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

       (a) AC = AB;

       (b) ÐBAC = 90^°.

View solution

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

View solution

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

(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

(b) Using (a), or otherwise, show that there exists a sequence { X_n } 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)

View solution

Solutions

Problem 1

1.

First solution. See Figure 1.1. Let |AB | = u. Then

|BD |+ |AD | = u sin100° sin60° + u sin20° sin60° = u sin60° (sin100° + sin20°) = 2usin60°cos40° sin60° = 2ucos40° = |BC | .

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

BC = BF + FC = BD + DF = BD + DE = BD + AD .

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.

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Problem 2.

2.

First solution. See Figure 2.1. Let AD and EF meet at H. Then

ÐDHF = ÐHEI + ÐHIE = ÐFEB + ÐAIE = ÐFCB + (ÐABI + ÐBAI) = 1 2 ÐACB + 1 2 ÐABC + 1 2 ÐBAC = 90°

as desired.

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

D ~ cos(b+ g) + i sin(b+ g) ,

E ~ cos(g+ p) + i sin(g+ p) = - [cosg+ i sing] ,

F ~ cosb+ i sinb .

The vector EF is given by

(cosb+ cosg) + i(sinb+ sing) = cos b- g 2 é ê ë cos b+ g 2 + i sin b+ g 2 ù ú û

and the vector AD by

(cos(b+ g) - 1) + i sin(b+ g) = 2 sin b+ g 2 é ê ë -sin b+ g 2 + i cos b+ g 2 ù ú û = 2 sin b+ g 2 é ê ë cos b+ g 2 + i sin b+ g 2 ù ú û i

from which it can be seen that they are perpendicular.

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.

Hence ÐAXY = ÐFXB = ÐFIB = ÐEIC = ÐEYC = ÐAYX, so that AX = XY. Since DAXY is isosceles and AD bisects ÐXAY, then AD ^FE.

2.

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.

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Problem 3.

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

DPCQ ~ DDAQ      and      DPBR ~ DDAR .

A rotation about Q followed by a dilatation takes C to A and P to D so that

CQ:PQ = AQ:DQ .

Since ÐCQA = ÐPQD, DCAQ ~ DPDQ so that AC:PD = AQ:DQ. Similarly, DBAR ~ DPDR and AB:PD = AR:DR. Since AQ = AR and DQ = DR, it follows that AC = AB.

(b) By the similar triangles identified in (a), ÐCAQ = ÐPDQ and ÐBAR = ÐPDR. Hence

ÐCAB = ÐQAR - (ÐCAQ + ÐBAR) = ÐQAR - (ÐPDQ + ÐPDR) = ÐQAR - ÐQDR = 150° - 60° = 90° .

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,

ÐBAC = ÐBAS + ÐSAC = ÐBAS + ÐRAB = ÐRAS = 90° .

3.

Third solution. [D. Brox] Note that sin75^° = cos15^° = (Ö6 + Ö2)/4 and sin15^° = (Ö6 - Ö2)/4. Let

a = æ ç è Ö6 - Ö2 2 ö ÷ ø æ ç è Ö2 2 ö ÷ ø = Ö3 - 1 2  .

We solve the problem using vectors in the complex plane. Let lower case letters correspond to the points in the plane given in upper case, so that a corresponds to A, et cetera.

Since |r - a | = |q - a | = |r - q |/2 cos15^° = ¹/₂(Ö6 -Ö2) |r - q |, we have that

q - a = Ö6 - Ö2 2 æ ç è Ö6 + Ö2 4 - Ö6 - Ö2 4 i ö ÷ ø (q - r) = a æ ç è Ö3 + 1 2 - Ö3 - 1 2 i ö ÷ ø ((q - p) - (r - p))

and

r - a = Ö6 - Ö2 2 æ ç è Ö6 + Ö2 4 + Ö6 - Ö2 4 i ö ÷ ø (r - q) = a æ ç è Ö3 + 1 2 + Ö3 - 1 2 i ö ÷ ø ((r - p) - (q - p)) .

Applying the sine law yields

|b - r | = |r - p |sin30°/sin105° = 1 2 (Ö6 - Ö2) |p - r |    and    |c - q | = 1 2 (Ö6 - Ö2)|q - p |

so that

b - r = Ö6 - Ö2 2 æ ç è Ö2 2 - Ö2 2 i ö ÷ ø (p - r) = a(i - 1)(r - p)

and

c - q = Ö6 - Ö2 2 æ ç è Ö2 2 + Ö2 2 i ö ÷ ø (p - q) = -a(i + 1)(q - p) .

Hence

c - a = (q - a) + (c - q) = a é ê ë æ ç è Ö3 - 1 2 - Ö3 + 1 2 i ö ÷ ø (q - p)- æ ç è Ö3 + 1 2 - Ö3 - 1 2 i ö ÷ ø (r - p) ù ú û = ia é ê ë - æ ç è Ö3 + 1 2 + Ö3 - 1 2 i ö ÷ ø (q - p) + æ ç è Ö3 - 1 2 + Ö3 + 1 2 i ö ÷ ø (r - p) ù ú û = i[(r - a) + (b - r)] = i(b - a)

from which it follows that AC and AB are perpendicular segments of equal length.

3.

Fourth solution. [D. Pritchard]

Lemma: For any triangle with angles a, b, g,

sin2 a+ sin2 g- 2sinasingcos(b+ 60°) = sin2 b+ sin2 g- 2 sinbsingcos(a+ 60°) .

Proof: Taking the difference between the two sides yields

sin2 a- sin2 b- sing[2 sinacos(b+ 60°) - 2 sinbcos(a+ 60°)] = 1 2 [cos2b- cos2a]- sin(a+ b)[ sin(a+ b+ 60°)+ sin(a- b- 60°)                         - sin(b+ a+ 60°) - sin(b- a- 60°)] = sin(a+ b) sin(a- b)- sin(a+ b)[sin(a- b)cos60° - cos(a-b) sin60°                         - sin(b- a)cos60° + cos(b- a) sin60°] = sin(a+ b) [sin(a- b) -2sin(a- b) cos60° ] = 0   .

(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

|AR | = sinP sin15° sin150° = Ö3 - 1 Ö2 sinP

|RB | = sinQ sin30° sin105° = Ö3- 1 Ö2 sinQ

|AB |2 = æ ç è Ö3 - 1 Ö2 ö ÷ ø 2   [sin2 P + sin2 Q - 2 sinP sinQ cos(R + 60°)]

Similarly

|AC |2 = æ ç è Ö3 - 1 Ö2 ö ÷ ø 2   [sin2 P + sin2 R - 2 sinP sinR cos(Q + 60°)]

Hence AB = AC.

(b) |BP | = sinQ sin45^°/sin105^° = (Ö3 - 1)sinQ, |CP | = (Ö3 - 1)sinR and

|BC |2 = (Ö3 - 1)2[(sin2 R + sin2 Q - 2sinR sinQ cos(P + 60°)]

from which |BC |² = 2|AB |², and the result follows.

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.

|PQ | = |QC |cos45°+ |CF |cot30° = 1 Ö2 (1 + Ö3)w

|PR | = 1 Ö2 (1 + Ö3)v

and

|QR | = 2ucos15° = 1 Ö2 (1 + Ö3)u .

Hence

cosa = v2 + w2 - u2 2vw

cosb = u2 + w2 - v2 2uw

cosg = u2 + v2 - w2 2uv  .

Then

|AC |2 = u2 + w2 - 2uw cos(b+ 60°) = 1 2 (u2 + v2 + w2) +Ö3 uw sinb

|AB |2 = u2 + v2 - 2uv cos(g+ 60°) = 1 2 (u2 + v2 + w2) +Ö3 uv sing  .

Now |PC | = |FC |csc 30^° = 2|FC | = Ö2w and |PB | = Ö2v. Hence

|BC |2 = 2v2 + 2w2 - 4vw cos(a+ 60°) = u2 + v2 + w2 + 2Ö3 vw sina .

Since

sina: sinb: sing = |QR |: |PR |: |PQ | = u : v : w  ,

uw sinb = uv sing = vw sina .

Thus |AC | = |AB | and |BC |² = |AC |² + |AB |², yielding the desired result.

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

a2 + b2 - 2ab cos(C + 60°) = b2 sin2 A sin2 B + b2 - 2b2 sinA sinB æ ç è cosC 2 - Ö3sinC 2 ö ÷ ø = b2 sin2 B [ sin2 A + sin2 B -sinA sinB cosC + Ö3sinA sinB sinC ] = b2 sin2 B [sin2 B + sin2 A -sinA cosC (sinA cosC + cosA sinC) +Ö3sinA sinB sinC ] = b2 sin2 B [sin2 B + sin2 A sin2 C- sinA cosA sinC cosC + Ö3sinA sinB sinC ] .

Since this is symmetric in A and C, it is equal to b² + c² - 2bc cos(A + 60^°), and, by further symmetry, to c² + a² - 2ac cos(B + 60^°).

(a) Let D be constructed as in the figure, and note that QC:QA = r:p by similar triangles. By the Law of Cosines,

AC2 = QC2 + AQ2 - 2QC·QA cos(Q +60°) = QA2 p2 [ r2 + p2 - 2pr cos(Q + 60°)]

and

AB2 = RA2 p2 [q2 + p2 - 2qp cos(R +60°) ]  .

An application of the preliminary result to DPQR yields AB = AC.

(b) PC:AD = r:p and PB:AD = q:p. Hence, by the Law of Cosines applied to DPCB,

BC2 = AD2 p2 [ r2 + q2 - 2qr cos(P +60°)] .

Since p = QR = 2QAcos15^° and

AD = QDcos30° - QAsin15° = p[cos30° - 1 2 tan15°] = p 2 [ Ö3 - (2 - Ö3)] = (Ö3- 1)p ,

so BC² = (4 - 2Ö3)[r² + q² - 2qrcos(P + 60^°)]. But

AB2 + AC2 = 2QA2 p2 [r2 + p2 - 2prcos(Q + 60°)] = 1 2 sec2 15°[r2 + q2 - 2qr cos(P + 60°)] = BC2

as desired.

3.

Seventh solution. [Y. Shen] (a) From the Sine Law,

AR = AQ = QR (2 sin15°)

BR = PR/2 sin75°

and

CQ = PQ/2 sin75° .

Hence

AB2 = AR2 + BR2 - 2AR·BR cos(60° + ÐPRQ) = 4 sin2 15° QR2 + PR2 2(1 - cos150°) -2 RQ ·PR tan15° æ ç è 1 2 · PR2 + QR2 - PQ2 2 PR ·QR - Ö3 2 ·sinÐPRQ ö ÷ ø = (2 - Ö3) QR2 + (2 - Ö3) PR2 - 1 2 (2 - Ö3)(PR2 + QR2 - PQ2) +(4 Ö3 - 6)[PQR] = 1 2 (2 - Ö3)(PQ2 + QR2 + PR2) +(4Ö3 - 6)[PQR] .

By symmetry, AC² = AB², so AC = AB.

(b)

BP = Ö2 2 · RP sin75°  ,       CP = Ö2 2 · PQ sin75°  .

Then

BC2 = BP2 + CP2 - 2BP ·CP cos(60° + ÐQPR) = 2 PR2 (2 - Ö3) + 2PQ2(2 -Ö3) - (2 - Ö3)(PR2 + PQ2 - QR2)+ (8Ö3 - 12)[PQR] = AB2 + AC2

so that ÐBAC = 90^°.

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Problem 5.

5.

First solution. See Figure 5.1. Since ÐO is common and OA:OB = OB¢:OA¢, triangles OAB and OB¢A¢ are similar. Hence

r:s:d = k2 s : k2 r :|A¢B¢|

from which it follows that |A¢B¢| = (k²d)/(rs).

(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 P_q be selected on \frakL with ÐP_qOF = 2arctanq. Then tanÐP_q OF = 2q(1 - q²)^-1 so that |FP_q | = 2q(1 - q²)^-1 and |OP_q | = (1 + q²)(1 - q²)^-1. Hence all pairs of points P_q are rational distance apart and each P_q 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 P_q 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,

d¢2 = k4 r2 + k4 s2 - 2k4 rs cosa = k4 r2 s2 [r2 + s2 - 2rscosa] = k4d2 r2 s2

as desired. Observe that if k, r, s are rational, then d¢ is rational if and only if d is rational.

(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 { U_n } of points on \frakL for which |OU_n | is rational and U_n has rational coordinates, and let { X_n } 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 è 1, 2n2 + 2n 2n + 1 ö ÷ ø

so that |OU_n | = (2n² + 2n + 1)/(2n + 1).

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,

l2 = cos2 a+ cos2 b-2cosacosb(cosacosb- sinasinb) = cos2 a(1 - cos2 b) +cos2 b(1 - cos2 a) + 2cosacosbsinasinb) = (cosasinb+ sinacosb)2

so that l = sinacosb+ cosasinb. On the other hand, if the points are on the same side of the diameter, then l² = cos² a+ cos² b- 2cosacosb(cos(a- b)) so that l = cosasinb-cosbsina. We can locate infinitely many points on the circumference of the circle for which the sine and cosine of the angle its chord makes with a given diameter are rational, and the distance between any pair of these points will be rational.

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