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- 1.
- ABC is an isosceles triangle with ÐA = 100° and AB = AC. The bisector of angle B meets AC in D. Show that BD + AD = BC.
- 2.
- Let I be the incentre of triangle ABC. Let the lines AI, BI and CI produced intersect the circumcircle of triangle ABC at D, E and F respectively. Prove that EF is perpendicular to AD.
- 3.
- Let PQR be an arbitrary triangle. Points A, B and C external to the triangle are determine for which
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(a) AC = AB;
(b) ÐBAC = 90°.
- 4.
- Let a and b be two positive real numbers. Suppose that ABC is a triangle and D a point in side AC for which
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- 5.
- Let \frakC be a circle with centre O and radius k. For each point P ¹ O, we define a mapping P ®P¢ where P¢ is that point on OP produced for which
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- (a) Suppose that A and B are two points in the plane for which |AB | = d, |OA | = r and |OB | = s, and let their respective images under the inversion be A¢ and B¢. Prove that
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- (b) Using (a), or otherwise, show that there exists a sequence { Xn } of distinct points in the plane with no three collinear for which all distances between pairs of them are rational.
- 6.
- Solve each of the following two systems of equations:
- (a) x + xy + y = 2 + 3Ö2 , x2 + y2 = 6;
- (b)
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- 1.
- First solution. See Figure 1.1. Let |AB | = u. Then
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- 1.
- Second solution. See Figure 1.2. Let E and F be located in BC so that ÐDEC = ÐDFB = 80°. Then DABD º DEBD while triangles DEF, DCF and DBF are isosceles. Hence
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- 1.
- Third solution. See Figure 1.3. Select G on BC so that BG = BD and H on AB so that HD || BC. Then, by filling in the angles, we see that triangles BDG, HBD, AHD and DGC are isosceles. Also triangles AHD and GCD are similar and DC = BH = HD. Hence DAHD º DGCD and AD = GC. Hence BC = BG + GC = BD + AD.
- 1.
- Fourth solution. See Figure 1.3. Define G as in the third solution. Triangles ABC and GCD are similar. Hence, using the angle bisection theorem, we have that GC:CD = AB:BC = AD:DC, so that GC = AD. Hence BC = BG + GC = BD + AD.
- 1.
- Fifth solution. See Figure 1.3. Define G and H as in the previous solution. ABGD is concyclic, so AD = DG since they are chords subtending equal angles of 20° at the circumference. It follows from the similarity of triangles AHD and GDC that they are congruent, so AD = DG = GC. Since also BD = BG, the result follows.
- 2.
- First solution. See Figure 2.1. Let AD and EF meet at H. Then
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- 2.
- Second solution. Identify points in the complex plane, with the circumcircle of DABC being the unit circle with centre 0. Let A ~ 1, B ~ cos2b+ isin2b, C ~ cos2g+ i sin2g. Observe that AD bisects the arc BC, BE the arc CA and CF the arc AB, so that
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- 2.
- Third solution. [D. Brox] See Figure 2.3. Since EFBC is concyclic, ÐIFX = ÐCFE = ÐCBE = ÐEBA = ÐIBX. Hence FBIX is concyclic and so ÐFXB = ÐFIB. Similarly ÐEYC = ÐEIC.
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Hence ÐAXY = ÐFXB = ÐFIB = ÐEIC = ÐEYC = ÐAYX, so that AX = XY. Since DAXY is
isosceles and AD bisects ÐXAY, then AD ^FE.
- 2.
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Fourth solution. [L. Lessard] See
Figure 2.4. Let O be the
circumcentre of the triangle. Since F bisects arc AB, then
OF right bisects AB. Also OE right bisects AC.
Since DOEF is isosceles, ÐOEF = ÐOFE
and so ÐEUC = ÐFVB. Hence ÐAUV = ÐAVU, so that DAUV is isosceles. Thus, the bisector of
angle A right bisects UV and the result follows.
- 2.
- Fifth solution. [H. Dong] See Figure 2.3. Using the fact that AFBCE is concyclic, we have that ÐAEF = ÐACF = ÐBCF = ÐBEF. Also ÐAFE = ÐCFE. Hence DAFE º DIFE (ASA) so that EA = EI. Thus DEAI is isosceles with apex angle AEI whose bisector EF must right bisect the base AI.
- 2.
- Sixth solution. See Figures 2.6 and 2.3. We first note a preliminary result: If P, R, Q, S are four points on a circle and PQ and RS intersect inside the circle at T, then ÐSTQ = ÐSPQ + ÐPSR, which is equal to half the sum of the angles subtended at the centre by arcs PR and SQ. Now, ÐABE = 1/2ÐABC, so that arc AE subtends an angle equal to ÐABC at the centre. Similarly BF subtends an angle equal to ÐACB at the centre and BD subtends an angle equal to ÐBAC at the centre. Hence FD subtends an angle equal to ÐACB + ÐBAC at the centre. By the preliminary result, ÐAHE is equal to half the sum of the angles subtended at the centre by arcs AE and FD, namely half of 180°. The result follows.
- 3.
- First solution. See Figure 3.1. (a) Let point D be selected on the same side of QR as A so that triangle QDR is equilateral. Then DA ^QR so that ÐQDA = ÐRDA = 30° and ÐDQA = ÐDRA = 45°. Hence
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- A rotation about Q followed by a dilatation takes C to A and P to D so that
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- (b) By the similar triangles identified in (a), ÐCAQ = ÐPDQ and ÐBAR = ÐPDR. Hence
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- 3.
- Second solution. See Figure 3.2. Let S be the image of R under a counterclockwise rotation about A through 90°. Since AS = AR, ÐASR = ÐARS = 45° so ÐQRS = 30°. Since QA = AR = AS, and since ÐQAS = 150° - 90° = 60°, ÐAQS = ÐASQ = 60° and so ÐSQR = 45°. Hence triangles CQP, BRP, SQR are similar, and CQ:PQ = QS:QR. Also ÐCQS = 45° ±ÐPQS = ÐPQR (± according as S lies inside or outside of DPQR). Hence DCQS ~ DPQR. Therefore ÐCSQ = ÐPRQ Þ ÐCSA = ÐCSQ + 60° = ÐPRQ + 60° = ÐARB. Also CS:RP = QC:QP = RB:RP Þ CS = RB. Since in addition SA = RA, DCSA º DBRA (SAS), so that AC = AB and ÐSAC = ÐRAB. Finally,
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- 3.
- Third solution. [D. Brox] Note that sin75° = cos15° = (Ö6 + Ö2)/4 and sin15° = (Ö6 - Ö2)/4. Let
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- Since |r - a | = |q - a | = |r - q |/2 cos15° = 1/2(Ö6 -Ö2) |r - q |, we have that
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- 3.
- Fourth solution. [D. Pritchard]
- Lemma: For any triangle with angles a, b, g,
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- Proof: Taking the difference between the two sides yields
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- (a) Wolog, we can let the lengths of PQ, PR and QR be sinR, sinQ and sinP, respectively, since by the Law of Sines, these lengths are proportional to these quantities. Then
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- (b) |BP | = sinQ sin45°/sin105° = (Ö3 - 1)sinQ, |CP | = (Ö3 - 1)sinR and
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- 3.
- Fifth solution. [D. Nicholson] See Figure 3.5. Let |AQ | = |AR | = u, |BR | = v, |CQ | = w, a = ÐQPR, b = ÐPQR and g = ÐPRQ. Let D, E, F be the respective feet of the perpendiculars from A, B, C to QR, PR and PQ.
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- 3.
- Sixth solution. [T. Costin] See Figure 3.1. First, we establish a preliminary result. Let ABC be an arbitrary triangle. Using the standard notation and applying the sine law, we find that
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- (a) Let D be constructed as in the figure, and note that QC:QA = r:p by similar triangles. By the Law of Cosines,
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- (b) PC:AD = r:p and PB:AD = q:p. Hence, by the Law of Cosines applied to DPCB,
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- 3.
- Seventh solution. [Y. Shen] (a) From the Sine Law,
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- (b)
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- 4.
- First solution. See Figure 4.1. q assumes its maximum value when the circumcircle of DADB is tangent to BC. For, if X is any other point on CB produced, X lies outside this circle and AD subtends at X a smaller angle than it subtends at the circumference of the circle.
- Let O be the centre of the circumcircle and r its radius. Then O lies on the right bisector of AD and two radii are OD and OB. Since BC is tangent, OB ^BC. Hence x2 + (a/2)2 = r2 = (b + (a/2))2 so that x = Ö[(b(a+b))]. Since q = arctan[(a+b)/x]- arctanb/x,
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- Comment. The angle can also be identified by noting that
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- 4.
- Second solution. See Figure 4.2. (using calculus) As in (a), for general position of B we can calculate
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- 4.
- Third solution. [D. Brox] As in the previous solution we find that
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- 4.
- Fourth solution. [K. Choi] Comparing two expressions for the area of triangle ABD, we have that
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- 4.
- Fifth solution. By the Law of Cosines,
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- 4.
- Sixth solution. [Y. Shen] As in the second solution,
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- 5.
- First solution. See Figure 5.1. Since ÐO is common and OA:OB = OB¢:OA¢, triangles OAB and OB¢A¢ are similar. Hence
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- (b) We first show that the inversion with respect to a circle \frakC of a line not passing through its centre O is a circle passing through O. Let F be the foot of the perpendicular from the point O to the line, and let P be any other point on the line. Let F¢ and P¢ be their respective images with respect to the inversion. As in (a), we have that DOPF ~ DOF¢P¢ whence ÐOP¢F¢ = ÐOFP = 90°. Hence P¢ lies on the circle with diameter OF¢.
- Note that applying the inversion twice yields the identity, so that the image of F¢ is F. Let P¢ be any point distinct from O on the circle of diameter OF¢. Its image P must satisfy ÐOFP = ÐOP¢F¢ = 90° and so it lies on the line perpendicular to OF. Hence the image of the line is the entire circle apart from O.
- To construct our example, let \frakL be a line at distance 1 from O, with F the foot of the perpendicular from O to the line. Let q be any rational number with 0 < q < 1 and let Pq be selected on \frakL with ÐPqOF = 2arctanq. Then tanÐPq OF = 2q(1 - q2)-1 so that |FPq | = 2q(1 - q2)-1 and |OPq | = (1 + q2)(1 - q2)-1. Hence all pairs of points Pq are rational distance apart and each Pq is a rational distance from O. Invert this line with respect to any circle with centre O and rational radius to obtain a circle through O. All images of points Pq lie on this circle, so no three are collinear. We can arrange these points in a sequence which satisfies the requirements.
- 5.
- Second solution. See Figure 5.1. (a) Suppose that |A¢B¢| = d¢ and let ÐAOB = a. By the Law of Cosines, d2 = r2 + s2 - 2rscosa. Since |OA¢| = k2/r and |OB¢| = k2/s,
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(b) In the cartesian plane, let \frakC be the circle
of radius 1 and centre (0, 0). Consider the line \frakL with
equation x = 1. A point (x, y) is on the image of this line
under inversion with respect to \frakC if and only if x > 0 and
the point (1, y/x) on the ray through (0, 0) and (x, y)
satisfies (x2 + y2)(1 + y2/x2) = 1. This is equivalent
to x2 + y2 = x or (x - 1/2)2 + y2 = 1/4.
Thus, the image of the line \frakL is the circle of radius
1/2 and centre (1/2, 0), and no three
points on this circle are collinear.
- To solve the problem, we select a sequence { Un } of points on \frakL for which |OUn | is rational and Un has rational coordinates, and let { Xn } be the images of these with respect to inversion in \frakC. But such a selection is possible since there are infinitely many rational pythagorean triples whose smallest number is 1. For example, we can take
Un ~ æ
ç
è1, 2n2 + 2n
2n + 1ö
÷
ø
- 5.
- Third solution. [Y. Shen] An alternative approach to the solution of the second part comes by the use of Ptolemy's Theorem. See Figure 5.3. Let a circle of diameter l be given and two additional points on the circle be given whose chords make angles a and b with the diameter. If the points are on the same side of the diameter and are distant l apart, we have, by the Law of Cosines,
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- 6.
- (a) First solution. Let u = x + y and v = xy. Then u + v = 2 + 3Ö2 and u2 - 2v = 6. Thus, u2 + 2u = 10 + 6Ö2 so that (u + 1)2 = 11 + 6Ö2 = (3 + Ö2)2. Therefore
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- 6.
- (a) Second solution. [D. Nicholson] Observe that (x + y + 1)2 = (x2 + y2) + 2(x + y + xy) + 1 = 11 + 6Ö2 = (3 + Ö2)2. Thus x + y = 2 + Ö2 or -4 - 2Ö2, and we can proceed as in the first solution.
- 6.
- (b) First solution. The first equation can be transformed to
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- Suppose that x + y = 1. Then the second equation becomes x2 - (1 - x) = 1 or 0 = x2 + x - 2 = (x + 2)(x - 1). Hence (x, y) = (-2, 3), (1, 0) and both these solutions check out.
- Suppose that (x2 + y2) + (x + y) = 0. Then x and y cannot be real. Otherwise, x + y = -(x2 + y2) £ 0. By the first equation, x + y ¹ 0 and by the second x + y > 0 and we obtain a contradiction. From the second equation
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- Let x2 + y2 + x + y = 0 and y = x2 - x. Then x2 = x + y = -(x2 + y2) Þ y2 = -2x2 Þy = ±iÖ2x Þ x2 = (1 ±iÖ2)x. Since x + y must be nonzero, x = 0 is inadmissible. Hence (x, y) = (1 + iÖ2, -2 + iÖ2), (1 - iÖ2,-2 - iÖ2), and both of these check out.
- Let x2 + y2 + x + y = 0 and y = x2 + x + 1. Then y2 + 2y - 1 = 0 or (y + 1)2 = 2. Hence y = -1 ±Ö2, so that x2 + x + (2 -±Ö2) = 0. Hence
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- 6.
- (b) Second solution. Let x + y = u2 so that u = x2 - y = x2 -(u2 - x), Therefore 0 = x2 - u2 + (x - u) = (x - u)(x + u + 1), whence y = x2 - x or y = x2 + x + 1.
- Plugging y = x2 - x into the first equation yields
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- Plugging y = x2 + x + 1 = (x+1)2 - x into the first equation and using x + y = (x + 1)2 yields
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- Comment. Since
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We can get a picture of the situation for real solutions.
See Figure 6(b). The equation Ö[(x+y)] = x2 - y
requires x + y ³ 0 and y ³ x2 to be viable for
a real solution. The locus of this pair of inequalities is
hatched. The first of the two given equations also requires
x + y ¹ 0. Hence there are no real solutions with
x2 + y2 + x + y = 0. However, when x + y = 1 there
is one real solution (x, y) = (1, 0) for which
y = x2 - x and one real solutions (x, y) = (-2, 3)
for which y = x2 + x + 1. The putative solutions
(x, y) = (-1, 2), (0, 1) lie outside the hatched region and
so are extraneous.
