Volume 21, Number 1 October 2016 – March 2017
Olympiad Corner
Below are the problems of the Final Round of the 65th Czech and Slovak
Math Olympiad (April 4-5, 2016). Problem 1. Let p>3 be a prime. Find the number of ordered sextuples (a,b,c,d,e,f) of positive integers, whose sum is 3p, and all the fractions
b a f e a f e d f e d c e d c b d c b a , , , , are integers.
Problem 2. Let r and ra be the radii of the inscribed circle and excircle opposite A of the triangle ABC. Show that if r+ra=|BC|, then the triangle is right-angled.
Problem 3. Mathematics clubs are very popular in certain city. Any two of them have at least one common member. Prove that one can distribute rulers and compasses to the citizens in such a way that only one citizen get both (compass and ruler) and any club has to his disposal both, compass and ruler, from its members.
(continued on page 4)
Miscellaneous Problems
Kin Y. Li
There are many Math Olympiad problems. Some are standard problems in algebra or in geometry or in number theory or in combinatorics, where there are some techniques for solving them. Then, there are problems that are not so standard, which cross two or more categories. In math problem books, they go under the category of miscellaneous problems. Some of these may arise due to curiosity. Then one may need to combine different facts to explain them. Below are some such problems we hope the readers will enjoy.
Example 1 (1995 USA Math Olympiad). A calculator is broken so that the only keys that still work are the sin, cos, tan, sin−1,cos−1,tan−1 buttons. The display
initially shows 0. Given any positive rational numbers q, show that pressing some finite sequence of buttons will yield q. Assume that the calculator does real number calculation with infinite precision. All functions are in terms of radians.
Solution. We will show that all numbers of the form m /n , where m, n are positive integers, can be displayed by doing induction on k=m+n. (Since r/s =
, / 2 2 s
r these include all positive
rational numbers.)
For k=2, pressing cos will display 1. Suppose the statement is true for integer less than k. Observe that if x is displayed, then letting θ=tan−1x, we see
. 1 2 tan 2 ) (sin cos1 x and x
So we can display 1/x=tan(cos−1(sin x)).
Therefore, to display m /n with k=m+n, we may assume m<n. By the induction step, since (n-m)+m = n < k,
m m
n )/
( can be displayed. Then using , / cos / ) ( tan1 nm m and m n
we can display m /n. This completes the induction.
Example 2 (1986 Brazilian Math Olympiad). A ball moves endlessly on a circular billiard table. When it hits the edge it is reflected. Show that if it passes through a point on the table three times, then it passes through it infinitely many times.
Solution. Suppose AB and BC are two successive chords of the ball’s path. By the reflection law, ∠ABO = ∠OBC. Now ΔOAB and ΔOBC are isosceles. So ∠AOB = ∠BOC. Hence, AB =BC. Then every chord of the path has the same length d.
We now claim that through any given point P inside the circle there are at most two chords with length d. Let AB and CD be a chord containing P, with AP=a and CP=b. The power of P with respect to the circle is PA·PB=PC·PD, which is a(d-a)=b(d-b). Hence, a=b or a+b=d. This means that P always divides the chord containing it in two segments of fixed lengths a and d-a. Now if three chords passes through P, the circle with center P and radius a would cut the circle of the billiard table three times, a contradiction.
Thus if the path passes through P more than twice, then on two occasions it must be moving along the same chord AB. That implies ∠AOB is a rational multiple of 2π and hence the path will traverse AB repeatedly.
Example 3. Is there a way to pack 250 1×1×4 bricks into a 10×10×10 box? Solution. Assign coordinate (x,y,z) to each of the cells, where x,y,z= 0,1,…,9. Let the cell (x,y,z) be given color x+y+z (mod 4). Note each 1×1×4 brick contain all 4 colors exactly once. If the packing is possible, then there are exactly 250 cells of each color. However, a direct counting shows there are 251 cells of color 0, a contradiction. So such packing is impossible.
(continued on page 2)
Editors: 高 子 眉 (KO Tsz-Mei)
梁 達 榮 (LEUNG Tat-Wing)
李 健 賢 (LI Kin-Yin), Dept. of Math., HKUST
吳 鏡 波 (NG Keng-Po Roger), ITC, HKPU Artist: 楊 秀 英 (YEUNG Sau-Ying Camille), MFA, CU Acknowledgment: Thanks to Elina Chiu, Math. Dept., HKUST for general assistance.
On-line: http://www.math.ust.hk/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 April 15, 2017.
For individual subscription for the next five issues for the 17-18 academic year, send us five stamped self-addressed envelopes. Send all correspondence to:
Dr. Kin-Yin LI, Math Dept., Hong Kong Univ. of Science and Technology, Clear Water Bay, Kowloon, Hong Kong
Fax: (852) 2358 1643 Email: [email protected]
© Department of Mathematics, The Hong Kong University of Science and Technology
Mathematical Excalibur, Vol. 21, No. 1, Oct. 16 – Mar. 17 Page 2 Example 4 (2013 Singapore Math
Olympiad). Six musicians gathered at a chamber music festival. At each scheduled concert some of the musicians played while the others listened as members of the audience. What is the least number of such concerts which would need to be scheduled so that every two musicians each must play for the other in some concert?
Solution. Let the musicians be A,B,C, D,E,F. We first show that four concerts are sufficient. The four concerts with the performing musicians: {A,B,C}, {A,D,E}, {B,D,F} and {C,E,F} satisfy the requirement. We shall now prove that three concerts are not sufficient. Suppose there are only three concerts. Since everyone must perform at least once, there is a concert where two of the musicians, say A, B, played. But they must also played for each other. Thus we have A played and B listened in the second concert and vice versa in the third. Now C,D,E,F must all perform in the second and third concerts since these are the only times when A and B are in the audience. It is not possible for them to perform for each other in the first concert. Thus the minimum is 4.
Example 5 (1999 Brazilian Math Olympiad). Prove that there is at least one nonzero digit between the 1,000,000th and the 3,000,000th
decimal digits of 2.
Solution. Let us suppose that all digits between the 1,000,000th and the
3,000,000th decimal digits of 2 are zeros. Then , 10 2 1,000n,000 (*) where n is a positive integer and ε > 0 satisfy . ) 10 ( 10 2 1,000,000 3101.000.000 and n
By squaring (*), we can get
. 10 10 2 10 2 000 . 000 . 2 2 000 , 000 , 1 2 000 , 000 , 2 n n
However, the left side is a positive integer and the right side is less than 1, which is a contradiction.
Example 6 (1995 Russian Math Olympiad). Is it possible to fill in the
cells of a 9×9 table with positive integers ranging from 1 to 81 in such a way that the sum of the elements of every 3×3 square is the same?
Solution. Place 0,1,2,3,4,5,6,7,8 on the first, fourth and seventh rows. Place 3,4,5,6,7,8,0,1,2 on the second, fifth and eigth rows. Place 6,7,8,0,1,2,3,4,5 on the third, sixth and ninth rows. Then every 3×3 square contains 0 to 8. Consider this table and its 90° rotation. For each cell, fill it with the number 9a+b+1, where a is the number in the cell originally and b is the number in the cell after the table is rotated by 90°. By inspection, 1 to 81 appears exactly once and every 3×3 square has sum 9×36+36+9=369.
Example 7. Can the positive integers be partitioned into infinitely many subsets such that each subset is obtained from any other subset by adding the same integer to each element of the other subset?
Solution. Yes. Let A be the set of all positive integers whose odd digit positions (from the right) are zeros. Let B be the set of all positive integers whose even digit positions (from the right) are zeros. Then A and B are infinite set and the set of all positive integers is the union of a+B={a+b: b∈B} as a range over the element of A. (For example, 12345 = 2040+10305 ∈ 2040+B.)
Example 8 (2015 IMO Shortlisted Problem proposed by Estonia). In Lineland there are n≥1 towns, arranged along a road running from left to right. Each town has a left bulldozer (put to the left of the town and facing left) and a right bulldozer (put to the right of the town and facing right). The sizes of the 2n bulldozers are distinct. Every time when a right and left bulldozer confront each other, the larger bulldozer pushes the smaller one off the road. On the other hand, the bulldozers are quite unprotected at their rears; so if a bulldozers reaches the rear-end of another one, the first one pushes the second one off the road, regardless of their sizes.
Let A and B be two towns, with B being to the right of A. We say that town A can sweep town B away if the right bulldozer of A can move over to B pushing off all bulldozers it meets. Similarly, B can sweep A away if the left bulldozer of B can move to A pushing off all bulldozers of the towns on its way.
Prove that there is exactly one town which cannot be swept away by any other one.
Solution. Let T1, T2,…,Tn be the towns
enumerated from left to right. Observe first that, if town Ta can sweep away town Tb, then Ta also can sweep away every town located between Ta and Tb. We prove by induction on n. The case n=1 is trivial. For the induction step, we first observe that the left bulldozer in T1 and the right bulldozer in Tn are
completely useless, so we may forget them forever. Among the other 2n-2 bulldozers, we choose the largest one. Without loss of generality, it is the right bulldozer of some town Tk with k<n. Surely, with this right bulldozer Tk can sweep away all towns to the right of it. Moreover, none of these towns can sweep Tk away; so they also cannot sweep away any town to the left of Tk. Thus, if we remove the towns Tk+1, Tk+2,…,Tn, none of the remaining towns would change its status of being (un)sweepable away by the others. Applying the induction hypothesis to the remaining towns, we find a unique town among T1,T2,…,Tk which cannot
be swept away. By the above reasons, it is also the unique such town in the initial situation. Thus the inductive step is established.
Example 9 (1991 Brazilian Math Olympiad). At a party every woman dances with at least one man, and no man dances with every woman. Show that there are men M and M’ and women W and W’ such that M dances with W, M’ dances with W’, but M does not dance with W’, and M’ does not dance with W.
Solution. Let M be one of the men who dance with the maximal number of women, W’ one of the women he doesn’t dance with, and M’ one of the men W’ dances with. If M’ were to dance with every woman that M dances with, then the maximality of the number of women that M dances with would be contradicted, so there is a woman W that dances with M but not with M’.
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 sending solutions is April 15, 2017.
Problem 496. Let a,b,c,d be real numbers such that a+sin b > c+sin d, b+sin a > d+sin c. Prove that a+b>c+d. Problem 497. Let there be three line segments with lengths 1, 2, 3. Let the segment of length 3 be cut into n≥2 line segments. Prove that among these n+2 segments, there exist three of them that can be put to form a triangle where each side is one of the three segments. Problem 498. Determine all integers n>2 with the property that there exists one of the numbers 1,2,…,n+1 such that after its removal, the n numbers left can be arranged as a1,a2,…,an with
no two of |a1-a2|, |a2-a3|, …, |an−1-an|,
|an-a1| being equal.
Problem 499. Let ABC be a triangle with circumcenter O and incenter I. Let Γ be the escribed circle of Δ ABC meeting side BC at L. Let line AB meet Γ at M and line AC meet Γ at N. If the midpoint of line segment MN lies on the circumcircle of ΔABC, then prove that points O, I, L are collinear. Problem 500. Determine all positive integers n such that there exist k≥2 positive rational numbers such that the sum and the product of these k numbers are both equal to n.
*****************
Solutions
****************
Problem 491. Is there a prime number p such that both p3+2008 and p3+2010
are prime numbers? Provide a proof.
Solution. Adnan ALI (Atomic Energy
Central School 4, Mumbai, India), Ioan Viorel CODREANU (Secondary School Satulung, Maramures, Romania), Prithwijit DE (HBCSE, Mumbai, India), EVGENIDIS Nikolaos (M. N. Raptou High School,
Palaiokastrou 10, Agia, Greece), Karaganda (Nazarbaev iIntellectual School, Nurligenov Temirlan - 9 grade student), Koopa KOO, KWOK Man Yi (Baptist Lui Ming Choi Secondary School, S6), Mark LAU, Toshihiro SHIMIZU (Kawasaki, Japan), Anderson TORRES, Titu ZVONARU (Comăneşti, Romania) and Neculai STANCIU (“George Emil Palade’’ Secondary School, Buzău, Romania).
Let p be a prime. If p≠7, then p3≡-1 or 1
(mod 7). Since 2008 ≡ -1 (mod 7) and 2010≡1 (mod 7), so either p3+2008 or
p3+2010 is divisible by 7, hence
composite. If p = 7, then p3+2010 = 2353
= 13×181 is composite. Therefore, there is no such prime.
Problem 492. In convex quadrilateral ADBE, there is a point C within ΔABE such that
∠EAD+∠CAB=180°=∠EBD+∠CBA. Prove that ∠ADE=∠BDC.
Solution. KWOK Man Yi (Baptist Lui
Ming Choi Secondary School, S6).
D
F
C
Let F be the second intersection of the circumcircle of ΔEAD and line EB. Then ∠DBF=180°-∠EBD=∠CBA. Moreover, ∠BDF = 180°-∠AEB-∠ADB
= 180°-(360°-∠EAD-∠EBD) = 180°-(∠CAB+∠CBA) =∠BCA. These two relations give ΔBDF∼ΔBCA. So BD/BF=BC/BA. Together with ∠DBF =∠CBA, we have ΔBDC∼ΔBFA. Then ∠ADE=∠AFE=∠BFA=∠BDC.
Other commended solvers: Toshihiro SHIMIZU (Kawasaki, Japan), Titu ZVONARU (Comăneşti, Romania) and Neculai STANCIU (“George Emil Palade’’ Secondary School, Buzău, Romania).
Problem 493. For n ≥ 4, prove that xn−xn−1−xn−2−⋯−x−1 cannot be factored into a product of two polynomials with rational coefficients, both with degree greater than 1.
Solution. Prithwijit DE (HBCSE,
Mumbai, India) and Toshihiro SHIMIZU (Kawasaki, Japan).
Let Pn(x) = xn−xn−1−xn−2−⋯−x−1 and Qn(x)= (x−1)Pn(x)= xn+1−2xn+1. The cases n = 2 or 3 follow directly from the rational root theorem. For n≥4, the Descartes’ rule of signs shows there is a positive root r. It is easy to check Pn( 3) < 0. So r > 3.
If Pn(s)=0 with |s|>1, then Qn(s)=0, which implies |s|n |s−2|=1. We get 2≤|s−2|+|s| = |s|−n+|s|. So Qn(|s|) ≥ 0. Since Qn(x)<0 for 1<x<r, we must have |s|≥r. On the other hand, if Pn(t)=0 and |t|<1, then 1=|t-2||t|n ≤ 3|t|n. It follows that the absolute value of the product of all roots t of Pn(x) with |t|<1 is at least 1/3. So r is the only root of Pn(x) with absolute value greater than 1.
Assume Pn(x)=f(x)g(x), where f(x), g(x) are monic polynomials with integer coefficients and f(r)=0. Then if g(x) has positive degree, its roots would have absolute value less than 1 and so |g(0)|<1. This contradicts the constant term of g(x), being g(0), must be ±1. Other commended solvers: Anderson TORRES.
Problem 494. In a regular n-sided polygon, either 0 or 1 is written at each vertex. By using non-intersecting diagonals, Bob divides this polygon into triangles. Then he writes the sum of the numbers at the vertices of each of these triangles inside the triangle. Prove that Bob can choose the diagonals in such a way that the maximal and minimal numbers written in the triangles differ by at most 1.
Solution. Adnan ALI (Atomic Energy
Central School 4, Mumbai, India) and Toshihiro SHIMIZU (Kawasaki, Japan).
If all numbers written at the vertices of the polygon are equal, then the claim holds trivially. Hence assume that there are both zeros and ones among the numbers at the vertices. We prove by induction that, for every convex polygon, the partition into triangles can be chosen in such a way that Bob writes either 1 or 2 to each triangle. If n=3, then this claim holds since the sum of the numbers at the vertices of a triangle can be neither 0 nor 3. If n=4, then draw the diagonal that connects
Mathematical Excalibur, Vol. 21, No. 1, Oct. 16 – Mar. 17 Page 4 the vertices where 0 and 1 are written,
respectively, or, if such a diagonal does not exist, then an arbitrary diagonal. In both cases, only sums 1 and 2 can arise. If n≥5, then choose two consecutive vertices with different labels and a third vertex P that is not neighbor to either of them. Irrespective of whether the label of P is 0 or 1, we can draw the diagonal from it to one of the two consecutive vertices chosen before so that the labels of its endpoints are different. Now the polygon is divided into two convex polygons with smaller number of vertices so that both 0 and 1 occur among their vertex labels. By the induction hypothesis, both polygons can be partitioned into triangles with sum of labels of vertices either 1 or 2. Other commended solvers: William FUNG.
Problem 495. The lengths of each side and diagonal of a convex polygon are rational. After all the diagonals are drawn, the interior of the polygon is partitioned into many smaller convex polygonal regions. Prove that the sides of each of these smaller convex polygons are rational numbers.
Solution. Adnan ALI (Atomic Energy
Central School 4, Mumbai, India), Toshihiro SHIMIZU (Kawasaki, Japan) and Anderson TORRES. We only need to show the quadrilateral case, since if this is showed, then the length of any segment of a diagonal connecting a vertex to an intersection point with other diagonal would be rational. Let ABCD be a quadrilateral with all sides and diagonals have rational lengths. Let α =∠ABD and β =∠DBC. Let P be the intersection of AC and BD. Since , 2 cos 2 2 2 BD AB AD BD AB
cos α is rational. Similarly, cos β and cos (α+β)=cos ∠ABC are rational. Then, since cos (α+β)= cos α cos β- sin α sin β, so sin α sin β is also rational. Also, sin2β=1-cos2β is rational. Thus,
sin α/ sin β= sin α sin β/sin2β is rational.
Then, AP/PC = area(ABD)/area(DBC) = (AB·BD sin α)/(BD·BC sin β) is rational. Therefore, AP and PC are rational. Similarly, PB and PD are rational.
Other commended solvers: Corneliu MĂNESCU-AVRAM (Transportation High School, Ploieşti, Romania).
Olympiad Corner
(Continued from page 1) Problem 4. For positive a, b, c, it holds (a+c)(b2+ac)=4a. Find the maximal
possible value of b+c and find all triples (a,b,c), for which the value is attained. Problem 5. There is |BC|=1 in a triangle ABC and there is a unique point D on BC such that |DA|2=|DB|·|DC|. Find all
possible values of the perimeter of ABC. Problem 6. There is a figure of a prince on a field of a 6×6 square chessboard. The prince can in one move jump either horizontally or vertically. The lengths of the jumps are alternately either one or two fields, and the jump on the next field is the first one. Decide whether one can choose the initial field for the prince, so that the prince visits in an appropriate sequence of 35 jumps every field of the chessboard.
Miscellaneous Problems
(Continued from page 2)
Example 10. Two triangles have the same incircle. If a circle passes through five of the six vertices of the two triangles, then must it also pass the sixth vertex?
Solution. Let ABC and DEF be the triangles. Let A, B, C, D, E be on the same circle Γ, with radius R and center O. Suppose that F does not belong to Γ. Let G≠D be the intersection of DF with Γ. Let θ =∠EDF=∠EDG. Let I and r be the common incenter and the inradius of Δ ABC and Δ DEF. Let J and s be the incenter and the inradius of ΔDEG.
s r G D J E M I A B C F
We will prove that the incircle of ΔABC and ΔDEG coincide. First, we prove that I=J by showing IM=JM. It is well known that IM = 2R sin(θ/2) = EM. From Euler’s formula, OI2 = R2-2Rr, which implies that
the power of I with respect to Γ is IM·ID = 2Rr. Since ID = r/sin(θ/2), we have IM = 2Rsin(θ/2) = JM. So I =J. This also proves r = s. Hence, the incircle of ΔABC and Δ DEG are the same. Then F=G follows.
Example 11 (1988 Brazilian Math Olympiad). A figure on a computer screen shows n points on a sphere, no four coplanar. Some pairs of points are joined by segments. Each segment is colored red or blue. For each point there is a key that switches the colors of all segments with that point as endpoint. For every three points there is a sequence of key presses that make all three segments between them red. Show that it is possible to make all the segments on the screen red. Find the smallest number of key presses that can turn all the segments red, starting from the worst case.
Solution. Consider three of the points. The parity of the number of blue segments of the triangle with these points as vertices doesn’t change while switching the keys. Since it is possible to make all three segments red, the number of blue segments in each triangle is even.
Let P be one of the n points. Let A be the set of points connected to P by red points and B be the set of points connected to P by blue segments. Let A1, A2∈A. So PA1 and PA2 are both red
and thus A1A2 is red. Now consider
B1B2∈B. Then PB1 and PB2 are both
blue and B1B2 is red. Finally consider
A∈A and B∈B. PA is red and PB is blue, so AB is blue. Put P in A. All this reasoning shows that segments in the same set are red and segments connect- ing points in different sets are blue. Switching all points in set A will make all segments red. Indeed, all segments in A will change twice, one time from each of its edges, all segments connecting points from A and B will change once, turning from blue to red and segments in B won’t change. This proves the first part.
For the second part, notice first that one needs to switch each point at most once. Let |A|=k and |B|=n-k. If we switch a point from A and b points from B, we change at most a(n-k)+bk blue segments. Suppose without loss of generality that k≤n-k, hence k≤[n/2]. Then k(n-k) ≤ a(n-k) + bk ≤ a(n-k) + b(n-k), hence k≤a+b. So the number of key presses is at most k and in the worst case, [n/2]. This number is needed to make all segments red if |A|=[n/2].
I. Let p > 3 be a prime. Find the number of ordered sextuples ( a, b, c, d, e, f) of positive integers, whose sum is 3p, and all the fractions
a+b b+c c+d d+e
c+d' d+e' e+f' f +a' e+f a+b
are integers. (Jaromir Simsa, Jaroslav Svrcek)
Solution. Taking the product of the 1st, the 3rd and the 5th fractions reveals that their value has to be 1, that is
a
+
b=
c + d=
e+
f=
p. (1)the form of the second and of the fourth fraction implies
f + a
I
d + e and d + eI
b + c. (2)that is first / + a is at most the arithmetic mean of its multiples,
f +a,:;;%((! +a)+ (d+e) + (b+c)) =p, (3)
and
f + a
I
(/+a) + (d + e) + (b + c)=
3p.Thus /+a divides 3p and is in the interval (2, p). Consequently either /+a = p or
f + a
=
3. We deal separately with these cases.(i) Let / + a
=
p. Because of (3) there is / + a=
d + e=
b + c=
p, which together with (1) gives p - l solutions of the form(a,b,c, d, e, /)
=
(a,p- a, a,p- a, a,p- a), where a E {1, 2, ... ,p- l}.(ii) Let/ +a= 3. Then {a,/}= {1,2}.
Firstly let a
=
l and f=
2. According to (1) then b=
p - 1 and e=
p - 2, and (2) has the form3ld+(p-2) and d+(p-2)i(p-l)+c. (4)
In analyzing (4) we distinguish between d = l and d ~ 2. If d
=
1 then c=
p-1 and (4) reads3
IP -
1 a P - 1I
2(p - 1).While the right relation always holds, the left one holds only for p = 3q + 1 (q is a suitable positive integer). For such prime numbers we get considering (1) solutions
(a,b,c,d,e,f) = (1,p-1,p-l,1,p-2,2).
If d ~ 2 we show first, that the right relation in (4) is satisfied if and only if d + (p - 2) = (p-1) +cord= c + 1. d ~ 2 namely implies c = p-d,:;; p-2, thus
d + (p- 2) ~ p and (p - 1) + c,:;; 2p - 3 < 2p
(a, b, c, d, e, /) = (1,p - 1, !(P - 1), !(p + 1),p - 2, 2). is a solution.
Finally a= 2 and f
=
1. In this case b = p - 2 a e = p - 1, and (2) reads3ld+(p-l) and d+(p-l)l(p-2)+c. (5)
Because
d + (p - 1) ~ p and (p - 2) + c < 2p,
the right relation in (5) holds if and only ifd+(p-1) = (p-2)+c, that is iff c=d+l. Together with c + d
=
p we get c=
!
(p + 1) and d =!
(p - 1), thus the right relation in (5) holds as well, and the last solution is(a, b, c, d, e, f) = (2,p - 2, !(p + 1), !(p-l),p - 1, 1).
Conclusion. All the solutions found are apparently mutually different and their number depends on p modul~ 3 (p > 3): If p
=
3q + 1 then there are p + 2 sextuples, if p=
3q + 2, there are p + l sextuples.2. Let r and r a be the radii of inscribed circle and excircle opposite A of the triangle ABC. Show, that if
r +ra = IBCI,
then the triangle is right-angled. (Michal Rolinek)
Solution. Let us use the standard notation of the inner angles of the triangle ABC,
further let I be the incenter and Ia be the excenter (of the excircle opposite A) and let D and E be in order the touching points of the thought circles. Since the bisectors
BI and Bla of the supplementary angles are perpendicular to each other (as well
as CJ and Ola), the points B, C, I a. Ia lie on the circle with the diameter JJ0 •
Thus D and E, the orthogonal projections of J and Ia onto the secant BC, are point reflections of each other with respect to the center of BC.
I
Fig.I
The right triangles BID and Ia BE are obviously similar and
IBDj: jJDI
=
IIaEI: jBEj or IBDI · IBEI=
IIDI · llaEIconsidering the mentioned point reflection also
urn wo equations 1mp1y that a pairs U1 UJ, J~lal) and (JBDJ, JBEI) are roots of the same quadratic equation, that is JIDJ
=
JBDJ or JIDJ=
JBEJ.IIDJ = IBDI means the right-angled triangle BID is isosceles, which is /3 = 90°. Similarly, if JIDJ = JBEJ that is JIDJ = JCDJ (D and E are point reflections in the mentioned reflection) means the right triangle CID is isosceles, that is 'Y
=
90°.In both cases the triangle ABC is right-angled.
3. Mathematics clubs are very popular in certain city. Any two of them have at least one common member. Prove, that one can distribute rulers and compasses to the citizens in such a way that only one citizen gets both {compass and ruler} and any club has to his disposal both, compass and ruler, from its members.
(Josef Tkadlec) Solution. Let us ·consider the club K with the least number of its members (in case there is more such clubs, we take any). We give to one of it's members {let us call him Jacob} both, a compass and a ruler. Each of the other members of the club will get a compass. Any other citizen will get a ruler. We show, that this distribution comply with the conditions of the _problem: Any club, which has Jacob as its member, has certainly both instruments.
If there is a club, where Jacob does not belong, then it has at least one common member with the club K, that is there is at least a compass at disoosal in the club. If . there were no ruler in the club, it would mean that it is a "subclub" of Kand therefore has at least one member (Jacob) Jess than K, whicli is a contradiction with the choice of K. The described distribution really satisfies the conditions of the problem.
4. For positive a, b, c it holds
(a+ c)(b2 + ac)
=
4a.Find the maximal possible value of b + c and find all triples (a,b,c}, for which
the value is attained. (Michal Rolinek}
Solution. We use the well know inequality a.2
+
b2 ;;;, 2ab to adjust the given one: 4a=
(a+ c)(b2 + ac)=
a(b2 + c2} + c(a2 + b2) ;;:, a(b2 + c2) + 2abc
=
a(b + c)2•We can see, that b+ c ,;;; 2, and also that the equality holds if and only if O < a = b < 2 a c
=
2 - b > O. Thats it.5. There is IBCJ = 1 in a triangle ABC and there is a unique point D on BC such
that IDAl2
=
IDBI· JDCI, Find all possible values of the perimeter of ABC.(Patrik Bak} Solution. Let us denote by E the second intersection of AD with the circumcircle k. The power of D with respect to k gives IDBI· JDCI
=
IDAI · IDBI, which together with the given condition JDAl2=
IDBI· IDCI yields IDAI=
JDEI. That is Elie theimage p of the line BC in the homothety with center A and a coefficient 2 (Fig. 1).
Vice versa, to any intersection of a line p with the circle k we reconstruct the point Don BC, whicli fulfills IDAJ2
=IDBI· IDCI,
If the reconstruction have to be unique, the line p has to touch p in E.
Fig.2 Fi.e:.3
Let us denote Sb and Sc in order the centers of AC and AB. The homothety with the center A and a coefficient! sends A, B, C, E {lying on the circle k) to A, Sc,
Sb, D which lie on the circle k' (Fig. 2), while the image of p is the tangent BC of k' in D. The powers of Band C with respect to k' give IBDl2
=
IBAI · IBScl=
!IBAl2 and jCDl2 = ICAI · ICSbl = !ICAl2• All together for the perimeter of ABC:IBCJ + IABI +JACI
=
IBCI + v'2(1BDI + ICDI)=
IBCI +\/2 ·
IBCI=
1 +\/2,
which is the only possible value.6. There is a figure of prince on a field of a 6 x 6 square chessboard. The prince
can in one move jump either horizontally or vertically. The lengths of the jumps are alternately either one or two fields, and the jump on the next field is the first one. Decide, whether one can chose the initial field for the prince, so that the prince visits in an appropriate sequence of S5 jumps every field of the chessboard.
(Peter Novotny)
Solution. Let us suppose, the appropriate sequence exists and let us enumerate the fields of the chessboard as follows:
1 2 3 4 1 2 2 3 4 1 2 3 3 4 1 2 3 4 4 1 2 3 4 1 1 2 3 4 1 2 2 3 4 1 2 3
The length one moves go from odd to even number and vice versa. The length two moves go from even to a different even number or from odd to a different odd number. If we denote Pi, P2, . .. , P3e the numbers of visited fields, then it follows, that among P2 , P3 , P4, A is each number {from 1 to 4} exactly once (P2 and Pa are different numbers with the same parity, and P4, Ps as well, only the parity is different). from the same reasons is any of the four numbers among P4k+2, P4k+3,
P4k+4, P41;;+5 for arbitrary k E {O, 1, ... , 7}. Between the numbers P2,P3, ... ,Paa is thus any of the numbers 1 to 4 exactly eight times.
The number 4 is on the chessboard just eight times, thus no from P1 , P34 , P35 ,
Pa6 can be 4. The numbers P34 and P35 have the same parity and are different {they
are the length two move apart) The number 4 is not among them, therefore both must be odd. Then P35 has to be even and Pi as well. Thus it has to be number 2.
The initial field (Pi) thus has to be one of the coloured fields on the left chess-board. One can repeat the arguments for the numbering of the right chessboard Qust a rotation of the left one). Since no field has number 2 on both chessboard, we came to the contradiction. The initial field cannot be chosen.
1 2 3 4 1 2 2 3 4 1 2 3 2 3 4 1 2 3 1 2 3 4 1 2 3 4 1 2 3 4 4 1 2 3 4 1 4 1 2 3 4 1 3 4 1 2 3 4 1 2 3 4 1 2 2 3 4 1 2 3 2 3 4 1 2 3 1 2 3 4 1 2
