CMS/SMC
Canadian Mathematical Society
www.cms.math.ca
Canadian Mathematical Society
  location: 
       
>

Solutions and comments.


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- x1 y3 - x2 y1 + x2 y3 )( x0 y0 - x0 y2 - x3 y0 + x3 y2 )=(1- x1 y2 - x3 y1 + x3 y2 )( x0 y0 - x0 y3 - x2 y0 + x2 y3 ).

Collecting up the terms in x0 y0 , x0 , y0 , and the rest, and simplifying, yields that x0 y0 =1, as desired.


98.
Let a1 , a2 ,, an+1 , b1 , b2 ,, bn be nonnegative real numbers for which
(i) a1 a2 an+1 =0,
(ii) 0 bk 1 for k=1,2,,n.
Suppose that m= b1 + b2 ++ bn +1. Prove that

k=1 n ak bk k=1 m ak .

Solution. Note that m-1 b1 + b2 ++ bm <m. We have that

a1 b1 + a2 b2 + + am bm + am+1 bm+1 ++ an bn a1 b1 + a2 b2 ++ am bm + am ( bm+1 + bm+2 ++ bn ) < a1 b1 + a2 b2 ++ am bm + am (m- b1 - b2 -- bm ) = a1 b1 + a2 b2 ++ am bm + am (1- b1 )+ am (1- b2 )++ am (1- bm ) a1 b1 + a2 b2 ++ am bm + a1 (1- b1 )+ a2 (1- b2 )++ am (1- bm ) = a1 + a2 ++ 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, DEA= 180ˆ -BED=DCF and DFC= 180ˆ -DFB=DAB. Since, also, AE=CF, ΔDAEΔDFC (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. CDF=CDE-FDE= 180ˆ -ABC-FDE=FDA-FDE=EDA and AED= 180ˆ -BED=BCD=FCD. Since AE=CF, ΔEADΔCFD (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 ABC.
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

( 2(m+a) m2 +1 , m2 +2am-1 m2 +1 )

and the circle of equation (x-b)2 + y2 = b2 +1 in the point

( 2(m+b) m2 +1 , m2 +2bm-1 m2 +1 )

The distance between these two points is the square root of

[ 2(a-b) m2 +1 ]2 + [ 2m(a-b) m2 +1 ]2 = 4(a-b)2 (1+ m2 ) ( m2 +1)2 = 4(a-b)2 m2 +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 m1 2 +1= m2 2 +1 or 0=( m1 - m2 )( m1 + m2 ). Since the sides are distinct, it follows that m1 =- m2 and so BD bisects ABC.


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 ABDIEH and BEAFHI, 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 DOU= 72ˆ , DAB= 36ˆ , ABU= 72ˆ , DUO=BUA= 72ˆ , 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 AOB= 36ˆ , DOB=OBA= 72ˆ , BOE= 108ˆ and OEB=OBE= 36ˆ . Also, EOV=EVO= 72ˆ and OE=EV=OA=OB. Hence, ΔDAB=ΔEVO (SAS), so that OV=AB. Since BVO= 108ˆ and BOV= 36ˆ , OBV= 36ˆ , 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 36ˆ , so its length is 2sin 18ˆ . Likewise, the length of a side of Q is 2sin 54ˆ . The difference between these is

2sin 54ˆ -2sin 18ˆ =2cos 36ˆ -2cos 72ˆ =2t-2(2 t2 -1)=2+2t-4 t2

where t=cos 36ˆ . Now

t =cos 36ˆ =-cos 144ˆ =1-2cos2 72ˆ =1-2(2 t2 -1)2 =-8 t4 +8 t2 -1,

so that

0 =8 t4 -8 t2 +t+1=(t+1)(8 t3 -8 t2 +1) =(t+1)(2t-1)(4 t2 -2t-1).

Since t is equal to neither -1 nor 1 2 , we must have that 4 t2 -2t=1. Hence

2sin 54ˆ -2sin 18ˆ =2-(4 t2 -2t)=1,

the radius of the circle.


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

a2 u3 + b2 v3 1.

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

2 a5 +3 u5 5 = a5 + a5 + u5 + u5 + u5 5 a10 u15 5= a2 u3

and, similarly,

2 b5 +3 v5 5 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 ΔACD 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 ΔADC and ΔABC 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 ΔACB is right, 1= t2 + s2 = t2 + t4 , whence s= t2 = 1 2 (-1+5)<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 ΔADB is folded flat partially covering ΔACB. 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 α=CAB and AB=1. By the condition on the acute angles of triangles ACB and ACD, ACB=ADB= 90ˆ , so that the triangles ΔACD and ΔADB, being similar and sharing a hypotenuse, are congruent.
Suppose, if possible, that BAD=α. Then AC=AD and so ΔACD must be isosceles with its right angle at A, contrary to hypothesis. So, ABD=α and BD=AC=cosα, AD=BC=sinα.
Consider ΔACD. Suppose that ACD= 90ˆ . If DAC=α, then ΔABCΔADC and 1=AB=AD=sinα, yielding a contradiction. Hence ADC=α, AD=AC/sinα=cosα/sinα and CD=ACcotα=cos2 α/sinα. Hence, looking at AD, we have that

cosα sinα =sinα0=cosα-sin2 α=cos2 α+cosα-1.

Therefore, cosα= 1 2 (5-1) and sin2 α=cosα.
Observe that BCsinα=sin2 α=cosα=BD and BCcosα=sinαcosα=cos2 α/sinα=CD, so that triangle BCD is right with CDB= 90ˆ 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,cosα,0)B~(sinα,0,0).

Since ACD= 90ˆ , D lies in the plane y=0 and so has coordinates of the form (x,0,z). Since CDB= 90ˆ , CDDB, so that

0=(x,0,z)·(x-sinα,0,z)- x2 + z2 -xsinα,

Now CD=cosαsinα forces cos2 αsin2 α= x2 + z2 . Hence

xsinα=cos2 αsin2 αx=cos2 αsinα.

Therefore

z2 =(cos2 α-cos4 α)sin2 α=cos2 αsin4 αz=cosαsin2 α,

Hence D~(cos2 αsinα,0,cosαsin2 α).
Thus, letting sinα=t= 1 2 (5-1), 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 (5-1)]3/2 =1:2+5.
We need to dispose of the other possibilities for ΔACD. By the given condition, DAC 90ˆ . If ADC= 90ˆ , then we have essentially the same situation as before with the roles of α 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.

© Canadian Mathematical Society, 2014 : https://cms.math.ca/