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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 $\mathrm{xy}=1$. Let the three vertices of the triangle be at $\left(a,1/a\right)$, $\left(b,1/b\right)$, $\left(c,1/c\right)$. The altitude to the points $\left(c,1/c\right)$ has slope $-\left(a-b\right)/\left({a}^{-1}-{b}^{-1}\right)=\mathrm{ab}$ and its equation is $y=\mathrm{abx}+\left(1/c\right)-\mathrm{abc}$. The altitude to the point $\left(a,1/a\right)$ has equation $y=\mathrm{bcx}+\left(1/a\right)-\mathrm{abc}$. These two lines intersect in the point $\left(-1/\mathrm{abc},-\mathrm{abc}\right)$ 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 $\left({x}_{i},{y}_{i}\right)$ ( $i=1,2,3$), and let the orthocentre be at $\left({x}_{0},{y}_{0}\right)$. Then

$\left({x}_{1}-{x}_{2}\right)\left({x}_{0}-{x}_{3}\right)=-\left({y}_{1}-{y}_{2}\right)\left({y}_{0}-{y}_{3}\right)$

and

$\left({x}_{1}-{x}_{3}\right)\left({x}_{0}-{x}_{2}\right)=-\left({y}_{1}-{y}_{3}\right)\left({y}_{0}-{y}_{2}\right) .$

Cross-multiplying these equations yields that

$\left({x}_{1}-{x}_{2}\right)\left({y}_{1}-{y}_{3}\right)\left({x}_{0}-{x}_{3}\right)\left({y}_{0}-{y}_{2}\right)=\left({x}_{1}-{x}_{3}\right)\left({y}_{1}-{y}_{2}\right)\left({x}_{0}-{x}_{2}\right)\left({y}_{0}-{y}_{3}\right) ,$

whence

$\left(1-{x}_{1}{y}_{3}-{x}_{2}{y}_{1}+{x}_{2}{y}_{3}\right)\left({x}_{0}{y}_{0}-{x}_{0}{y}_{2}-{x}_{3}{y}_{0}+{x}_{3}{y}_{2}\right)=\left(1-{x}_{1}{y}_{2}-{x}_{3}{y}_{1}+{x}_{3}{y}_{2}\right)\left({x}_{0}{y}_{0}-{x}_{0}{y}_{3}-{x}_{2}{y}_{0}+{x}_{2}{y}_{3}\right) .$

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 $m=⌊{b}_{1}+{b}_{2}+\dots +{b}_{n}⌋+1$. Prove that

$\sum _{k=1}^{n}{a}_{k}{b}_{k}\le \sum _{k=1}^{m}{a}_{k} .$

Solution. Note that $m-1\le {b}_{1}+{b}_{2}+\dots +{b}_{m}. We have that

$\begin{array}{cc}{a}_{1}{b}_{1}+{a}_{2}{b}_{2}+\dots \hfill & +{a}_{m}{b}_{m}+{a}_{m+1}{b}_{m+1}+\dots +{a}_{n}{b}_{n}\hfill \\ \multicolumn{0}{c}{}& \le {a}_{1}{b}_{1}+{a}_{2}{b}_{2}+\dots +{a}_{m}{b}_{m}+{a}_{m}\left({b}_{m+1}+{b}_{m+2}+\dots +{b}_{n}\right)\hfill \\ \multicolumn{0}{c}{}& <{a}_{1}{b}_{1}+{a}_{2}{b}_{2}+\dots +{a}_{m}{b}_{m}+{a}_{m}\left(m-{b}_{1}-{b}_{2}-\dots -{b}_{m}\right)\hfill \\ \multicolumn{0}{c}{}& ={a}_{1}{b}_{1}+{a}_{2}{b}_{2}+\dots +{a}_{m}{b}_{m}+{a}_{m}\left(1-{b}_{1}\right)+{a}_{m}\left(1-{b}_{2}\right)+\dots +{a}_{m}\left(1-{b}_{m}\right)\hfill \\ \multicolumn{0}{c}{}& \le {a}_{1}{b}_{1}+{a}_{2}{b}_{2}+\dots +{a}_{m}{b}_{m}+{a}_{1}\left(1-{b}_{1}\right)+{a}_{2}\left(1-{b}_{2}\right)+\dots +{a}_{m}\left(1-{b}_{m}\right)\hfill \\ \multicolumn{0}{c}{}& ={a}_{1}+{a}_{2}+\dots +{a}_{m} .\hfill \end{array}$

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, $\angle \mathrm{DEA}={180}^{ˆ}-\angle \mathrm{BED}=\angle \mathrm{DCF}$ and $\angle \mathrm{DFC}={180}^{ˆ}-\angle \mathrm{DFB}=\angle \mathrm{DAB} .$ 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. $\angle \mathrm{CDF}=\angle \mathrm{CDE}-\angle \mathrm{FDE}={180}^{ˆ}-\angle \mathrm{ABC}-\angle \mathrm{FDE}=\angle \mathrm{FDA}-\angle \mathrm{FDE}=\angle \mathrm{EDA}$ and $\angle \mathrm{AED}={180}^{ˆ}-\angle \mathrm{BED}=\angle \mathrm{BCD}=\angle \mathrm{FCD}$. 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 $\left(0,-1\right)$ and D the point $\left(0,1\right)$. The centres of both circles are on the right bisector of $\mathrm{BD}$, namely the $x-$axis. Let the two circles have equations $\left(x-a\right){}^{2}+{y}^{2}={a}^{2}+1$ and $\left(x-b\right){}^{2}+{y}^{2}={b}^{2}+1$. Suppose that $y=\mathrm{mx}-1$ is a line through $B$; this line intersects the circle of equation $\left(x-a\right){}^{2}+{y}^{2}={a}^{2}+1$ in the point

$\left(\frac{2\left(m+a\right)}{{m}^{2}+1},\frac{{m}^{2}+2\mathrm{am}-1}{{m}^{2}+1}\right)$

and the circle of equation $\left(x-b\right){}^{2}+{y}^{2}={b}^{2}+1$ in the point

$\left(\frac{2\left(m+b\right)}{{m}^{2}+1},\frac{{m}^{2}+2\mathrm{bm}-1}{{m}^{2}+1}\right)$

The distance between these two points is the square root of

$\left[\frac{2\left(a-b\right)}{{m}^{2}+1}{\right]}^{2}+\left[\frac{2m\left(a-b\right)}{{m}^{2}+1}{\right]}^{2}=\frac{4\left(a-b\right){}^{2}\left(1+{m}^{2}\right)}{\left({m}^{2}+1\right){}^{2}}=\frac{4\left(a-b\right){}^{2}}{{m}^{2}+1} .$

Now suppose that the side $\mathrm{AB}$ of the triangle has equation $y={m}_{1}x-1$ and the side $\mathrm{BC}$ the equation $y={m}_{2}x-1$, so that $\left(A,E\right)$ and $\left(C,F\right)$ 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 $0=\left({m}_{1}-{m}_{2}\right)\left({m}_{1}+{m}_{2}\right)$. 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 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 $\mathrm{ABCDEFGHIJ}$. 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 $\angle \mathrm{DOU}={72}^{ˆ}$, $\angle \mathrm{DAB}={36}^{ˆ}$, $\angle \mathrm{ABU}={72}^{ˆ}$, $\angle \mathrm{DUO}=\angle \mathrm{BUA}={72}^{ˆ}$, 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 $\angle \mathrm{AOB}={36}^{ˆ}$, $\angle \mathrm{DOB}=\angle \mathrm{OBA}={72}^{ˆ}$, $\angle \mathrm{BOE}={108}^{ˆ}$ and $\angle \mathrm{OEB}=\angle \mathrm{OBE}={36}^{ˆ}$. Also, $\angle \mathrm{EOV}=\angle \mathrm{EVO}={72}^{ˆ}$ and $\mathrm{OE}=\mathrm{EV}=\mathrm{OA}=\mathrm{OB}$. Hence, $\Delta \mathrm{DAB}=\Delta \mathrm{EVO}$ (SAS), so that $\mathrm{OV}=\mathrm{AB}$. Since $\angle \mathrm{BVO}={108}^{ˆ}$ and $\angle \mathrm{BOV}={36}^{ˆ}$, $\angle \mathrm{OBV}={36}^{ˆ}$, 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 ${36}^{ˆ}$, so its length is $2\mathrm{sin}{18}^{ˆ}$. Likewise, the length of a side of $Q$ is $2\mathrm{sin}{54}^{ˆ}$. The difference between these is

$2\mathrm{sin}{54}^{ˆ}-2\mathrm{sin}{18}^{ˆ}=2\mathrm{cos}{36}^{ˆ}-2\mathrm{cos}{72}^{ˆ}=2t-2\left(2{t}^{2}-1\right)=2+2t-4{t}^{2}$

where $t=\mathrm{cos}{36}^{ˆ}$. Now

$\begin{array}{cc}t\hfill & =\mathrm{cos}{36}^{ˆ}=-\mathrm{cos}{144}^{ˆ}=1-2\mathrm{cos}{}^{2}{72}^{ˆ}\hfill \\ \multicolumn{0}{c}{}& =1-2\left(2{t}^{2}-1\right){}^{2}=-8{t}^{4}+8{t}^{2}-1 ,\hfill \end{array}$

so that

$\begin{array}{cc}0\hfill & =8{t}^{4}-8{t}^{2}+t+1=\left(t+1\right)\left(8{t}^{3}-8{t}^{2}+1\right)\hfill \\ \multicolumn{0}{c}{}& =\left(t+1\right)\left(2t-1\right)\left(4{t}^{2}-2t-1\right) .\hfill \end{array}$

Since $t$ is equal to neither $-1$ nor $\frac{1}{2}$, we must have that $4{t}^{2}-2t=1$. Hence

$2\mathrm{sin}{54}^{ˆ}-2\mathrm{sin}{18}^{ˆ}=2-\left(4{t}^{2}-2t\right)=1 ,$

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

${a}^{2}{u}^{3}+{b}^{2}{v}^{3}\le 1 .$

[With thanks to Ross Honsberger.]
Solution. By the arithmetic-geometric 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,

$\frac{2{b}^{5}+3{v}^{5}}{5}\ge {b}^{2}{v}^{3} .$

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 $‖\mathrm{AC}‖=‖\mathrm{BD}‖=s. 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 $r=‖\mathrm{CD}‖$. 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 $s={t}^{2}=\frac{1}{2}\left(-1+\sqrt{5}\right)<1$. 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

$‖\mathrm{CP}‖=‖\mathrm{DQ}‖={t}^{3} , ‖\mathrm{AP}‖=‖\mathrm{QB}‖={t}^{4} , ‖\mathrm{AQ}‖=‖\mathrm{PB}‖={t}^{2} ,$

and

$‖\mathrm{CD}‖=‖\mathrm{PQ}‖={t}^{2}-{t}^{4}=\left({t}^{4}+{t}^{6}\right)-{t}^{4}={t}^{6}<{t}^{3} .$

When $C$ and $D$ are located, we have $‖\mathrm{AB}‖=1$, $‖\mathrm{AD}‖=‖\mathrm{BC}‖=t$, $‖\mathrm{AC}‖=‖\mathrm{BD}‖={t}^{2}$ and $‖\mathrm{CD}‖={t}^{3}$. 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 $‖\mathrm{AB}‖=1$. By the condition on the acute angles of triangles $\mathrm{ACB}$ and $\mathrm{ACD}$, $\angle \mathrm{ACB}=\angle \mathrm{ADB}={90}^{ˆ}$, 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 $‖\mathrm{BD}‖=‖\mathrm{AC}‖=\mathrm{cos}\alpha$, $‖\mathrm{AD}‖=‖\mathrm{BC}‖=\mathrm{sin}\alpha$.
Consider $\Delta \mathrm{ACD}$. Suppose that $\angle \mathrm{ACD}={90}^{ˆ}$. If $\angle \mathrm{DAC}=\alpha$, then $\Delta \mathrm{ABC}\equiv \Delta \mathrm{ADC}$ and $1=‖\mathrm{AB}‖=‖\mathrm{AD}‖=\mathrm{sin}\alpha$, yielding a contradiction. Hence $\angle \mathrm{ADC}=\alpha$, $‖\mathrm{AD}‖=‖\mathrm{AC}‖/\mathrm{sin}\alpha =\mathrm{cos}\alpha /\mathrm{sin}\alpha$ and $‖\mathrm{CD}‖=‖\mathrm{AC}‖\mathrm{cot}\alpha =\mathrm{cos}{}^{2}\alpha /\mathrm{sin}\alpha$. Hence, looking at $‖\mathrm{AD}‖$, we have that

$\frac{\mathrm{cos}\alpha }{\mathrm{sin}\alpha }=\mathrm{sin}\alpha ⇒0=\mathrm{cos}\alpha -\mathrm{sin}{}^{2}\alpha =\mathrm{cos}{}^{2}\alpha +\mathrm{cos}\alpha -1 .$

Therefore, $\mathrm{cos}\alpha =\frac{1}{2}\left(\sqrt{5}-1\right)$ and $\mathrm{sin}{}^{2}\alpha =\mathrm{cos}\alpha$.
Observe that $‖\mathrm{BC}‖\mathrm{sin}\alpha =\mathrm{sin}{}^{2}\alpha =\mathrm{cos}\alpha =‖\mathrm{BD}‖$ and $‖\mathrm{BC}‖\mathrm{cos}\alpha =\mathrm{sin}\alpha \mathrm{cos}\alpha =\mathrm{cos}{}^{2}\alpha /\mathrm{sin}\alpha =‖\mathrm{CD}‖$, so that triangle $\mathrm{BCD}$ is right with $\angle \mathrm{CDB}={90}^{ˆ}$ and similar to the other three faces.
We need to check that this set-up is feasible. Using spatial coordinates, take

$C~\left(0,0,0\right) A~\left(0,\mathrm{cos}\alpha ,0\right) B~\left(\mathrm{sin}\alpha ,0,0\right) .$

Since $\angle \mathrm{ACD}={90}^{ˆ}$, $D$ lies in the plane $y=0$ and so has coordinates of the form $\left(x,0,z\right)$. Since $\angle \mathrm{CDB}={90}^{ˆ}$, $\mathrm{CD}\perp \mathrm{DB}$, so that

$0=\left(x,0,z\right)·\left(x-\mathrm{sin}\alpha ,0,z\right)-{x}^{2}+{z}^{2}-x\mathrm{sin}\alpha ,$

Now $‖\mathrm{CD}‖=\mathrm{cos}\alpha \mathrm{sin}\alpha$ forces $\mathrm{cos}{}^{2}\alpha \mathrm{sin}{}^{2}\alpha ={x}^{2}+{z}^{2}$. Hence

$x\mathrm{sin}\alpha =\mathrm{cos}{}^{2}\alpha \mathrm{sin}{}^{2}\alpha ⇒x=\mathrm{cos}{}^{2}\alpha \mathrm{sin}\alpha .$

Therefore

${z}^{2}=\left(\mathrm{cos}{}^{2}\alpha -\mathrm{cos}{}^{4}\alpha \right)\mathrm{sin}{}^{2}\alpha =\mathrm{cos}{}^{2}\alpha \mathrm{sin}{}^{4}\alpha ⇒z=\mathrm{cos}\alpha \mathrm{sin}{}^{2}\alpha ,$

Hence $D~\left(\mathrm{cos}{}^{2}\alpha \mathrm{sin}\alpha ,0,\mathrm{cos}\alpha \mathrm{sin}{}^{2}\alpha \right)$.
Thus, letting $\mathrm{sin}\alpha =t=\frac{1}{2}\left(\sqrt{5}-1\right)$, we have $A~\left(0,{t}^{2},0\right)$, $B~\left(t,0,0\right)$, $C~\left(0,0,0\right)$, $D~\left({t}^{5},0,{t}^{4}\right)$ with ${t}^{4}+{t}^{2}-1=0$, and $‖\mathrm{AB}‖=1$, $‖\mathrm{AD}‖=‖\mathrm{BC}‖=t$, $‖\mathrm{BD}‖=‖\mathrm{AC}‖={t}^{2}$ and $‖\mathrm{CD}‖={t}^{3}$. [Exercise: Check that the coordinates give the required distances and similar right triangles.] The ratio of largest to smallest edges is $1:{t}^{3}=1:\left[\frac{1}{2}\left(\sqrt{5}-1\right)\right]{}^{3/2}=1:\sqrt{2+\sqrt{5}}$.
We need to dispose of the other possibilities for $\Delta \mathrm{ACD}$. By the given condition, $\angle \mathrm{DAC}\ne {90}^{ˆ}$. If $\angle \mathrm{ADC}={90}^{ˆ}$, 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.