience of
applied some magic trick to get to the SINN Ming Chun (STFA Leung Kau Kui interested readers, the Chinese translation
opposite point of the Earth. We know College, Form 4), SIU Ho Chung
that he travelled a total distance less (Queen’s College, Form 5), TONG Yiu of this book can be found in many
Wai (Queen Elizabeth School), YAU Chi bookstore.–Ed) LEE Man Fui and
than half of the length of the equator Keung (CNC Memorial College, Form 7) Achilleas PORFYRIADIS gave a proof
altogether during his westward moves. and YIM Wing Yin (South Tuen Mun of the cotangent relation and use it to
Government Secondary School, Form 4).
Mathematical Excalibur, Vol. 9, No. 1, Jan 04- Apr 04 Page 4
derive the cosecant relation, which is the (Kiangsu-Chekiang College Shatin, Teacher), There are n
C pigeonholes. For each
equation in the problem, by trigonometric Helder Oliveira de CASTRO 3
manipulations. (ITA-Aeronautic Institute of Technology, Sao segment joining a pair of endpoints, that
Paulo, Brazil), LEE Tsun Man Clement (St. segment will be in n − 2 pigeonholes. So
Problem 193. Is there any perfect square, Paul’s College), LING Shu Dung, POON if x(n − 2) ≥ 2Cn +1, that is
which has the same number of positive Ming Fung (STFA Leung Kau Kui College, 3
Form 6), SIU Ho Chung (Queen’s College, n
divisors of the form 3k + 1 as of the form Form 5), YEUNG Wai Kit (STFA Leung Kau 2C +1 n(n−1)(n−2)+3
Kui College), Yufei ZHAO (Don Mills x ≥ 3 = ,
3k + 2? Give a proof of your answer. Collegiate Institute, Toronto, Canada, Grade 10) n−2 3(n−2)
Solution 1. K.C. CHOW (Kiangsu-Chekiang and the proposer. then by the pigeonhole principle, there is
College Shatin, Teacher), LEE Tsun Man If S, N, T are collinear, then triangles SO N at least one triangle having these line
Clement (St. Paul’s College), SIU Ho Chung 1
(Queen’s College, Form 5) and Yufei ZHAO and SOT are isosceles and share the segments as edges.
(Don Mills Collegiate Institute, Toronto, common angle OST, which imply they are If f (n) = (n(n − 1)(n − 2) + 3) / (3(n − 2))
Canada, Grade 10).
∠SON = ∠SOT and so
similar. Thus 1 is an integer, then 3(n − 2) f (n) = n(n −
2 a
No. For a perfect square m , let m = 3 b lines O N and OT are parallel. Similarly,
2 2a 2 1 1)(n − 2) + 3 implies 3 is divisible by n −
with b not divisible by 3. Then m = 3 b . lines O2N and OS are parallel. Hence, 2. Since n > 3, we must have n = 5. Then
Observe that divisors of the form 3k + 1 OONO is a parallelogram and OO =
2 2 1 2 2 f (5) = 7. For the five vertices A, B, C, D,
or 3k + 2 for m and for b consist of the ON = O S, OO = O N = O T. Therefore,
same numbers because they cannot have 1 1 1 2 2 E of a regular pentagon, if we connected
SO /OO = OO /TO . Conversely, if
2 1 1 2 2 the six segments BC, CD, DE, EA, AC,
any factor of 3. Since b has an odd SO /OO = OO /TO , then using OO = OS
number of divisors and they can only be 1 1 2 2 1 BE, then there is no triangle. So a
− O S and OO = OT − O T, we get
of the form 3k + 1 or 3k + 2, so the 1 2 2 minimum of f (5) = 7 segments is needed
OS OT−OT to get a triangle.
number of divisors of the form 3k + 1 1 = 2 ,
OS−OS O T
cannot be the same as the number of 1 2 K. C. CHOW
Other commended solvers:
divisors of the form 3k + 2. Therefore, which reduces to O S + O T = OS. Then (Kiangsu-Chekiang College Shatin,
1 2 Teacher) and
the same is true for m2. OO = OS − OS = O T = O N and OO = POON Ming Fung (STFA
1 1 2 2 2 Leung Kau Kui College, Form 6).
Solution 2. Helder Oliveira de CASTRO OT − O T = O S = O N. Hence OO NO is
(ITA-Aeronautic Institute of Technology, Sao 2 1 1 1 2
Paulo, Brazil), again a parallelogram. Then
LEE Man Fui (CUHK, Year
1), LING Shu Dung, POON Ming Fung ∠O NS +∠O NO +∠O NT
(STFA Leung Kau Kui College, Form 6), 1 1 2 2 Olympiad Corner
=∠OSN+∠O NO +∠O TN
Achilleas P. PORFYRIADIS (American 1 1 2 2 (continued from page 1)
College of Thessaloniki “Anatolia”, = 1∠OO N +∠O NO +1∠OO N
Thessaloniki, Greece), Alan T.W. WONG 2 1 1 2 2 2
(Markham, Ontario, Canada) and YIM Wing =180°. Problem 4. Find, with reasons, all
Yin (South Tuen Mun Government Secondary integers a, b, and c such that
School, Form 4). Therefore, S, N, T are collinear. 1 3
(a + b) (b + c) (c + a) + (a + b + c)
No. For a perfect square, its prime Other commended solver: TONG Yiu 2
factorization is of the form Wai (Queen Elizabeth School). = 1 – abc.
2e 2e 2e
2 13 25 3 ···. Let x, y, z be the number Problem 195. (Due to Fei Zhenpeng,
of divisors of the form 3k, 3k + 1, 3k + 2 Yongfeng High School, Yancheng City,
for this perfect square respectively. Then Jiangsu Province, China) Given n (n > 3) Geometry via Complex Numbers
x + y + z = (2e + 1) (2e + 1) (2e + 1) ··· is points on a plane, no three of them are (continued from page 2)
1 2 3 collinear, x pairs of these points are
odd. Now divisor of the form 3k has at connected by line segments. Prove that if
least one factor 3, so x = (2e + 1) (2e )
1 2 By Simson’s theorem, P, Q, R are
(2e + 1) ··· is even. Then y + z is odd. n(n −1)(n − 2) + 3
3 x ≥ , collinear. So PQ = QR if and only if Q =
Therefore y cannot equal z. 3(n − 2) (P + R)/2. In terms of A, B, C, D, this may
then there is at least one triangle having be simplified to
Other commended solvers: CHENG Tsz
Chung (La Salle College, Form 5) and YEUNG these line segments as edges. Find all C+A−2B=(2CA−AB−BC)D.
Wai Kit (STFA Leung Kau Kui College). possible values of integers n > 3 such that
n(n − 1)( n − 2) + 3 In terms of B, D, θ, this is equivalent to
Problem 194. (Due to Achilleas Pavlos is an integer and (2cosθ − 2B)D = 2 − 2Bcos θ. This is
PORFYRIADIS, American College of 3(n − 2) easily checked to be the same as
Thessaloniki “Anatolia”, Thessaloniki, the minimum number of line segments
Greece) A circle with center O is guaranteeing a triangle in the above 2cosθ − B−1 = 2Dcosθ + D−1,
internally tangent to two circles inside it, situation is this integer. B−1 D+1
with centers O and O , at points S and T
1 2 Solution. SIU Ho Chung (Queen’s i.e. Z = Z'.
respectively. Suppose the two circles College, Form 5), Yufei ZHAO (Don
inside intersect at points M, N with N Mills Collegiate Institute, Toronto, Comments: The official solution by
closer to ST. Show that S, N, T are Canada, Grade 10) and the proposer.
pure geometry is shorter, but it takes a
collinear if and only if SO /OO = For every three distinct points A, B, C,
1 1 form a pigeonhole containing the three fair amount of time and cleverness to
OO/TO .
2 2 segments AB, BC, CA. (Each segment discover. Using complex numbers as
Solution. CHENG Tsz Chung (La Salle may be in more than one pigeonholes.) above reduces the problem to straight
College, Form 5), K. C. CHOW computations.
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