Solutions

325.

Solve for positive real values of x, y, t:

(x2 + y2)2 + 2tx(x2 + y2) = t2y2 .

Are there infinitely many solutions for which the values of x, y, t are all positive integers? What is the smallest value of t for a positive integer solution?

Comment. L. Fei gives the solution set

326.

In the triangle ABC with semiperimeter s = ¹/₂(a +b + c), points U, V, W lie on the respective sides BC, CA, AB. Prove that

s < |AU |+ |BV |+ |CW | < 3s .

Give an example for which the sum in the middle is equal to 2s.

Again, by the triangle inequality, |AU |+ |BU | > |AB | and |AU |+ |CU | > |AC |. Adding these inequalities gives 2 |AU | > |AB |+ |AC |- |BC |. Adding this to analogous inequalities for BV and CW and dividing by 2 yields that |AU |+ |BV |+ |CW | > s.

Comment. For the last part, most students gave a degenerate example in which the points U, V, W coincided with certain vertices of the triangle. A few gave more interesting examples. However, it was necessary to make clear that the stated lengths assigned to AU, BV and CW were indeed possible, i.e. they were at least as great as the altitudes. The nicest example came from D. Dziabenko: |AB | = 6, |BC | = 8, |CA | = 10, |AU | = 8, |BV | = 7 |CW | = 9.

327.

Let A be a point on a circle with centre O and let B be the midpoint of OA. Let C and D be points on the circle on the same side of OA produced for which ÐCBO = ÐDBA. Let E be the midpoint of CD and let F be the point on EB produced for which BF = BE.

(a) Prove that F lies on the circle.

(b) What is the range of angle EAO?

Let H lie on OA produced so that OA = AH. Since OB ·OH = OA², the inversion in the given circle with centre O interchanges B and H, fixes C and D, and carries the circle OBDC (which passes through the centre O of inversion) to a straight line passing through H, D, C. Thus C, D, H are collinear.

This means that CD always passes through the point H on OA produced for which OA = AH. Since E is the midpoint of CD, a chord of the circle with centre O, ÐOEH = ÐOED = 90^°. Hence E lies on the circle with centre A and radius OA.

Consider the reflection in the point B (the dilation with centre B and factor -1). This takes the circle with centre O and radius OA to the circle with centre A and the same radius, and also interchanges E and F. Since E is on the latter circle, F is on the given circle.

Ad (b), E lies on the arc of the circle with centre A and radius OA that joins O to the point R of intersection of this circle and the given circle. Since RB ^OA, and OA = OR = RA, ÐRAO = 60^°. It can be seen that ÐEAO ranges from 0^° (when CD is a diameter) to 60^° (when C = D = R).

Solution 2. [A. Wice] We first establish a Lemma.

Lemma. Let UZ be an angle bisector of triangle UVW with Z on VW. Then

Proof. By the Cosine Law,

Let R be a point on the circle with BR ^OA, S be the intersection of CD in OA produced, and D¢ be the reflection of D in OA. (Wolog, ÐCBO < 90^°.) Since SB is an angle bisector of triangle SCD¢, from the Lemma, we have that

Now

Consider the reflection in the point B (dilation in B with factor -1). It interchanges E and F, interchanges O and A, and switches the circles ADRC and OERS. Since E lies on the latter circle, F must lie on the former circle, and the desired result (a) follows.

Ad (b), the locus of E is that part of the circle with centre A that lies within the circle with centre O. Angle EAO is maximum when E coincides with R, and minimum when D coincides with A. Since triangle ORA is equilateral, the maximum angle is 60^° and the minimum angle is 0^°.

Solution 3. [M. Elqars] Let the radius of the circle be r. Let ÐCBO = ÐDBA = a, ÐDOB = b and ÐCOF = g. By the Law of Sines, we have that

Observe that sin(a- b) : sina = OB : OD = 1:2, so that sin(a- b) = ¹/₂sina.

Solution 4. Assign coordinates: O ~ (0, 0), B ~ (¹/₂, 0), C ~ (1, 0), and let the slope of the lines BC and BD be respectively -m and m. Then C ~ (¹/₂ + s, -ms) and D ~ (¹/₂ + t,mt) for some s and t. Using the fact that the coordinates of C and D satisfy x² + y² = 1, we find that

Solution 5. [P. Shi] Assign the coordinates O ~ (0, -1), B ~ (0, 0), A ~ (0, 1). Taking the coordinates of C and D to be of the form (x, y) = (r cosq,r sinq) and (x, y) = (s cosq, - s sinq) and using the fact that both lie of the circle of equation x² + (y + 1)² = 4, we find that

328.

Let \frakC be a circle with diameter AC and centre D. Suppose that B is a point on the circle for which BD ^AC. Let E be the midpoint of DC and let Z be a point on the radius AD for which EZ = EB.

Prove that

(a) The length c of BZ is the length of the side of a regular pentagon inscribed in \frakC.

(b) The length b of DZ is the length of the side of a regular decagon (10-gon) inscribed in \frakC.

(c) c² = a² + b² where a is the length of a regular hexagon inscribed in \frakC.

(d) (a + b):a = a:b.

Solution. [J. Park] Wolog, suppose that the radius of the circle is 2.

(a) Select W on the arc of the circle joining A and B such that BW = BZ. We have that |BE | = |EZ | = Ö5, b = |ZD | = Ö5 - 1, c² = |BZ|² +|BW |² = 10 - 2Ö5 and, by the Law of Cosines applied to triangle BDW,

(b) The angle at each vertex of a regular decagon is 144^°. Thus, the triangle formed by the side of an inscribed regular pentagon and two adjacent sides of an inscribed regular decagon has angles 144^°, 18^°, 18 ^°. Conversely, if a triangle with these angles has its longest side equal to that of an inscribed regular pentagon, then its two equal sides have lengths equal to those of the sides of a regular decagon inscribed in the same circle.

Now b = Ö5 - 1, c² = 10 - 2Ö5, so 4b² - c² = 14 - 2Ö{45} > 0. Thus, c < 2b, and we can construct an isosceles triangle with sides b, b, c. By the Law of Cosines, the cosine of the angle opposite c is equal to (2b² - c²)/(2b²) = -¹/₄(Ö5 - 1). This angle is equal to 144^°, and so b is the side length of a regular inscribed decagon.

(c) The side length of a regular inscribed hexagon is equal to the radius of the circle. We have that

(d) Since b² + 2b - 4 = (b + 1)² - 5 = 0, a² = 4 = (2 + b)b = (a + b)b, whence (a + b): a = a:b.

329.

Let x, y, z be positive real numbers. Prove that

Ö   x2 - xy + y2   + Ö   y2 - yz + z2   ³ Ö   x2 + xz + z2    .

Solution 2. [B. Braverman; B.H. Deng] Note that x² - xy + y², y² - yz + z² and z² + xz + z² are always positive [why?]. By squaring, we see that the given inequality is equivalent to

Comment. The above write-up proceeds by a succession of equivalent inequalities to one that is trivial, a working-backwards from the result. The danger of this approach is that one may come to a step where the reasoning is not necessarily reversible, so that instead of a chain of equivalent statements, you get to a stage where the logical implication is in the wrong direction. The possibility that xy + yz + zx < 2y² complicates the argument a litle and needs to be dealt with. To be on the safe side, you could frame the solution by starting with the observation that (xy + yz - zx)² ³ 0 and deducing that

Solution 3. [L. Fei] Let x = ay and z = by for some positive reals a and b. Then the given inequality is equivalent to

We have that

Comment. The student who produced this solution worked backwards down to the obvious inequality (ab - a - b)² ³ 0, However, for a proper argument, you need to show how by logical steps you can go in the other direction, i.e. from the obvious inequality to the desired one. You will notice that this has been done in the write-up above; the price that you pay is that the evolution from (ab - a - b)² ³ 0 seems somewhat artificial. There is a place where care is needed. It is conceivable that ab + a + b - 2 is negative, so that 2Ö{a² - a + 1}Ö{b² - b + 1} ³ ab + a + b - 2 is always true, even when 4(a² - a + 1)(b² - b + 1) ³ (ab + a + b - 1)² fails. The inequality A² ³ B² is equivalent to A ³ B only if you know that A and B are both positive.

330.

At an international conference, there are four official languages. Any two participants can communicate in at least one of these languages. Show that at least one of the languages is spoken by at least 60% of the participants.

Finally, suppose that every participant speaks at least three languages. Let p% speak E, F and G (and possibly, but not necessarily I), q% speak E, F, I but not G, r% speak E, G, I but not F, s% speak F, G, I but not E. Then p + q + r + s = 100 and so

Solution 2. [D. Rhee] As in Solution 1, we take the languages to be E, F, G, I, and can dispose of the case where someone speaks exactly one of these. Let (A¼C) denote the set of people who speaks the languages A ¼C and no other language, and suppose that each person speaks at least two languages.

We first observe that in each of the pairs of sets, {(EF),(GI)}, {(EG),(FI)}, {(EI),(FG)}, at least one of the sets in each pair is empty. So either there is one language that is spoken by everyone speaking exactly two languages, or else there are only three languages spoken among those that speak exactly two languages. Thus, wolog, we can take the two language sets among the participants to be either { (EF), (EG), (EI) } or { (EF), (FG), (EG) }.

Case 1. Everyone is in exactly one of the language groups

Case 2. Everyone is in exactly one of the language groups

331.

Some checkers are placed on various squares of a 2m ×2n chessboard, where m and n are odd. Any number (including zero) of checkers are placed on each square. There are an odd number of checkers in each row and in each column. Suppose that the chessboard squares are coloured alternately black and white (as usual). Prove that there are an even number of checkers on the black squares.

Solution 2. [F. Barekat] Suppose that the (1, 1) square on the chessboard is black. The set of black squares is contained in the union U of the odd-numbered columns along with the union V of all the even-numbered rows. Note that U contains an odd number of columns and V an odd number of rows. Since each row and each column contains an odd number of checkers, U has an odd number u of checkers and V has an odd number v of checkers. Thus u + v is even. Note that each white square belongs either to both of U and V, or to neither of them. Thus, the u + v is equal to the number of black checkers plus twice the number of checkers on the white squares common to U and V. The result follows.