Sección B1 – Power
20 Prueba Individual
FIRST SELECTION TEST
Problem 1. Call a positive integer balanced if the number of its distinct prime factors is equal to the number of its digits in the decimal representation; for example, the number 385 = 5 · 7 · 11 is balanced, while 275 = 52· 11 is not. Prove that there exist only a finite number of balanced numbers.
Solution. Let p1 = 2, p2 = 3, p3 = 5, . . . be the sequence of primes. Any balanced number a with n digits satisfies a ≥ p1p2· · · pn. Since p1p2· · · p11 = 2· 3 · 5 · · · 29 · 31 => 1011and pk > 10, for any k > 11, it follows that there are no balanced numbers having more than 10 digits.
Problem 2. Consider a convex pentagon A0A1A2A3A4so that the rays (AiAi+1
and (Ai+3Ai+2 meet at Bi+4, for each i = 0, 1, 2, 3, 4 – indices are considered modulo 5. Show that
4 i=0
AiBi+3=
4 i=0
AiBi+2.
Solution. Let Ci be the projection of Bi on the line Ai+2Ai+3, i = 0, 1, 2, 3, 4. Notice that the right–angled triangles (possibly degenerate) AiBi+3Ci+3 and AiBi+2Ci+2 are similar, for angles Ai are vertical. Hence AiBi+3
AiBi+2
= Bi+3Ci+3
Bi+2Ci+2
, implying
4 i=0
AiBi+3
AiBi+2
=
4 i=0
Bi+3Ci+3
Bi+2Ci+2
= 1, as claimed.
47
Remark. The claim holds for any n-gon with n ≥ 5 sides, within the given condi-tions.
Problem 3. Let n be a positive integer and let x1, x2, . . . , xn and y1, y2, . . . , yn
be real numbers. Prove that there exists a number i, i = 1, 2, . . . , n, such that
n j=1
|xi− xj| ≤
n j=1
|xi− yj|.
Solution. Without the loss of generality, suppose x1 ≤ x2≤ . . . ≤ xn. For each k = 1, 2, . . . , n we have |x1− xk| + |xn− xk| = |x1− xn| ≤ |x1− yk| + |xn− yk|, hence
n k=1
|x1− xk| +
n k=1
|xn− xk| ≤
n k=1
|x1− yk| +
n k=1
|xn− yk|.
The claim holds for i = 1 or i = n.
Problem 4. Let k and n be integer numbers with 2 ≤ k ≤ n − 1. Consider a set Aof n real numbers such that the sum of any k distinct elements of A is a rational number. Prove that all elements of the set A are rational numbers.
Solution. The difference of any two elements from A is a rational number. To show this, let x = y ∈ A and choose other k − 1 elements of A – the choice can be made, for k − 1 ≤ n − 2. Denote s the sum of the k − 1 elements and apply the hypothesis to infer that x + s and y + s are both rational numbers. Subtracting w get x− y ∈ Q, as claimed.
Now let α ∈ A. Rewrite the elements of A as α + qi, qi ∈ Q. The sum of k distinct elements is equal to kα + qi∈ Q, hence α ∈ Q, as needed.
Problem 5. Consider n persons, each of them speaking at most 3 languages. From any 3 persons there are at least two which speak a common language.
i) For n ≤ 8, exhibit an example in which no language is spoken by more than two persons.
ii) For n ≥ 9, prove that there exists a language which is spoken by at least three persons.
Solution. i) Split the 8 persons in two groups of 4. Set any pair of persons in each group to speak a different language for a total of 6 + 6 = 12 languages, each spoken by 2 persons, each person speaking 3 languages.
For n ≤ 7, just remove 8 − n persons.
ii) Assume by contrary that each language is spoken by at most two persons. Then each person A can speak with at most three others, for otherwise, by pigeon–hole prin-ciple, there exists a language spoken by other two persons besides A, a contradiction.
Let B, C, D the persons which whom A can speak. Likewise, E can speak with (at most) three others, namely F, G, H. There is left at least another person, say Z, and in the group A, E, Z no language is spoken in common, a contradiction.
SECOND SELECTION TEST
Problem 6. Determine a) the smallest number b) the biggest number
n≥ 3 of non-negative integers x1, x2, ... , xn, having the sum 2011 and satisfying:
x1≤ | x2− x3| , x2≤ | x3− x4| , ... , xn−2≤ | xn−1− xn| , xn−1≤ | xn− x1| and xn ≤ | x1− x2| .
Solution. Let x1, x2, ... , xnnon-negative integers satisfying the conditions from the statement. Let us imagine them as being arranged on a circle, in this order. We denote with M the biggest of the numbers x1, ... , xn. Without loss of generality, we may suppose that x1= M. Obviously M = 0. From x1≤ | x2− x3| and the choice of x1it follows that {x2, x3} = {M, 0}. Therefore any M is followed by an M and a 0 on the circle, either in the succession M − M − 0 or the succession M − 0 − M.
By induction after the last M known on the circle, we find that on the circle there are only the numbers 0 and M. Moreover
• there are no the neighboring zeros (or else the number preceding them would also be a 0 and, inductively, all the numbers on the circle would be zeros, which contradicts the fact that their sum is 2011)
• there are no three consecutive M -s. Since 2011 is prime and x1+x2+...+xn= kM, where k is the number of the M -s, we find that M = 1 and k = 2011.
a) The smallest value of n is obtained when the number of zeros is the smallest possible. Out of three consecutive numbers on the circle, at least one is 0, therefore we have at least 1006 zeros, which means at least 3017 numbers. A possible configuration fulfilling the conditions of the statement is x3k = 0, k = 1, 2, ..., 1005, x3017 = 0, and the rest of the xj= 1.
b) The biggest n is obtained when the number of zeros is the largest possible.
Since there can be no neighboring zeros, we have at most 2011 zeros. This happens if we place alternatively 0 and 1 on the circle, configuration that satisfies the conditions from the statement. Thus, the biggest value of n is 4022.
Problem 7. We consider an n × n (n ∈ N, n ≥ 2) square divided into n2 unit squares. Determine all the values of k ∈ N for which we can write a real number in each of the unit squares such that the sum of the n2 numbers is a positive number, while the sum of the numbers from the unit squares of any k × k square is a negative number.
Solution. We will prove that the desired numbers k are those that are not factors of n.
If k | n that we can tile the n × n square with k × k squares and the total sum should simultaneously be positive and negative.
If k | n then n = kq + r, where 0 < r < k. We fill the unit squares with a (to be chosen conveniently later on) in the positions (ik, jk) with i, j = 1, ..., q and with 1 in the other positions. Every k × k square contains exactly one unit square of the form (ik, jk), therefore the sum in every k × k square is a + k2− 1. The total sum is q2a + n2− q2. We will choose a arbitrarily from the non-empty interval
1−nq22, 1− k2 .
Remark: For the case k | n there are many other ways of choosing the numbers from the unit squares. Another choice is to fill all the unit squares of the columns jk, j = 1, ..., q, with a convenient a and the other ones with 1.
Problem 8. a) Prove that if the sum of the non-zero digits a1, a2, ... , an is a multiple of 27, then it is possible to permute these digits in order to obtain an n-digit number that is a multiple of 27.
b) Prove that if the non-zero digits a1, a2, ... , anhave the property that every n-digit number obtained by permuting these n-digits is a multiple of 27, then the sum of these digits is a multiple of 27.
Andrei Eckstein Solution. a) Obviously n ≥ 3. If n = 3 then a1 = a2 = a3 = 9and 999... 27.
Suppose n ≥ 4. Having the sum of its digits a multiple of 9, each of the numbers formed with the digits a1, a2, ... , an is a multiple of 9, therefore division with 27 will yield one of the remainders 0, 9 or 18.
• If all the digits give the same remainder r when divided by 3: the sum of its digits a multiple of 9, therefore the numberN
3 is a multiple of 9, hence
give the same remainder at the division by 27.
• If not all of the digits a1, a2, ... , an have the same reminder when divided by 3, then we choose three of them such that their sum is not a multiple of 3. Let a, b, c be these three digits. Then the numbers x = ...abc, y = ...bca and z = ...cab give different reminders at the division by 27. Indeed, x − y = abc − bca = 108a − 81b − 9(a+b+c) = M27−9(a+b+c) = M27. Since the possible reminders at the division by 27 are only 0, 9 and 18, it follows that (exactly) one of the numbers x, y, z is a multiple of 27.
b) The differece between two numbers obtained one from the other by swapping two neighboring digits a and b is 9(a − b) · 10kand needs to be a multiple of 27. We obtain that each of the digits a1, ..., anhas to give the same remainder when divided by 3. From here one continues as bya) in order to show that, in this case, the numbers a1a2...anand a1+ a2+ ... + angive the same remainder when divided by 27.
Problem 9. The measure of the angle Aof the acute triangle ABC is 60◦, and HI = HB, where I and H are the incenter and the orthocenter of the triangle ABC.
Find the measure of the angle B.
80◦.It is easy to see that the case m(∠B) ≤ 60◦is not possible.
THIRD SELECTION TEST
Problem 10. It is said that a positive integer n > 1 has the property (p) if in its prime factorization
n = pα11· ... · pαjj
at least one of the prime factors p1, ... , pjhas the exponent equal to 2.
a) Find the largest number k for which there exist k consecutive positive integers that do not have the property (p).
b) Prove that there is an infinite number of positive integers n such that n, n + 1 and n + 2 have the property (p).
Dorel Mihet¸
Solution. a) Among any 8 consecutive integers there exists one of the form 8j +4.
This number has the property (p) because the factor 2 from its prime factorization has the exponent 2. Therefore there can be at most 7 consecutive positive integers that do not have the property (p). Since none of the numbers 29, 30, 31, 32, 33, 34, 35 has the property (p), the largest k is k = 7.
b) Since the numbers 98 = 2·72, 99 = 32·11, and 100 = 22·52have the property (p), so will the numbers 98 + (7 · 3 · 2)3· k, 99 + (7 · 3 · 2)3· k, and 100 + (7 · 3 · 2)3· k.
Alternative solution. By the Chinese remainder theorem, there exist an infinite num-ber of solutions for the system of simultaneous congruences: n ≡ 4 (mod 8), n ≡ 8 (mod 27), n ≡ 23 (mod 125). Then n, n + 1, and n + 2 all have the property (p) because the factor 2, 3, and 5, respectively, has the exponent 2 in their prime factorizations.
Problem 11. Find all the finite sets A of real positive numbers having at least two elements, with the property that a2+ b2∈ A for every a, b ∈ A with a = b.
Solution. If a1 < a2 < ... < an are the elements of A, then a21+ a22, a21+ a23, a21+ a2n, a22+ a2n,..., a2n−1+ a2nbelong to A, and a21+ a22< a21+ a23< ... < a21+ a2n<
a22+ a2n < ... < a2n−1+ a2n, which implies that 2n − 3 ≤ n; hence A has at most 3 elements.
If n = 3 and A = {a, b, c} with a < b < c, then a2+ b2 = a, b2+ c2 =
2,which contradicts a < b.
Hence n = 2, and A = {a, b} with a2+ b2 = a, i.e., for a ∈ (0, 1), we get
Problem 12. Let ABC be a triangle, Iathe center of the excircle at side BC, and Mits reflection across BC. Prove that AM is parallel to the Euler line of the triangle BCIa.
Solution. Let I be the incenter of ABC, Hathe orthocenter of IaBC, and Oathe midpoint of the segment [IIa]. Then BOa = IOa = IaOa = COa, therefore Oa is the circumcenter of triangle IaBC. Moreover, IB CHa(both are perpendicular on BIa) and, similarly, IC BHa, hence BICHais a parallelogram.
Let P denote the projection of Iaonto BC and T be the midpoint of [AIa]. If r, ra
are the inradius and the radius of the excircle at side BC, respectively, then we have:
IA parallel to the Euler line of the triangle IaBC.
Problem 13. Let m be a positive integer. Determine the smallest positive integer nfor which there exist real numbers x1, x2, . . . , xn∈ (−1, 1) such that |x1| + |x2| +
choice for n = m+1 is x1= ... = xm+1
2 =− m
m + 1, xm+1
2 +1= ... = xn= m m + 1. In conclusion, the smallest value of n for which the equation has solutions is
n = m + 1if m is odd and n = m + 2 if m is even.
FOURTH SELECTION TEST
Problem 14. For every positive integer n let τ(n) denote the number of its positive factors. Determine all n ∈ N that satisfy the equality τ(n) = n
3. Solution. If n ∈ N satisfies the condition τ(n) = n
3,then 3 | n. Put n = 3k, k ∈ N. If k is even then k
2 = n
6 is a factor of n. Even if all the positive numbers smaller than n
6 are factors of n and the numbers n 5, n
4, ..., n
1 are also factors of n, we haven
3 = τ (n)≤n
6 + 5. Hence n ≤ 30. Checking the numbers 6, 12, 18, 24 and 30 we find that 18 and 24 satisfy the desired condition.
In the case when k is odd we can proceed similarly, obtaining n = 9, or, alterna-tively, we can use the fact that a number having an odd number of positive factors is a perfect square. If n = m2, then n has at most m + 1 factors, hence m + 1 ≥ m2
3 . We obtain that m ≤ 3 and the conclusion.
In conclusion, the problem admits three solutions: 9, 18 and 24.
Alternative solution: Since 3 | n, it follows that n has a prime factorization of the form n = 3a· pα11 · ... · pαjj and the number of his positive factors is τ(n) = (a + 1)(α1+ 1)...(αj+ 1). The condition τ(n) = n
3 leads to 3a−1· pα11· ... · pαjj = (a + 1)(α1+ 1)...(αj+ 1). Since pαii ≥ 2αi ≥ αi+ 1, in order for n to satisfy the equation from the statement, it is necessary that a + 1 ≥ 3a−1, hence a = 1 or a = 2.
If in the prime factorization of n there is a prime pi> 3, then pαii> 4αi ≥ 2αi+2 and the equality τ(n) = n
3 can not take place. If a = 1 then n = 3 · 2mand the equality τ(n) = n
3 reduces to 2(m + 1) = 2m. We find m = 3 (for m ≥ 4 we have 2m> 2m + 2), hence n = 24.
Similarly, if a = 2 then n = 32· 2m and from 3(m + 1) = 3 · 2mwe obtain m∈ {0, 1}, i.e. n ∈ {9, 18}.
Problem 15. Let ABC be a triangle with circumcenter O. The points P and Q
are interior points of the sides CA and AB respectively. Let K, L and M be the midpoints of the segments BP , CQ and P Q, respectively, and let Γ be the circle passing through K, L and M. Suppose that the line P Q is tangent to the circle Γ.
Prove that OP = OQ.
Sergei Berlov, Russia, IMO 2009 Solution. From AB KM and AC LM we obtain that ∠KMQ ≡ ∠AQP and∠LMP ≡ ∠APQ. If PQ is tangent to the circumcircle of the triangle KLM then∠KLM ≡ ∠KMQ, hence ∠KLM ≡ ∠AQP. Then ∆APQ ∼ ∆MKL, hence AQ
M L = AP
M K.It follows thatAQ P C = AP
BQ,i.e. AQ · BQ = AP · P C. This means that P and Q have equal powers with respect to the circumcircle of the triangle ABC. As they are both situated inside the circle, they are at equal distance from the center of the circle, O.
Problem 16. a) Find the largest possible value of the number x1x2+ x2x3+ ... + xn−1xn,
if x1, x2, ... , xn(n ≥ 2) are non-negative integers and their sum is 2011.
b) Find the numbers x1, x2, ... , xnfor which the maximum value determined at a) is obtained.
Dorel Mihet¸
Solution. a) Let x1, x2, ... , xnbe non-negative integers satisfying the conditions from the statement. We put M = max
1≤i≤nxi. If xj = M then x1x2+ x2x3+ ... + xn−1xn≤ x1xj+ x2xj+ ... + xj−1xj+ xjxj+1+ xjxj+2+ ... + xjxn= xj(2011− xj) = M (2011− M) ≤ 1005 · 1006. Indeed, the last inequality comes to (M − 1005)(M− 1006) ≥ 0 which is true for any integer M. The largest possible value is 1005· 1006 because this value can be obtained by choosing, for example, x1= 1005, x2= 1006and xk = 0for k ≥ 3.
b) For n = 2 we have x1x2 = 1005· 1006 ⇔ x1(2011− x1) = 1005· 1006 ⇔ (x1− 1005)(x1− 1006) = 0 ⇔ (x1, x2)∈ {(1005, 1006), (1006, 1005)}.
For n = 3, x1x2+ x2x3 = 1005· 1006 ⇔ x2(x1+ x3) = 1005· 1006 comes, as above, to x2 = 1005, x1+ x3 = 1006or x2 = 1006, x1+ x3 = 1005. We obtain (x1, x2, x3) ∈ {(k, 1005, 1006 − k) | k = 0, 1, ..., 1006} ∪ {(k, 1006, 1005 − k) | k = 0, 1, ..., 1005}.
For n ≥ 4, we denote by j the smallest index for which xj > 0. Then, repla-cing xj by 0 and xj+2 by xj+2 + xj, increases the value of the sum by xjxj+3. Using this remark it is easy to see that if x1x2+ x2x3+ ... + xn−1xn = 1005· 1006, then at most three of the terms can be non-zero. We obtain (x1, ..., xn) ∈ {(0, ..., 0, k, 1005, 1006−k, 0, ..., 0) | k = 0, 1, ..., 1006}∪{(0, ..., 0, k, 1006, 1005−
k, 0, ..., 0) | k = 0, 1, ..., 1005}, where the group of the three non-zero components can be located anywhere.
Problem 17. Show that there is an infinite number of positive integers t such that none of the equations x2+ y6= t, x2+ y6= t + 1, x2− y6= t, x2− y6= t + 1 has solutions (x, y) ∈ Z × Z.
Dorel Mihet¸
Solution. If x is a positive integer, then either x12≡ 0 (mod 13) or x12≡ 1 (mod 13), hence x6is congruent with −1, 0 or 1 modulo 13. Therefore, if t is congruent with 6 (mod 13), then ±x6+ tis congruent with 5, 6 or 7 (mod 13), while ±x6+ t + 1 is congruent with 6, 7 or 8 (mod 13).
On the other hand, perfect squares are congruent with 0, 1, 4, 9, 3, 12 or 10 (mod 13). Therefore a perfect square can not be equal to a number of the form ±x6+ tor
±x6+ t + 1if t ≡ 6 (mod 13). In conclusion, all the numbers that are congruent with 6 modulo 13 have the required property.
Ias¸i, Romania, May 4th– May 8th, 2011
Problem 1. Let ABCD be a cyclic quadrilateral which is not a trapezoid and whose diagonals meet at E. The midpoints of AB and CD are F and G respectively and is the line through G parallel to AB. The feet of the perpendiculars from E onto and CD are H and K, respectively. Prove that the lines EF and HK are perpendicular.
United Kingdom Solution. The points E, K, H, G are on the circle of diameter GE, so the angles EHKand EGK are equal.
57
Also, from∠DCA = ∠DBA and CE/CD = BE/BA follows CE
CG = 2CE
CD =2BE
BA =BE
BF,
so the triangles CGE and BF E are similar. In particular, the angles EGC and BF E are equal, and therefore so are the angles EHK and BF E.
But the lines EH and BF are perpendicular and so, since EF and HK are ob-tained by rotations of these lines by the same (directed) angle, the lines EF and HK also perpendicular.
Problem 2. Given three real numbers x, y, z such that x + y + z = 0, show that x(x + 2)
2x2+ 1 +y(y + 2)
2y2+ 1 +z(z + 2) 2z2+ 1 ≥ 0.
When does equality hold?
Greece Solution. The inequality is clear if xyz = 0, in which case equality holds if and only if x = y = z = 0.
Henceforth assume xyz = 0 and rewrite the inequality as (2x + 1)2
2x2+ 1 +(2y + 1)2
2y2+ 1 +(2z + 1)2 2z2+ 1 ≥ 3.
Notice that (exactly) one of the products xy, yz, zx is positive, say yz > 0, to get (2y + 1)2
2y2+ 1 +(2z + 1)2
2z2+ 1 ≥2(y + z + 1)2
y2+ z2+ 1 (by Jensen)
= 2(x− 1)2
x2− 2yz + 1 (for x + y + z = 0)
≥2(x− 1)2
x2+ 1 . (for yz > 0) Here equality holds if and only if x = 1 and y = z = −1/2. Finally, since
(2x + 1)2
2x2+ 1 +2(x− 1)2
x2+ 1 − 3 = 2x2(x− 1)2
(2x2+ 1)(x2+ 1) ≥ 0, x ∈ R,
the conclusion follows. Clearly, equality holds if and only if x = 1, so y = z = −1/2.
Therefore, if xyz = 0, equality holds if and only if one of the numbers is 1, and the other two are −1/2.
Solution 2. Rewrite the inequality as in Solution 1, (2x + 1)2
2x2+ 1 +(2y + 1)2
2y2+ 1 +(2z + 1)2 2z2+ 1 ≥ 3,
and use the condition x + y + z = 0 to replace 3 in the right-hand member by (2x + 1)2
4
3(x2+ y2+ z2) + 1+ (2y + 1)2
4
3(x2+ y2+ z2) + 1+ (2z + 1)2
4
3(x2+ y2+ z2) + 1, and thereby deduce that it is sufficient to show that
(2x + 1)2
2x2+ 1 ≥ (2x + 1)2
4
3(x2+ y2+ z2) + 1 (∗) and the like. Unless x = −1/2, in which case equality holds trivially, use again the condition x + y + z = 0 to transform (∗) into the equivalent and obvious inequality (y−z)2≥ 0, which yields the equality case y = z. This proves the required inequality and shows that equality holds if and only if the variables all vanish simultaneously or two equal −1/2 and the third equals 1.
Problem 3. Let S be a finite set of positive integer numbers which has the fol-lowing property: if x is a member of S, then so are all positive divisors of x. Call a subset T of S good (respectively, bad ) if it is non-empty and, whenever x and y are members of T , and x < y, the ratio y/x is (respectively, is not) a power of a prime number; agree that a singleton (one-element) subset is both good and bad. Show that a maximal good subset of S has as many elements as a minimal partition of S into bad subsets.
Bulgaria Solution. Notice first that a bad subset of S contains at most one element from a good one, to deduce that a partition of S into bad subsets has at least as many members as a maximal good subset.
Notice further that the elements of a good subset of S must be among the terms a geometric sequence whose ratio is a prime: if x < y < z are elements of a good subset of S, then y = xpαand z = yqβ= xpαqβfor some primes p and q and some positive integers α and β, so p = q for z/x to be a power of a prime.
Next, let P = {2, 3, 5, 7, 11, · · · } denote the set of all primes, let m = max{exppx : x∈ S and p ∈ P },
where exppxis the exponent of the prime p in the canonical decomposition of x, and notice that a maximal good subset of S must be of the form {a, ap, · · · , apm} for some prime p and some positive integer a which is not divisible by p. Consequently, a maximal good subset of S has m + 1 elements, so a partition of S into bad subsets has at least m + 1 members.
Finally, notice by maximality of m that the sets Sk ={x : x ∈ S and
p∈P
exppx≡ k (mod m + 1)}, k = 0, 1, · · · , m,
form a partition of S into m + 1 bad subsets. The conclusion follows.
Problem 4. The opposite sides of a convex hexagon of unit area are pairwise parallel. The lines of support of three alternate sides meet in pairs to determine the vertices of a triangle. Similarly, the lines of support of the other three alternate sides meet in pairs to determine the vertices of another triangle. Show that the area of at least one of these two triangles is greater than or equal to 3/2.
Bulgaria Solution. Unless otherwise stated, throughout the proof indices take on values from 0 to 5 and are reduced modulo 6. Label the vertices of the hexagon in circular order, A0, A1, · · · , A5, and let the lines of support of the alternate sides AiAi+1and Ai+2Ai+3meet at Bi. To show that the area of at least one of the triangles B0B2B4, B1B3B5is greater than or equal to 3/2, it is sufficient to prove that the total area of the six triangles Ai+1BiAi+2is at least 1:
5 i=0
area Ai+1BiAi+2≥ 1.
To begin with, reflect each Bi through the midpoint of the segment Ai+1Ai+2
to get the points B i. We shall prove that the six triangles Ai+1Bi Ai+2 cover the hexagon. To this end, reflect A2i+1 through the midpoint of the segment A2iA2i+2
to get the points A 2i+1, i = 0, 1, 2. The hexagon splits into three parallelograms, A2iA2i+1A2i+2A 2i+1, i = 0, 1, 2, and a (possibly degenerate) triangle, A 1A 3A 5. Notice first that each parallelogram A2iA2i+1A2i+2A 2i+1 is covered by the pair of triangles (A2iB2i+5 A2i+1, A2i+1B2i A2i+2), i = 0, 1, 2. The proof is completed by showing that at least one of these pairs contains a triangle that covers the trian-gle A 1A 3A 5. To this end, it is sufficient to prove that A2iB 2i+5 ≥ A2iA 2i+5 and
A2j+2B2j ≥ A2j+2A 2j+3 for some indices i, j ∈ {0, 1, 2}. To establish the first inequality, notice that
A2iB2i+5 = A2i+1B2i+5, A2iA 2i+5= A2i+4A2i+5, i = 0, 1, 2, A1B5
A4A5
=A0B5
A5B3
and A3B1
A0A1
= A2A3
A0B5
, to get
2 i=0
A2iB2i+5 A2iA 2i+5 = 1.
Similarly,
2 j=0
A2j+2B2j A2j+2A 2j+3 = 1, whence the conclusion.
Solution 2. With reference to Solution 1, suppose, if possible, that the six tri-angles AiAi+1Bi+5 fail to cover the hexagon A0A1· · · A5. Then some point X of the hexagon falls outside each of those triangles. Up to an affine transformation, we
Solution 2. With reference to Solution 1, suppose, if possible, that the six tri-angles AiAi+1Bi+5 fail to cover the hexagon A0A1· · · A5. Then some point X of the hexagon falls outside each of those triangles. Up to an affine transformation, we