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Notes. A rectangular hyperbola is an hyperbola whose asymmptotes are at right angles.

97.
A triangle has its three vertices on a rectangular hyperbola. Prove that its orthocentre also lies on the hyperbola.
Solution 1. A rectangular hyperbola can be represented as the locus of the equation xy = 1. Let the three vertices of the triangle be at (a, 1/a), (b, 1/b), (c, 1/c). The altitude to the points (c, 1/c) has slope -(a - b)/(a-1 - b-1) = ab and its equation is y = abx + (1/c) - abc. The altitude to the point (a, 1/a) has equation y = bcx + (1/a) - abc. These two lines intersect in the point (-1/abc, -abc) and the result follows.

Solution 2. [R. Barrington Leigh] Suppose that the equation of the rectangular hyperbola is xy = 1. Let the three vertices be at (xi, yi) (i = 1, 2, 3), and let the orthocentre be at (x0, y0). Then

 (x1 - x2)(x0 - x3) = -(y1 - y2)(y0 - y3)
and

 (x1 - x3)(x0 - x2) = -(y1 - y3)(y0 - y2) .
Cross-multiplying these equations yields that

 (x1 - x2)(y1 - y3)(x0 - x3)(y0 - y2) = (x1 - x3)(y1 - y2)(x0 - x2)(y0 - y3) ,
whence

 (1 - x1y3 - x2y1 + x2y3)(x0y0 - x0y2 - x3y0 + x3y2) = (1 - x1y2 - x3y1 + x3y2)(x0y0 - x0y3 - x2y0 + x2y3) .
Collecting up the terms in x0y0, x0, y0, and the rest, and simplifying, yields that x0y0 = 1, as desired.

98.
Let a1, a2, ¼, an+1, b1, b2, ¼, bn be nonnegative real numbers for which
(i) a1 ³ a2 ³ ¼ ³ an+1 = 0,

(ii) 0 £ bk £ 1 for k = 1, 2, ¼, n.

Suppose that m = ëb1 + b2 + ¼+ bn û+ 1. Prove that

 nå k=1 ak bk £ må k=1 ak .
Solution. Note that m-1 £ b1 + b2 + ¼+ bm < m. We have that

 a1b1 + a2b2 + ¼
 + am bm+ am+1 bm+1 + ¼+ an bn
 £ a1 b1 + a2 b2 + ¼+ am bm + am (bm+1 + bm+2 + ¼+ bn)
 < a1 b1 + a2 b2 + ¼+ am bm+ am (m - b1 - b2 - ¼- bm)
 = a1 b1 + a2 b2 + ¼+ am bm + am (1 - b1)+ am (1 - b2) + ¼+ am (1 - bm)
 £ a1 b1 + a2 b2 + ¼+ am bm +a1 (1 - b1) + a2 (1 - b2) + ¼+ am (1 - bm)
 = a1 + a2 + ¼+ am  .

99.
Let E and F be respective points on sides AB and BC of a triangle ABC for which AE = CF. The circle passing through the points B, C, E and the circle passing through the points A, B, F intersect at B and D. Prove that BD is the bisector of angle ABC.
Solution 1. Because of the concyclic quadrilaterals, ÐDEA = 180° - ÐBED = ÐDCF and ÐDFC = 180° - ÐDFB = ÐDAB . Since, also, AE = CF, DDAE º DDFC (ASA) so that AD = DF. In the circle through ABFD, the equal chords AD and DF subtend equal angles ABD and FBD at the circumference. The result follows.

Solution 2. ÐCDF = ÐCDE - ÐFDE = 180° - ÐABC - ÐFDE = ÐFDA - ÐFDE = ÐEDA and ÐAED = 180° - ÐBED = ÐBCD = ÐFCD. Since AE = CF, DEAD º DCFD (ASA). The altitude from D to AE is equal to the altitude from D to FC, and so D must be on the bisector of ÐABC.

Solution 3. Let B be the point (0, -1) and D the point (0, 1). The centres of both circles are on the right bisector of BD, namely the x-axis. Let the two circles have equations (x - a)2 + y2 = a2 + 1 and (x - b)2 + y2 = b2 + 1. Suppose that y = mx - 1 is a line through B; this line intersects the circle of equation (x - a)2 + y2 = a2 + 1 in the point

 æç è 2(m+a)m2 + 1 , m2 + 2am - 1m2 + 1 ö÷ ø
and the circle of equation (x - b)2 + y2 = b2 + 1 in the point

 æç è 2(m+b)m2 + 1 , m2 + 2bm - 1m2 + 1 ö÷ ø
The distance between these two points is the square root of

 éê ë 2(a-b)m2 + 1 ùú û 2 + éê ë 2m(a-b)m2 + 1 ùú û 2 = 4(a-b)2(1 + m2)(m2 + 1)2 = 4(a-b)2m2 + 1 .

Now suppose that the side AB of the triangle has equation y = m1 x - 1 and the side BC the equation y = m2 x - 1, so that (A, E) and (C, F) are the pairs of points where the lines intersect the circles. Then, from the foregoing paragraph, we must have m12 + 1 = m22 + 1 or 0 = (m1 - m2)(m1 + m2). Since the sides are distinct, it follows that m1 = -m2 and so BD bisects ÐABC.

100.
If 10 equally spaced points around a circle are joined consecutively, a convex regular inscribed decagon P is obtained; if every third point is joined, a self-intersecting regular decagon Q is formed. Prove that the difference between the length of a side of Q and the length of a side of P is equal to the radius of the circle. [With thanks to Ross Honsberger.]
Solution 1. Let the decagon be ABCDEFGHIJ. Let BE and DI intersect at K and let AF and DI intersect at L. Observe that AB || DI || EH and BE || AF || HI, so that ABKL and KIHE are parallelograms. Now AB is a side of P and HE is a side of Q, and the length of the segment IL is the difference of the lengths of EH = IK and AB = KL. Since L, being the intersection of the diameters AF and DI, is the centre of the circle, the result follows.

Solution 2. [R. Barrington Leigh] Use the same notation as in Solution 1. Let O be the centre of P. Now, AB is an edge of P, AD is an edge of Q, DO is a radius of the circle and BG a diameter. Let AD and BO intersect at U. Identify in turn the angles ÐDOU = 72°, ÐDAB = 36°, ÐABU = 72°, ÐDUO = ÐBUA = 72°, whence AU = AB, DU = DO and AD - AB = AD - AX = DX = DO, as desired.

Solution 3. Label the vertices of P as in Solution 1. Let O be the centre of P, and V be a point on EB for which EV = OE. We have that ÐAOB = 36°, ÐDOB = ÐOBA = 72°, ÐBOE = 108° and ÐOEB = ÐOBE = 36°. Also, ÐEOV = ÐEVO = 72° and OE = EV = OA = OB. Hence, DDAB = DEVO (SAS), so that OV = AB. Since ÐBVO = 108° and ÐBOV = 36°, ÐOBV = 36°, and so BV = OV = AB. Hence BE - AB = EV + BV - AB = EV = OE, the radius.

Solution 4. Let the circumcircle of P and Q have radius 1. A side of P is the base of an isosceles triangle with equal sides 1 and apex angle 36°, so its length is 2sin18°. Likewise, the length of a side of Q is 2sin54°. The difference between these is

 2sin54° - 2sin18° = 2cos36°- 2cos72° = 2t - 2(2t2 - 1) = 2 + 2t - 4t2
where t = cos36°. Now

 t
 = cos36° = - cos144° = 1 - 2cos2 72°
 = 1 - 2(2t2 - 1)2 = -8t4 + 8t2 - 1 ,
so that

 0
 = 8t4 - 8t2 + t + 1 = (t + 1)(8t3 - 8t2 + 1)
 = (t + 1)(2t - 1)(4t2 - 2t - 1) .
Since t is equal to neither -1 nor 1/2, we must have that 4t2 - 2t = 1. Hence

 2sin54° - 2sin18° = 2 - (4t2 - 2t) = 1 ,

101.
Let a, b, u, v be nonnegative. Suppose that a5 + b5 £ 1 and u5 + v5 £ 1. Prove that

 a2u3 + b2v3 £ 1 .
[With thanks to Ross Honsberger.]
Solution. By the arithmetic-geometric means inequality, we have that

 2a5 + 3u55 = a5 + a5 + u5 + u5 + u55 ³ Ö[5 ]a10u15 = a2u3
and, similarly,

 2b5 + 3v55 ³ b2 v3 .
Adding these two inequalities yields the result.

102.
Prove that there exists a tetrahedron ABCD, all of whose faces are similar right triangles, each face having acute angles at A and B. Determine which of the edges of the tetrahedron is largest and which is smallest, and find the ratio of their lengths.
Solution 1. Begin with AB, a side of length 1. Now construct a rectangle ACBD with diagonal AB, so that |AC | = |BD | = s < t = |AD | = |BC |. The requisite values of s and t will be determined in due course. We want to show that we can fold up D and C from the plane in which AB lies (like folding up the wings of a butterfly) in such a way that we can obtain the desired tetrahedron.

When the triangles ADB and ACB lie flat, we see that C and D are distance 1 apart. Suppose that, when we have folded up C and D to get the required tetrahedron, they are distance r apart. Then ACD should be a right triangle similar to ABC. The hypotenuse of DACD cannot be AC as AC < AD. Nor can it be CD, for then, we would have AD = BC, AC = AC, and CD would have to have length 1, possible only when ABCD is coplanar. So the hypotenuse must be AD. The similarity of DADC and DABC would require that

 1 : t : s = t : s : r
where r = |CD |. Thus, 1/t = t/s or s = t2 and t/s = s/r or r = s2/t = t3. So we must fold C and D until they are distance t3 apart.

Is this possible? Since DACB is right, 1 = t2 + s2 = t2 + t4, whence s = t2 = 1/2(-1 + Ö5) < 1. Hence r < 1. To arrange that we can make the distance between C and D equal to r, we must show that r exceeds the minimum possible distance between C and D, which occurs when DADB is folded flat partially covering DACB. Suppose this has been done, with ABCD coplanar and C, D both on the same side of AB. Let P and Q be the respective feet of the perpendiculars to AB from C and D. Then

 |CP | = |DQ | = t3 ,   |AP | = |QB | = t4 ,   |AQ | = |PB | = t2 ,
and

 |CD | = |PQ | = t2 - t4 = (t4 + t6) - t4 = t6 < t3 .

When C and D are located, we have |AB | = 1, |AD | = |BC | = t, |AC | = |BD | = t2 and |CD | = t3. Since all faces of the tetrahedron ABCD have sides in the ratio 1 : t : t2, all are similar right triangles and AB : CD = 1 : t3.

Solution 2. Let a = ÐCAB and |AB | = 1. By the condition on the acute angles of triangles ACB and ACD, ÐACB = ÐADB = 90°, so that the triangles DACD and DADB, being similar and sharing a hypotenuse, are congruent.

Suppose, if possible, that ÐBAD = a. Then AC = AD and so DACD must be isosceles with its right angle at A, contrary to hypothesis. So, ÐABD = a and |BD | = |AC | = cosa, |AD | = |BC | = sina.

Consider DACD. Suppose that ÐACD = 90°. If ÐDAC = a, then DABC º DADC and 1 = |AB | = |AD | = sina, yielding a contradiction. Hence ÐADC = a, |AD | = |AC |/sina = cosa/sina and |CD | = |AC |cota = cos2 a/sina. Hence, looking at |AD |, we have that

 cosasina = sinaÞ 0 = cosa- sin2 a = cos2 a+ cosa- 1 .
Therefore, cosa = 1/2(Ö5 - 1) and sin2 a = cosa.

Observe that |BC |sina = sin2 a = cosa = |BD | and |BC |cosa = sinacosa = cos2 a/sina = |CD |, so that triangle BCD is right with ÐCDB = 90° and similar to the other three faces.

We need to check that this set-up is feasible. Using spatial coordinates, take

 C ~ (0, 0, 0)      A ~ (0, cosa, 0)     B ~ (sina, 0, 0) .
Since ÐACD = 90°, D lies in the plane y = 0 and so has coordinates of the form (x, 0, z). Since ÐCDB = 90°, CD ^DB, so that

 0 = (x, 0, z)·(x - sina, 0, z) -x2 + z2 - xsina ,
Now |CD | = cosasina forces cos2 asin2 a = x2 + z2. Hence

 xsina = cos2 asin2 aÞ x = cos2 asina .
Therefore

 z2 = (cos2 a- cos4 a)sin2 a = cos2 asin4 aÞz = cosasin2 a ,
Hence D ~ (cos2 asina, 0, cosasin2 a).

Thus, letting sina = t = 1/2(Ö5 - 1), we have A ~ (0, t2, 0), B ~ (t, 0, 0), C ~ (0, 0, 0), D ~ (t5, 0, t4) with t4 + t2 - 1 = 0, and |AB | = 1, |AD | = |BC | = t, |BD | = |AC | = t2 and |CD | = t3. [Exercise: Check that the coordinates give the required distances and similar right triangles.] The ratio of largest to smallest edges is 1 : t3 = 1 : [1/2(Ö5 - 1)]3/2 = 1 : Ö{2 + Ö5}.

We need to dispose of the other possibilities for DACD. By the given condition, ÐDAC ¹ 90°. If ÐADC = 90°, then we have essentially the same situation as before with the roles of a and its complement, and of C and D switched.

Comment. Another way in that was used by several solvers was to note that there are four right angles involved among the four sides, and that at most three angles can occur at a given vertex of the tetrahedron. It is straightforward to argue that it is not possible to have three of the right angles at either C or D. Since all right angles occur at these two vertices, then there must be two at each. As an exercise, you might want to complete the argument from this beginning.

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