>
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
$\mathrm{xy}=1$. 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
$(ab)/({a}^{1}{b}^{1})=\mathrm{ab}$ and its equation is
$y=\mathrm{abx}+(1/c)\mathrm{abc}$. The altitude to
the point
$(a,1/a)$ has equation
$y=\mathrm{bcx}+(1/a)\mathrm{abc}$. These
two lines intersect in the point
$(1/\mathrm{abc},\mathrm{abc})$ and the result
follows.
Solution 2. [R. Barrington Leigh] Suppose that the equation
of the rectangular hyperbola is
$\mathrm{xy}=1$. Let the three vertices
be at
$({x}_{i},{y}_{i})$ (
$i=1,2,3$), and let the orthocentre be
at
$({x}_{0},{y}_{0})$. Then
$({x}_{1}{x}_{2})({x}_{0}{x}_{3})=({y}_{1}{y}_{2})({y}_{0}{y}_{3})$
and
$({x}_{1}{x}_{3})({x}_{0}{x}_{2})=({y}_{1}{y}_{3})({y}_{0}{y}_{2})\hspace{1em}.$
Crossmultiplying these equations yields that
$({x}_{1}{x}_{2})({y}_{1}{y}_{3})({x}_{0}{x}_{3})({y}_{0}{y}_{2})=({x}_{1}{x}_{3})({y}_{1}{y}_{2})({x}_{0}{x}_{2})({y}_{0}{y}_{3})\hspace{1em},$
whence
$(1{x}_{1}{y}_{3}{x}_{2}{y}_{1}+{x}_{2}{y}_{3})({x}_{0}{y}_{0}{x}_{0}{y}_{2}{x}_{3}{y}_{0}+{x}_{3}{y}_{2})=(1{x}_{1}{y}_{2}{x}_{3}{y}_{1}+{x}_{3}{y}_{2})({x}_{0}{y}_{0}{x}_{0}{y}_{3}{x}_{2}{y}_{0}+{x}_{2}{y}_{3})\hspace{1em}.$
Collecting up the terms in
${x}_{0}{y}_{0}$,
${x}_{0}$,
${y}_{0}$, and the rest,
and simplifying, yields that
${x}_{0}{y}_{0}=1$, as desired.

98.

Let
${a}_{1},{a}_{2},\dots ,{a}_{n+1},{b}_{1},{b}_{2},\dots ,{b}_{n}$ be nonnegative real numbers for which
(i)
${a}_{1}\ge {a}_{2}\ge \dots \ge {a}_{n+1}=0$,
(ii)
$0\le {b}_{k}\le 1$ for
$k=1,2,\dots ,n$.


Suppose that
$m=\lfloor {b}_{1}+{b}_{2}+\dots +{b}_{n}\rfloor +1$. Prove that
$\sum _{k=1}^{n}{a}_{k}{b}_{k}\le \sum _{k=1}^{m}{a}_{k}\hspace{1em}.$
Solution. Note that
$m1\le {b}_{1}+{b}_{2}+\dots +{b}_{m}<m$.
We have that
$\begin{array}{cc}{a}_{1}{b}_{1}+{a}_{2}{b}_{2}+\dots \hfill & +{a}_{m}{b}_{m}+{a}_{m+1}{b}_{m+1}+\dots +{a}_{n}{b}_{n}\hfill \\ \multicolumn{0}{c}{}& \le {a}_{1}{b}_{1}+{a}_{2}{b}_{2}+\dots +{a}_{m}{b}_{m}+{a}_{m}({b}_{m+1}+{b}_{m+2}+\dots +{b}_{n})\hfill \\ \multicolumn{0}{c}{}& <{a}_{1}{b}_{1}+{a}_{2}{b}_{2}+\dots +{a}_{m}{b}_{m}+{a}_{m}(m{b}_{1}{b}_{2}\dots {b}_{m})\hfill \\ \multicolumn{0}{c}{}& ={a}_{1}{b}_{1}+{a}_{2}{b}_{2}+\dots +{a}_{m}{b}_{m}+{a}_{m}(1{b}_{1})+{a}_{m}(1{b}_{2})+\dots +{a}_{m}(1{b}_{m})\hfill \\ \multicolumn{0}{c}{}& \le {a}_{1}{b}_{1}+{a}_{2}{b}_{2}+\dots +{a}_{m}{b}_{m}+{a}_{1}(1{b}_{1})+{a}_{2}(1{b}_{2})+\dots +{a}_{m}(1{b}_{m})\hfill \\ \multicolumn{0}{c}{}& ={a}_{1}+{a}_{2}+\dots +{a}_{m}\hspace{1em}\hspace{1em}.\hfill \end{array}$

99.

Let
$E$ and
$F$ be respective points on sides
$\mathrm{AB}$ and
$\mathrm{BC}$ of a triangle
$\mathrm{ABC}$ for which
$\mathrm{AE}=\mathrm{CF}$. 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
$\mathrm{BD}$ is the bisector of angle
$\mathrm{ABC}$.
Solution 1. Because of the concyclic quadrilaterals,
$\angle \mathrm{DEA}={180}^{\u02c6}\angle \mathrm{BED}=\angle \mathrm{DCF}$ and
$\angle \mathrm{DFC}={180}^{\u02c6}\angle \mathrm{DFB}=\angle \mathrm{DAB}\hspace{1em}.$
Since, also,
$\mathrm{AE}=\mathrm{CF}$,
$\Delta \mathrm{DAE}\equiv \Delta \mathrm{DFC}$ (ASA)
so that
$\mathrm{AD}=\mathrm{DF}$. In the circle through
$\mathrm{ABFD}$, the
equal chords
$\mathrm{AD}$ and
$\mathrm{DF}$ subtend equal angles
$\mathrm{ABD}$ and
$\mathrm{FBD}$ at the circumference. The result follows.
Solution 2.
$\angle \mathrm{CDF}=\angle \mathrm{CDE}\angle \mathrm{FDE}={180}^{\u02c6}\angle \mathrm{ABC}\angle \mathrm{FDE}=\angle \mathrm{FDA}\angle \mathrm{FDE}=\angle \mathrm{EDA}$ and
$\angle \mathrm{AED}={180}^{\u02c6}\angle \mathrm{BED}=\angle \mathrm{BCD}=\angle \mathrm{FCD}$. Since
$\mathrm{AE}=\mathrm{CF}$,
$\Delta \mathrm{EAD}\equiv \Delta \mathrm{CFD}$ (ASA). The altitude from
$D$ to
$\mathrm{AE}$ is equal to
the altitude from
$D$ to
$\mathrm{FC}$, and so
$D$ must be on the bisector of
$\angle \mathrm{ABC}$.
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
$\mathrm{BD}$, namely the
$x$axis.
Let the two circles have equations
$(xa){}^{2}+{y}^{2}={a}^{2}+1$ and
$(xb){}^{2}+{y}^{2}={b}^{2}+1$.
Suppose that
$y=\mathrm{mx}1$ is a line
through
$B$; this line intersects the circle of equation
$(xa){}^{2}+{y}^{2}={a}^{2}+1$ in the point
$(\frac{2(m+a)}{{m}^{2}+1},\frac{{m}^{2}+2\mathrm{am}1}{{m}^{2}+1})$
and the circle of equation
$(xb){}^{2}+{y}^{2}={b}^{2}+1$
in the point
$(\frac{2(m+b)}{{m}^{2}+1},\frac{{m}^{2}+2\mathrm{bm}1}{{m}^{2}+1})$
The distance between these two points is the square root of
$[\frac{2(ab)}{{m}^{2}+1}{]}^{2}+[\frac{2m(ab)}{{m}^{2}+1}{]}^{2}=\frac{4(ab){}^{2}(1+{m}^{2})}{({m}^{2}+1){}^{2}}=\frac{4(ab){}^{2}}{{m}^{2}+1}\hspace{1em}.$
Now suppose that the side
$\mathrm{AB}$ of the triangle has equation
$y={m}_{1}x1$ and the side
$\mathrm{BC}$ the equation
$y={m}_{2}x1$,
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
${m}_{1}^{2}+1={m}_{2}^{2}+1$ or
$0=({m}_{1}{m}_{2})({m}_{1}+{m}_{2})$.
Since the sides are distinct, it follows that
${m}_{1}={m}_{2}$
and so
$\mathrm{BD}$ bisects
$\angle \mathrm{ABC}$.

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
selfintersecting 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
$\mathrm{ABCDEFGHIJ}$. Let
$\mathrm{BE}$ and
$\mathrm{DI}$ intersect at
$K$ and let
$\mathrm{AF}$ and
$\mathrm{DI}$
intersect at
$L$. Observe that
$\mathrm{AB}\mathrm{DI}\mathrm{EH}$ and
$\mathrm{BE}\mathrm{AF}\mathrm{HI}$, so that
$\mathrm{ABKL}$ and
$\mathrm{KIHE}$ are parallelograms. Now
$\mathrm{AB}$ is a side of
$P$ and
$\mathrm{HE}$ is a side of
$Q$, and the length of the segment
$\mathrm{IL}$ is the difference of the lengths of
$\mathrm{EH}=\mathrm{IK}$ and
$\mathrm{AB}=\mathrm{KL}$.
Since
$L$, being the intersection of the diameters
$\mathrm{AF}$ and
$\mathrm{DI}$,
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,
$\mathrm{AB}$ is an edge of
$P$,
$\mathrm{AD}$ is an edge of
$Q$,
$\mathrm{DO}$ is a radius of the circle and
$\mathrm{BG}$ a diameter. Let
$\mathrm{AD}$ and
$\mathrm{BO}$ intersect at
$U$. Identify
in turn the angles
$\angle \mathrm{DOU}={72}^{\u02c6}$,
$\angle \mathrm{DAB}={36}^{\u02c6}$,
$\angle \mathrm{ABU}={72}^{\u02c6}$,
$\angle \mathrm{DUO}=\angle \mathrm{BUA}={72}^{\u02c6}$, whence
$\mathrm{AU}=\mathrm{AB}$,
$\mathrm{DU}=\mathrm{DO}$ and
$\mathrm{AD}\mathrm{AB}=\mathrm{AD}\mathrm{AX}=\mathrm{DX}=\mathrm{DO}$, 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
$\mathrm{EB}$ for which
$\mathrm{EV}=\mathrm{OE}$. We have that
$\angle \mathrm{AOB}={36}^{\u02c6}$,
$\angle \mathrm{DOB}=\angle \mathrm{OBA}={72}^{\u02c6}$,
$\angle \mathrm{BOE}={108}^{\u02c6}$ and
$\angle \mathrm{OEB}=\angle \mathrm{OBE}={36}^{\u02c6}$.
Also,
$\angle \mathrm{EOV}=\angle \mathrm{EVO}={72}^{\u02c6}$ and
$\mathrm{OE}=\mathrm{EV}=\mathrm{OA}=\mathrm{OB}$. Hence,
$\Delta \mathrm{DAB}=\Delta \mathrm{EVO}$ (SAS), so that
$\mathrm{OV}=\mathrm{AB}$.
Since
$\angle \mathrm{BVO}={108}^{\u02c6}$ and
$\angle \mathrm{BOV}={36}^{\u02c6}$,
$\angle \mathrm{OBV}={36}^{\u02c6}$, and so
$\mathrm{BV}=\mathrm{OV}=\mathrm{AB}$.
Hence
$\mathrm{BE}\mathrm{AB}=\mathrm{EV}+\mathrm{BV}\mathrm{AB}=\mathrm{EV}=\mathrm{OE}$, 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
${36}^{\u02c6}$, so its length is
$2\mathrm{sin}{18}^{\u02c6}$. Likewise, the length of a side of
$Q$ is
$2\mathrm{sin}{54}^{\u02c6}$. The difference between these is
$2\mathrm{sin}{54}^{\u02c6}2\mathrm{sin}{18}^{\u02c6}=2\mathrm{cos}{36}^{\u02c6}2\mathrm{cos}{72}^{\u02c6}=2t2(2{t}^{2}1)=2+2t4{t}^{2}$
where
$t=\mathrm{cos}{36}^{\u02c6}$.
Now
$\begin{array}{cc}t\hfill & =\mathrm{cos}{36}^{\u02c6}=\mathrm{cos}{144}^{\u02c6}=12\mathrm{cos}{}^{2}{72}^{\u02c6}\hfill \\ \multicolumn{0}{c}{}& =12(2{t}^{2}1){}^{2}=8{t}^{4}+8{t}^{2}1\hspace{1em},\hfill \end{array}$
so that
$\begin{array}{cc}0\hfill & =8{t}^{4}8{t}^{2}+t+1=(t+1)(8{t}^{3}8{t}^{2}+1)\hfill \\ \multicolumn{0}{c}{}& =(t+1)(2t1)(4{t}^{2}2t1)\hspace{1em}.\hfill \end{array}$
Since
$t$ is equal to neither
$1$ nor
$\frac{1}{2}$, we must have
that
$4{t}^{2}2t=1$. Hence
$2\mathrm{sin}{54}^{\u02c6}2\mathrm{sin}{18}^{\u02c6}=2(4{t}^{2}2t)=1\hspace{1em},$
the radius of the circle.

101.

Let
$a,b,u,v$ be nonnegative. Suppose that
${a}^{5}+{b}^{5}\le 1$ and
${u}^{5}+{v}^{5}\le 1$. Prove that
${a}^{2}{u}^{3}+{b}^{2}{v}^{3}\le 1\hspace{1em}.$
[With thanks to Ross
Honsberger.]
Solution. By the arithmeticgeometric means inequality,
we have that
$\frac{2{a}^{5}+3{u}^{5}}{5}=\frac{{a}^{5}+{a}^{5}+{u}^{5}+{u}^{5}+{u}^{5}}{5}\ge \sqrt[5]{{a}^{10}{u}^{15}}={a}^{2}{u}^{3}$
and, similarly,
$\frac{2{b}^{5}+3{v}^{5}}{5}\ge {b}^{2}{v}^{3}\hspace{1em}.$
Adding these two inequalities yields the result.

102.

Prove that there exists a tetrahedron
$\mathrm{ABCD}$, 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
$\mathrm{AB}$, a side of length 1. Now construct a
rectangle
$\mathrm{ACBD}$ with diagonal
$\mathrm{AB}$, so that
$\Vert \mathrm{AC}\Vert =\Vert \mathrm{BD}\Vert =s<t=\Vert \mathrm{AD}\Vert =\Vert \mathrm{BC}\Vert $.
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
$\mathrm{AB}$ lies (like folding up the wings of
a butterfly) in such a way that we can obtain the desired
tetrahedron.
When the triangles
$\mathrm{ADB}$ and
$\mathrm{ACB}$ 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
$\mathrm{ACD}$ should be a right triangle
similar to
$\mathrm{ABC}$. The hypotenuse of
$\Delta \mathrm{ACD}$ cannot be
$\mathrm{AC}$ as
$\mathrm{AC}<\mathrm{AD}$. Nor can it be
$\mathrm{CD}$, for then, we would have
$\mathrm{AD}=\mathrm{BC}$,
$\mathrm{AC}=\mathrm{AC}$, and
$\mathrm{CD}$ would have to have length 1,
possible only when
$\mathrm{ABCD}$ is coplanar. So the hypotenuse must
be
$\mathrm{AD}$. The similarity of
$\Delta \mathrm{ADC}$ and
$\Delta \mathrm{ABC}$
would require that
$1:t:s=t:s:r$
where
$r=\Vert \mathrm{CD}\Vert $. Thus,
$1/t=t/s$ or
$s={t}^{2}$
and
$t/s=s/r$ or
$r={s}^{2}/t={t}^{3}$. So we must fold
$C$ and
$D$ until they are distance
${t}^{3}$ apart.
Is this possible? Since
$\Delta \mathrm{ACB}$ is right,
$1={t}^{2}+{s}^{2}={t}^{2}+{t}^{4}$, whence
$s={t}^{2}=\frac{1}{2}(1+\sqrt{5})<1$. 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
$\Delta \mathrm{ADB}$ is folded flat partially covering
$\Delta \mathrm{ACB}$.
Suppose this has been done, with
$\mathrm{ABCD}$ coplanar and
$C$,
$D$
both on the same side of
$\mathrm{AB}$. Let
$P$ and
$Q$ be the respective
feet of the perpendiculars to
$\mathrm{AB}$ from
$C$ and
$D$. Then
$\Vert \mathrm{CP}\Vert =\Vert \mathrm{DQ}\Vert ={t}^{3}\hspace{1em},\hspace{1em}\hspace{1em}\hspace{1em}\Vert \mathrm{AP}\Vert =\Vert \mathrm{QB}\Vert ={t}^{4}\hspace{1em},\hspace{1em}\hspace{1em}\hspace{1em}\Vert \mathrm{AQ}\Vert =\Vert \mathrm{PB}\Vert ={t}^{2}\hspace{1em},$
and
$\Vert \mathrm{CD}\Vert =\Vert \mathrm{PQ}\Vert ={t}^{2}{t}^{4}=({t}^{4}+{t}^{6}){t}^{4}={t}^{6}<{t}^{3}\hspace{1em}.$
When
$C$ and
$D$ are located, we have
$\Vert \mathrm{AB}\Vert =1$,
$\Vert \mathrm{AD}\Vert =\Vert \mathrm{BC}\Vert =t$,
$\Vert \mathrm{AC}\Vert =\Vert \mathrm{BD}\Vert ={t}^{2}$ and
$\Vert \mathrm{CD}\Vert ={t}^{3}$. Since all
faces of the tetrahedron
$\mathrm{ABCD}$ have sides in the ratio
$1:t:{t}^{2}$, all are similar right triangles and
$\mathrm{AB}:\mathrm{CD}=1:{t}^{3}$.
Solution 2. Let
$\alpha =\angle \mathrm{CAB}$ and
$\Vert \mathrm{AB}\Vert =1$.
By the condition on the acute angles of triangles
$\mathrm{ACB}$ and
$\mathrm{ACD}$,
$\angle \mathrm{ACB}=\angle \mathrm{ADB}={90}^{\u02c6}$, so that the
triangles
$\Delta \mathrm{ACD}$ and
$\Delta \mathrm{ADB}$, being similar and
sharing a hypotenuse, are congruent.
Suppose, if possible, that
$\angle \mathrm{BAD}=\alpha $. Then
$\mathrm{AC}=\mathrm{AD}$ and so
$\Delta \mathrm{ACD}$ must be isosceles with its
right angle at
$A$, contrary to hypothesis.
So,
$\angle \mathrm{ABD}=\alpha $ and
$\Vert \mathrm{BD}\Vert =\Vert \mathrm{AC}\Vert =\mathrm{cos}\alpha $,
$\Vert \mathrm{AD}\Vert =\Vert \mathrm{BC}\Vert =\mathrm{sin}\alpha $.
Consider
$\Delta \mathrm{ACD}$. Suppose that
$\angle \mathrm{ACD}={90}^{\u02c6}$.
If
$\angle \mathrm{DAC}=\alpha $, then
$\Delta \mathrm{ABC}\equiv \Delta \mathrm{ADC}$ and
$1=\Vert \mathrm{AB}\Vert =\Vert \mathrm{AD}\Vert =\mathrm{sin}\alpha $,
yielding a contradiction. Hence
$\angle \mathrm{ADC}=\alpha $,
$\Vert \mathrm{AD}\Vert =\Vert \mathrm{AC}\Vert /\mathrm{sin}\alpha =\mathrm{cos}\alpha /\mathrm{sin}\alpha $ and
$\Vert \mathrm{CD}\Vert =\Vert \mathrm{AC}\Vert \mathrm{cot}\alpha =\mathrm{cos}{}^{2}\alpha /\mathrm{sin}\alpha $. Hence, looking at
$\Vert \mathrm{AD}\Vert $,
we have that
$\frac{\mathrm{cos}\alpha}{\mathrm{sin}\alpha}=\mathrm{sin}\alpha \Rightarrow 0=\mathrm{cos}\alpha \mathrm{sin}{}^{2}\alpha =\mathrm{cos}{}^{2}\alpha +\mathrm{cos}\alpha 1\hspace{1em}.$
Therefore,
$\mathrm{cos}\alpha =\frac{1}{2}(\sqrt{5}1)$ and
$\mathrm{sin}{}^{2}\alpha =\mathrm{cos}\alpha $.
Observe that
$\Vert \mathrm{BC}\Vert \mathrm{sin}\alpha =\mathrm{sin}{}^{2}\alpha =\mathrm{cos}\alpha =\Vert \mathrm{BD}\Vert $ and
$\Vert \mathrm{BC}\Vert \mathrm{cos}\alpha =\mathrm{sin}\alpha \mathrm{cos}\alpha =\mathrm{cos}{}^{2}\alpha /\mathrm{sin}\alpha =\Vert \mathrm{CD}\Vert $,
so that triangle
$\mathrm{BCD}$ is right with
$\angle \mathrm{CDB}={90}^{\u02c6}$
and similar to the other three faces.
We need to check that this setup is feasible. Using spatial
coordinates, take
$C~(0,0,0)\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}A~(0,\mathrm{cos}\alpha ,0)\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}B~(\mathrm{sin}\alpha ,0,0)\hspace{1em}.$
Since
$\angle \mathrm{ACD}={90}^{\u02c6}$,
$D$ lies in the plane
$y=0$
and so has coordinates of the form
$(x,0,z)$. Since
$\angle \mathrm{CDB}={90}^{\u02c6}$,
$\mathrm{CD}\perp \mathrm{DB}$, so that
$0=(x,0,z)\xb7(x\mathrm{sin}\alpha ,0,z){x}^{2}+{z}^{2}x\mathrm{sin}\alpha \hspace{1em},$
Now
$\Vert \mathrm{CD}\Vert =\mathrm{cos}\alpha \mathrm{sin}\alpha $ forces
$\mathrm{cos}{}^{2}\alpha \mathrm{sin}{}^{2}\alpha ={x}^{2}+{z}^{2}$. Hence
$x\mathrm{sin}\alpha =\mathrm{cos}{}^{2}\alpha \mathrm{sin}{}^{2}\alpha \Rightarrow x=\mathrm{cos}{}^{2}\alpha \mathrm{sin}\alpha \hspace{1em}.$
Therefore
${z}^{2}=(\mathrm{cos}{}^{2}\alpha \mathrm{cos}{}^{4}\alpha )\mathrm{sin}{}^{2}\alpha =\mathrm{cos}{}^{2}\alpha \mathrm{sin}{}^{4}\alpha \Rightarrow z=\mathrm{cos}\alpha \mathrm{sin}{}^{2}\alpha \hspace{1em},$
Hence
$D~(\mathrm{cos}{}^{2}\alpha \mathrm{sin}\alpha ,0,\mathrm{cos}\alpha \mathrm{sin}{}^{2}\alpha )$.
Thus, letting
$\mathrm{sin}\alpha =t=\frac{1}{2}(\sqrt{5}1)$,
we have
$A~(0,{t}^{2},0)$,
$B~(t,0,0)$,
$C~(0,0,0)$,
$D~({t}^{5},0,{t}^{4})$ with
${t}^{4}+{t}^{2}1=0$, and
$\Vert \mathrm{AB}\Vert =1$,
$\Vert \mathrm{AD}\Vert =\Vert \mathrm{BC}\Vert =t$,
$\Vert \mathrm{BD}\Vert =\Vert \mathrm{AC}\Vert ={t}^{2}$ and
$\Vert \mathrm{CD}\Vert ={t}^{3}$. [Exercise: Check that the
coordinates give the required distances and similar right
triangles.] The ratio of largest to smallest edges is
$1:{t}^{3}=1:[\frac{1}{2}(\sqrt{5}1)]{}^{3/2}=1:\sqrt{2+\sqrt{5}}$.
We need to dispose of the other possibilities for
$\Delta \mathrm{ACD}$.
By the given condition,
$\angle \mathrm{DAC}\ne {90}^{\u02c6}$.
If
$\angle \mathrm{ADC}={90}^{\u02c6}$, then we have essentially the
same situation as before with the roles of
$\alpha $ 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.