FAU/Stuyvesant Alumni Mathematics Competition2006–2007

(1) There are two isosceles triangles. One has with top angleα, equal sides of length a, base b. The other has baseangle α, equal sides of length b and base a. Assumea �= b. Find the ratio a

b and the size of the angle α.

(2) Let N be the set of positive integers, and f : N → N astrictly increasing function satisfying f ◦ f(n) = 3n forevery n ∈ N.(a) Find f(2007).(b) Is there an integer n with f(n) = 2007?

(3) Let x and y be integers such that x2+x = 2y2+y. Provethat x − y, x + y + 1 and 2x + 2y + 1 are all squares ofintegers.

(4) Find the smallest positive integer value of n for which2n−6833n+641 is not a reduced fraction. (A reduced fraction isone whose numerator and denominator do not have com-mon divisors).

(5) Solve the simultaneous equations xy = z + t, zt = x+yin positive integers.

(6) Let M be the midpoint of the hypotenuse AB of a righttriangle ABC . Prove that the line joining the incentersof triangles ACM and BCM bisects the area of triangleABC .

Name: Solution

Problem 1.There are two isosceles triangles. One has with top angle α,equal sides of length a, base b. The other has base angle α,equal sides of length b and base a. Assume a �= b. Find theratio a

band the size of the angle α.

Solution. Form a quadrilateral ABCD with isosceles trianglesABC and DAC satisfying the given condition. The quadrilat-eral is an isosceles trapezoid since AB//CD and AD = BC .Therefore, it is cyclic.

(1) 12(180◦ − α) = 2α ⇒ α = 36◦.

(2) By Ptolemy’s theorem, AC ·BD = AB ·CD+AD ·BC ,a2 = ab + b2, a

b = 1+√

52 .

Name: Solution

Problem 2.Let N be the set of positive integers, and f : N → N a strictlyincreasing function satisfying f ◦ f(n) = 3n for every n ∈ N.(a) Find f(2007).(b) Is there an integer n with f(n) = 2007?

Solution. Write positive integers in base 3. Note that the left-most digit is either 1 or 2. The base 3 expansion of f(n) isobtained from that of n by switching the leftmost digit between1 and 2, and appending an extra 0 as the rightmost digit whenthe leftmost digit of n is 2. Thus, f(110123) = 210123 andf(210123) = 1101203.(a) f(2007) = f(22021003) = 120210003 = 3834.(b) f(1278) = f(12021003) = 22021003 = 2007.

Name: Solution

Problem 3.Let x and y be integers such that x2 + x = 2y2 + y. Prove thatx − y, x + y + 1 and 2x + 2y + 1 are all squares of integers.

Solution. From x2 + x = 2y2 + y, we easily deduce

y2 =(x − y)(x + y + 1), (1)

x2 =(x − y)(2x + 2y + 1). (2)

We claim that x − y and x + y + 1 are relatively prime. If not,they contains a common prime divisor p. From (1), p|y. Sincep|x − y, we must have p|x, and p|x + y. Finally, p|((x + y +1)− (x+ y)) = 1, a contradiction. From (1), each of x− y andx + y + 1 is a square. From (2), 2x + 2y + 1 is also a square.

Name: Solution

Problem 4.Find the smallest positive integer value of n for which 2n−683

3n+641is not a reduced fraction. (A reduced fraction is one whosenumerator and denominator do not have common divisors).

Solution. Let d > 1 be a common divisor of 2n − 683 and3n+641. Then d|(2(3n+641)− 3(2n− 683)) = 3331, whichis a prime number. Therefore, d = 3331. If the given fraction isnot reduced, then 2n − 683 = 3331k for some integer k. Withk = 1, we have 2n = 3331 + 683 = 4014, n = 2007. Withthis, 2n−683

3n+641 = 33316662 = 1

2 is not reduced. The smallest value of nfor which the fraction is not reduced is n = 2007.

Name: Solution

Problem 5.Solve the simultaneous equations xy = z + t, zt = x + y inpositive integers.

Solution. (1) y = t2+xtx−1 , z = t+x2

tx−1 .(2) If tx − 1 = 1, then t = 1, x = 2, y = 3, z = 5. This is theonly solution with t = 1.(3) Assume 2 ≤ t ≤ x. Note

y =t2 + x

tx − 1= 1+

(t2 + x) − (tx − 1)

tx − 1≤ 1+

x + 1

tx − 1≤ 1+

x + 1

2x − 1.

(3a) If x = 2, then t = x = y = z = 2.(3b) If x ≥ 3, then y < 2. We must have y = 1. From (1),x = t2+1

t−1 = t + 1 + 2t−1 . This is an integer only if t = 2, x = 5,

and z = 3.(5) Interchanging t and x, we obtain two more solutions from(2) and (3b). The only solutions are

(t, x, y, z) = (1, 2, 3, 5), (2, 1, 5, 3), (2, 2, 2, 2), (2, 5, 1, 3), (5, 2, 3, 1).

Name: Solution

Problem 6.Let M be the midpoint of the hypotenuse AB of a right trian-gle ABC . Prove that the line joining the incenters of trianglesACM and BCM bisects the area of triangle ABC .

Solution. Let P , Q be the incenters of triangles ACM andBCM , which touch the sides AC and BC respectively at Yand Z. Let the line PQ intersect AC and BC at H and Krespectively.

K

H

Z

Q

Y

P

MAB

C

Since MA = MC , the points M , P , Y are collinear. Simi-larly, M , Q, Z are collinear. In fact Y and Z are the midpointsof AC and BC respectively. The lines MY and MZ are per-pendicular to AC and BC respectively. CY MZ is a rectangle,and the right triangles PHY , PQM , and KQZ are similar.Since AP bisects angle A, PY

PM = AYAM = cos A. Similarly,

QZQM = cos B. It follows that

�PHY + �KQZ

�PQM= cos2 A + cos2 B = 1.

It follows that

�HCK =area CY PMZ + �PHY + �KQZ

=area CY PMZ + �PQM

=area CY MZ

=1

2�ABC.