Here is a graph.

2a - The diameter can't go through arc BC because then it doesn't pass through the orthocenter. But wait, it's acute so you can't fit it in one arc so it passes through both and you're done. 2b - I don't think you can do this without trig, but if D is the midpoint of arc BC you get APDQ is a rhombus with one diagonal being 2Rcos something and height of ABC is something R cos something so then everything cancels out nicely and you get a rational linear function which is very easy to find the range of so hoo rah. Anyone got a "nice" bary, complex, or calc solution?

This is a condensation of my fairly rigorous proof. EDIT: Oops, then I wasted the whole JMO. I was just thinking, "$\angle BOC=120^\circ$ doesn't uniquely define $O$" but thanks to your explanation it turned out to be sufficient to solve the problem. Hopefully I will be more productive on my birthday.

uhh how $BOC=BHC=120$ since $A=60$...?

ABCDE wrote: 2b - I don't think you can do this without trig I got pretty close by continuing coordinate bashing set point A to be (0,0), C=(1,0), and b=(b,b*sqrt3). solve for slope of OH=-sqrt3 so APQ is equilateral with area s=(2b+1)^2*sqrt3/36. s/t is then (2b+1)^2/(18b-(2b+1)^2). I couldn't prove that it had range (4/5,1), unfortunately, although I think I could've calculus bashed it to find b=1/2 is the only local minimum.

ABCDE wrote: uhh how $BOC=BHC=120$ since $A=60$...? $\angle BOC = \angle OBA + \angle BAO + \angle OCA + \angle OAC = 2(\angle BAO + \angle OAC) $ $= 2A = 120^o$ $\angle BHC = 180^o - A = 120^o$

It should say $60^{\circ}$ How can you prove 2b? I got the answer, but how??

You can WLOG it and scale some length to 1 for convenience and pick some angle or length $l$ to vary (some work out better than others), but in the end $s/t$ becomes some function $f(l)$ and you know what domain you need so you get the range. I personally set the circumradius as 1 and varied $60-B$ and it worked out very nicely.

So for part a I coordinate bashed. Then just now, I read Geo Revisited to look at special properties of Euler lines and I see this: If ABC has the special property that the Euler line is parallel to side BC, then $\tan B \tan C = 3$. ... Great timing -_- It basically instant-kills the problem. One angle is 60, $\tan 60 = \sqrt{3}$, so therefore the tangent of the other angle is $\sqrt{3}$, so it is 60 degrees. However, the triangle is not equilateral, so contradiction.

Somewhat bad way for 2b: (close to) equilateral, then (close to) 30-60-90 right, then assume those are the boundaries

Is it OK if I proved that, according to mathboy's diagram, $\angle AOH < \angle AOB$ and $180^{\circ} - \angle AOH < \angle AOC$ for part 2a?

vincenthuang75025 wrote: So for part a I coordinate bashed. Then just now, I read Geo Revisited to look at special properties of Euler lines and I see this: If ABC has the special property that the Euler line is parallel to side BC, then $\tan B \tan C = 3$. ... Great timing -_- It basically instant-kills the problem. One angle is 60, $\tan 60 = \sqrt{3}$, so therefore the tangent of the other angle is $\sqrt{3}$, so it is 60 degrees. However, the triangle is not equilateral, so contradiction. Does it? It's segments AB and AC not lines.

yeah that doesn't seem to be right.

Can someone explain 2a to me? sorry if this is a really dumb question

(maybe read the thread for an explanation)

I am just beginning to realize how extravagantly I failed this question....

oh oops I did some really nasty thing finding the line through O and H and showing the intersection of this line with $AB$ and $AC$ has all positive coordinates

coordinate bash is not that bad...

mymathboy wrote: DaChickenInc wrote: Subject: Let's bash some Euler Lines I am pretty sure this is synthetic. Also, I don't see how $APQ$ is equilateral immediately follows from $BCOH$ is cyclic.... Sorry for the ambiguity. I have changed the order of arguments to make it clear. $BCOH$ is cyclic, $\triangle BPO$ and $\triangle OQC$ are congruent. so, $\triangle APC$ is equalaterial (as we get two more $60^o$ angles there), I am missing something very obvious, why are does are $\triangle BPO$ and $\triangle OQC$ congruent?

How does one rigorously solve part (a)?

Thanks to LC for helping me with this proof!

We begin with some synthetic work. Let $I$ denote the incenter, and recall (``fact 5'') that the arc midpoint $M$ is the center of $(BIC)$, which we denote by $\gamma$. Now we have that \[ \angle BOC = \angle BIC = \angle BHC = 120^{\circ}. \]Since all three centers lie inside $ABC$ (as it was acute), and hence on the opposite side of $\overline{BC}$ as $M$, it follows that $O$, $I$, $H$ lie on minor arc $BC$ of $\gamma$. We note this implies (a) already, as line $OH$ meets line $BC$ outside of segment $BC$. [asy][asy] pair A = dir(50); pair B = dir(210); pair C = dir(330); pair M = dir(270); pair I = incenter(A, B, C); pair O = circumcenter(A, B, C); pair H = orthocenter(A, B, C); pair P = extension(O, H, A, B); pair Q = extension(O, H, A, C); filldraw(A--P--Q--cycle, invisible, deepgreen); filldraw(B--P--Q--C--cycle, invisible, deepgreen); filldraw(circumcircle(A, B, C), invisible, red); filldraw(CP(M, I), invisible, dotted+blue); draw(A--M, red); filldraw(A--O--M--H--cycle, invisible, lightred); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$M$", M, dir(M)); dot("$I$", I, dir(20)); dot("$O$", O, dir(90)); dot("$H$", H, dir(50)); dot("$P$", P, dir(P)); dot("$Q$", Q, dir(Q)); /* TSQ Source: A = dir 50 B = dir 210 C = dir 330 M = dir 270 I = incenter A B C R20 O = circumcenter A B C R90 H = orthocenter A B C R50 P = extension O H A B Q = extension O H A C A--P--Q--cycle 0.2 lightgreen / deepgreen B--P--Q--C--cycle 0.2 lightcyan / deepgreen circumcircle A B C 0.1 yellow / red CP M I 0.1 lightblue / dotted blue A--M red A--O--M--H--cycle 0.1 orange / lightred */ [/asy][/asy] Claim: Triangle $APQ$ is equilateral with side length $\frac{b+c}{3}$. Proof. Let $R$ be the circumradius. We have $R = OM = OA = MH$, and even $AH = 2R \cos A = R$, so $AOMH$ is a rhombus. Thus $\overline{OH} \perp \overline{AM}$ and in this way we derive that $\triangle APQ$ is isosceles, hence equilateral. Finally, since $\angle PBH = 30^{\circ}$, and $\angle BPH = 120^{\circ}$, it follows that $\triangle BPH$ is isosceles and $BP = PH$. Similarly, $CQ = QH$. So $b+c = AP + BP + AQ + QC = AP + AQ + PQ$ as needed. $\blacksquare$ Finally, we turn to the boring task of extracting the numerical answer. We have \[ \frac{s}{s+t} = \frac{[APQ]}{[ABC]} = \frac{\frac{\sqrt3}4 \left( \frac{b+c}{3} \right)^2} {\frac{\sqrt3}4 bc} = \frac{b^2+bc+c^2}{9bc} = \frac19 \left( 1 + \frac bc + \frac cb \right). \] So the problem is reduced to analyzing the behavior of $b/c$. For this, we imagine fixing $\Gamma$ the circumcircle of $ABC$, as well as the points $B$ and $C$. Then as we vary $A$ along the ``topmost'' arc of measure $120^{\circ}$, we find $b/c$ is monotonic with values $1/2$ and $2$ at endpoints, and by continuity all values $b/c \in (1/2,2)$ can be achieved. So \[ \frac{1}{2} < \frac bc < 2 \implies 4/9 < \frac{s}{s+t} < 1/2 \implies 4/5 < \frac st < 1 \]as needed.

WLOG $C\le 60$. Since $\angle BHC=180-A=120$ and $\angle BOC=2A=120$, $O\in (BHC)$. Let $f(X)$ be the reflection of $X$ over $BC$. Then $f(ABC)=(BHC)$, and let $f(H)=H' \in (ABC)$, and $f(O)=O'$, the arc midpoint of minor arc $\widehat{BC}$ in $(ABC)$. Note that this finishes part (a), since $H$ lies on the minor arc $\widehat{BC}$ in $(BHC)$, and since $O$ is the arc midpoint, clearly $OH$ cannot intersect segment $BC$. Hence, $OH$ must intersect $AB$ and $AC$. Then, \[ \angle HOB = \tfrac12 \widehat{BH} = \tfrac12 \widehat{BH'} = \angle BAH' = 90-B.\]Now, \[ \angle BHO = 180-\angle HBO-\angle HOB = 180-(60-C)-(90-B) = 150, \]so \[ \angle AHO = 360-\angle BHA - 150 = 360-(180-C)-150 = 150-B.\]Therefore, \[ \angle APH = 180-\angle PAH - (180-\angle AHO) = 60. \]This means that $\angle APQ=60$, hence $\triangle APQ$ is equilateral. Note that $AH=2R\cos A = R$. By Law of Sines on $\triangle AQH$, we have \[ \frac{AQ}{\sin \angle AHQ} = \frac{AH}{\sin 60} \implies AQ = R\frac{\sin(150-B)}{\sqrt3/2}=\frac{2R}{\sqrt3}\sin(30+B). \]Now, we find the areas. We have \begin{align*} [APQ] &= \frac{\sqrt3}{4} AQ^2 \\ &= \frac{\sqrt3}{4} \cdot \left(\frac{2R\sin (30+B)}{\sqrt3}\right)^2 \\ &= \frac{R^2\sin^2(30+B)}{\sqrt3}. \end{align*}And we know \[ [ABC] = \frac12 bc\sin 60 = \frac{\sqrt3}{4}(2R\sin B)(2R\sin C) = R^2 \sqrt3 \sin B \sin(120-B). \]Therefore, \begin{align*} \frac{s}{s+t} &= \frac{[APQ]}{[ABC]} \\ &= \frac{\sin^2(30+B)/\sqrt3}{\sqrt3 \sin B \sin(120-B)} \\ &= \frac{\sin^2(30+B)}{3\sin B\sin (60+B) } \\ &= \frac{\sin^2x}{3\sin (x-30) \sin (x+30)} \end{align*}where we let $x=30+B$, so that $90\le x\le 120$ and $3/4 \le \sin^2 x \le 1$. Using product-to-sum, this simplifies to \[ \frac{\sin^2x}{3(\sin^2x-1/4)} = \frac{a}{3(a-1/4)}\]for $a=\sin^2 x$ and $3/4\le a\le 1$. It is not hard to show that the above function is decreasing and continuous for $a\ge 3/4$, so

Note : This is very similar to $\text{Romania JBMO TST 2016}$ Proving $\triangle APQ$ is equilateral the problem gets trivial , so I'll show that it holds. My solution: This is a very well-known property that if $\angle A=60^{\circ}$ then $\bar{AH}=\bar{AO}$(really easy to prove) Thus $\angle AHP=\angle AOE$ ,but also : $\angle BAH=\angle CAO$ . Thus $\triangle AHP\cong{\triangle AOQ}$ , so $\angle A=\angle APQ=\angle AQP=60^{\circ}$ , done.

I totally did not mess up the calculations for an hour. Let $D$ be the arc midpoint of $BC.$ We claim that $OH$ is the perpendicular bisector of $AD.$ We prove this with complex numbers. Let the unit circle be the circumcircle of $\triangle ABC,$ let $B=-\frac{\sqrt{3}}{2}-\frac{1}{2}i$ and $C=\frac{\sqrt{3}}{2}-\frac{1}{2}i.$ Then note $H=A+B+C=A-i,$ and $D=-i.$ Since $A+D=H=H+O,$ $AODH$ is a parallelogram. Since $AO=OH,$ it follows that $DH=HA$ and $HO$ is the perpendicular bisector of $AD.$ The first part is equivalent to proving the perpendicular bisector of $AD$ intersects $(ABC)$ at points on major arc $BAC.$ Note $OH$ bisects arcs $ABD$ and $ACD.$ Since $\angle B,\angle C<90^{\circ},$ $60^{\circ}<\angle AOB,\angle AOC<180^{\circ}.$ Since $\angle BOD=\angle COD=60^{\circ},$ $\angle AOB>\angle BOD$ and $\angle AOC>\angle BOC.$ We claim the answer to the second part is the interval $(\frac{4}{5},1).$ Note that $\triangle APQ$ is equilateral since $\angle PAD=\angle QAD=30^{\circ}$ and $PQ\perp AD.$ Without loss of generality, let $AB>AC.$ Then note $\angle AOD=\angle AOB+\angle BOD=2\angle C+60^{\circ},$ implying that $\angle ODA=60^{\circ}-\angle C.$ Thus $AD=2\cos(60^{\circ}-C)=2\sin(30^{\circ}+C).$ Also note that $[APQ]=\frac{AD^2\sqrt{3}}{12}=\frac{\sin^2(C+30^{\circ})\sqrt{3}}{3}$ and $[ABC]=\frac{1}{2}\cdot AC\cdot BC\cdot \sin C=\frac{\sqrt{3}}{2}\cdot AC\sin C=\sqrt{3}\sin C\sin B,$ where the last step follows from the Extended Law of Sines. Also note $\sin B = \sin(120^{\circ}-C)=\sin(C+60^{\circ}).$ Let $\theta=C+30^{\circ}.$ Then \[\frac{[APQ]}{[ABC]}=\frac{\frac{\sqrt{3}}{3}\sin^2\theta}{\sqrt{3}\sin (\theta-30)\sin(\theta+30)}=\frac{1}{3}\cdot \frac{\sin^2\theta}{(\sin\theta\cos 30^{\circ})^2-(\cos\theta\sin 30^{\circ})^2}=\frac{1}{3}\cdot \frac{\sin^2\theta}{\frac{3}{4}\sin^2\theta-\frac{1}{4}\cos^2\theta}=\frac{1}{3}\cdot \frac{1}{\frac{3}{4}-\frac{1}{4}\cot^2\theta},\]where $\theta\in (60^{\circ},90^{\circ})$ since $\angle C\in (30^{\circ},60^{\circ}).$ Since this is a decreasing continuous function, we can plug in the minimum and maximum of $\cot \theta.$ The minimum is $\cot 90^{\circ}=0,$ and the maximum is $\cot 60^{\circ}=\frac{\sqrt{3}}{3}.$ This gives us $\frac{[APQ]}{[ABC]}\in (\frac{4}{9},\frac{1}{2}),$ giving us $\frac{s}{t}\in (\frac{4}{5},1),$ as desired.

Part a) follows from the fact that $BCOH$ is cyclic and $O,H$ lie on the same side of $BC$. It's not hard to see (say, by isogonality of $O,H$ or complex numbers) that $AH=AO$, so $APQ$ is equilateral. WLOG scale so $AO=\frac{1}{2}$, implying $[ABC]=\frac{1}{2} \sin(60)\sin(B)\sin(120-B)$. Noting that $\angle APQ = 60$ and $\angle AOP = 150-B$, by law of sines we have $$AP=\frac{\tfrac{1}{2}}{\sin(60)} \cdot \sin(B+30) = \frac{\sin(B+30)}{\sqrt{3}}$$. Thus we compute \begin{align*} \frac{[APQ]}{[ABC]} &= \frac{AP^2 \cdot \sqrt{3}/4}{\sin(B)\sin(120-B) \cdot \sqrt{3}/4} \\ &=\frac{1}{3} \cdot \frac{\sin^2(B+30)}{\sin(B)\sin(120-B)} \\ &=\frac{1}{3} \cdot \frac{\sin^2(x)}{\sin(x-30)\sin(x+30)} \\ &=\frac{1}{3} \cdot \frac{\sin^2(x)}{\frac{3}{4}\sin^2(x)-\frac{1}{4}\cos^2(x)} \\ &=\frac{1}{3} \cdot \left( 1+ \frac{1}{4\sin^2(x)-1} \right) \\ \end{align*}where we set $x=B+30$. Since $30 < B < 90$ we have $\frac{3}{4} < \sin^2(x) \leq 1$, giving us $$\frac{[APQ]}{[ABC]} \in (\frac{4}{3},\frac{3}{2})$$. It's easy to see that every value in this range can indeed be achieved by varying $A$ along major arc $BC$ while keeping $\Delta ABC$ acute (since $\Delta ABC$ cannot be equilateral, it cannot achieve the endpoint $\frac{4}{3}$). Thus we can compute the answer as $\frac{[APQ]}{[BCQP]} \in (\frac{4}{5},1)$.

(a) We suppose that $OH$ intersect $BC$, and if we can derive a contradiction, we are done. There are two possible configurations, that with $H$ to the left of $O$ and that with $H$ to the right of $O$. We have that $\angle ABH=30^{\circ}$, and it is easy to see that $\angle BOC=120^{\circ}, \angle OBC=\angle OCB = 30^{\circ}$. Similarly, it is well known that $\angle ACH=\angle OCB=30^{\circ},$ and this is obviously a contradiction, because we get that $\angle HCO=\angle HBO=0^{\circ},$ forcing $O$ and $H$ to be the same point, which is clearly not true because the triangle is given to be non-equilateral. (b) We first have a claim. Claim: $\triangle APQ$ is equilateral. Pf: It is easy to see that $\angle BOC=\angle BHC=120^{\circ},$ meaning BHOC is cyclic. We have $\angle CHO=\angle CBO=30^{\circ},$ so $\angle PHB=180^{\circ}-120-30=30,$ meaning $\angle APQ=60^{\circ}$. WLOG, assume $\triangle APQ$ has side length $1$. Let $BP=PH=x,$ then $HQ=1-x$. Notice $\triangle HQC$ is isosceles so $QC=1-x$. Therefore, $\frac{s}{t}=\frac{\sqrt{3}/4}{\sqrt{3}/4 \cdot ((x+1)(2-x)-1)}=\frac{1}{-x^2+x+1}$. It is easy to see by completing the square that for $0\leq x\leq 1,$(which is range of achievable $x$ on segment $PQ$), the min and max are $\frac{4}{5}$ to $1,$ and since quadratics are continuous, this is a continuous range and we are done.

Solution. Let $D,E,F$ be the foot of altitude from $A$, $B$, and $C$ respectively. Let $AB = c, AC=b$. WLOG $b+c=3$. Claim 1. $AFHE$ is cyclic Proof. $\angle AFH + \angle AEH = 90^{\circ} + 90^{\circ} = 180^{\circ}$. $\blacksquare$. Claim 2. WLOG, assume $ABC$ in a way such that $BHOC$ is convex. $BHOC$ is cyclic. Proof. From Claim 1. $\angle BHC = \angle FHE = 120^{\circ} = \angle BOC$. $\blacksquare$. (a). It follows since $BHOC$ is cyclic. $\blacksquare$. (b). By angle-chasing, $\triangle APQ$ is equilateral. Note that, $$BF = AB-AF = c- \frac{b}{2}$$$$FH = \frac{c}{\sqrt{3}} - \frac{b}{2\sqrt{3}}$$$$HP = \frac{2c}{3} - \frac{b}{3}.$$By symmetry, $HQ = \frac{2b}{3}-\frac{c}{3}$ thus $PQ = \frac{b+c}{3}$. Now note that $$\frac{[APQ]}{[BPQC]} = \frac{AP \cdot AQ}{AB \cdot AC - AP \cdot AQ} = \frac{\left(\frac{b+3}{3}\right)^2}{bc-\left(\frac{b+3}{3}\right)^2} = \frac{1}{b(3-b)-1} \ge \frac{4}{5}$$WLOG $b \ge c$. Note that $\triangle ABC$ is acute is equivalent to $2c>b\ge c$ by law of cosines. $2c > b \implies b < 2$. $b \ge c \implies b \ge \frac{3}{2}$. So the condition is only $\frac{3}{2} \leq b < 2$. Thus the max is $1$. $$\frac{4}{5} \leq\frac{s}{t} < 1$$

Why trig? For part (a), note that by the angle condition $BCOH$ is cyclic, so $\overline{OH}$ doesn't intersect $\overline{BC}$, thus it must intersect $\overline{AB}$ and $\overline{AC}$ as any line passing through the interior of $\triangle ABC$ intersects at least two of its sides. Now for part (b), note that $$\angle PBH=\angle ABH=90^\circ-\angle A=30^\circ,$$and (abusing notation), $$\measuredangle PHB=\measuredangle OHB=\measuredangle OCB=30^\circ,$$hence $\angle PBH=\angle PHB=30^\circ$, thus $PB=PH$, and $\angle APQ=180^\circ-\angle BPH=60^\circ$. Similarly, $QC=QH$, and $\angle AQP=60^\circ$, thus $\triangle APQ$ is equilateral. WLOG let $AP=AQ=PQ=1$, so as $PQ=PH+QH=PB+QC=1$, and by the sine area formula we have $$\frac{s}{t}=\frac{1}{(1+PB)(1+QC)-1}=\frac{1}{1+PB\cdot QC}.$$Since any valid choice of $(PB,QC)$ with $PB+QC=1$ yields a valid $\triangle ABC$ unless $PB=QC=\tfrac{1}{2}$ (which gives $\triangle ABC$ equilateral) or $PB\cdot QC=0$ (which gives $\triangle ABC$ right), and we have $0 < PB\cdot QC < \tfrac{1}{4}$ by AM-GM, it follows that $$\frac{s}{t} \in \left(\frac{4}{5},1\right).~\blacksquare$$

Wow, @above's solution is so clean! Part A: Because $ABC$ is acute and $\angle A = 60^{\circ}$, we know $BHOC$ is cyclic. Furthermore, since $O$ and $H$ are both inside of $ABC$ and $OB = OC$, we know $HO$ is the external bisector of $\angle BHC$. Thus, $OH$ meets $BC$ at a point not between $B$ and $C$, so the line must intersect segments $AB$ and $AC$, as none of $A, B, C$ lie on $OH$. Part B: WLOG, assume that $\angle B > \angle C$, and let the internal bisector of $\angle BAC$ meet $PQ$ at $M$. The angle condition implies $AO = AH$, so isogonality yields $$AM \perp OH \equiv PQ$$which means $AP = AQ$, i.e. $APQ$ is equilateral. Now, LoS allows us to compute $$AP = \sin AHP \cdot \frac{AH}{\sin APH} = \sin (APH + PAH) \cdot \frac{AO}{\sin 60^{\circ}}$$$$= \sin(60^{\circ} + (90^{\circ} - B)) \cdot \frac{R}{\frac{\sqrt{3}}{2}} = \sin(B + 30^{\circ}) \cdot \frac{2R}{\sqrt{3}}$$so $$[APQ] = \frac{AP^2 \sqrt{3}}{4} = \sin^2 (B + 30^{\circ}) \cdot \frac{R^2}{\sqrt{3}}.$$Now, the Extended LoS gives $$[ABC] = \frac{abc}{4R} = \frac{(2R \cdot \sin 60^{\circ}) \cdot (2R \cdot \sin B) \cdot (2R \cdot \sin C)}{4R}$$$$= R^2 \sqrt{3} \cdot \sin B \cdot \sin C = R^2 \sqrt{3} \cdot \sin B \cdot \sin (120^{\circ} - B).$$Hence, $$\frac{s}{t} = \frac{[APQ]}{[BPQC]} = \frac{[APQ]}{[ABC] - [APQ]} = \frac{\sin^2 (B + 30^{\circ})}{3 \cdot \sin B \cdot \sin (120^{\circ} - B) - \sin^2 (B + 30^{\circ})}.$$ Claim: We have $$\sin^2 (B + 30^{\circ}) = \frac{1}{4} + \sin B \cdot \sin(120^{\circ} - B).$$ Proof. The Sum Formula yields $$\sin^2 (B + 30^{\circ}) = \left(\sin B \cdot \frac{\sqrt{3}}{2} + \cos B \cdot \frac{1}{2} \right)^2$$$$= \frac{3}{4} \cdot \sin^2 B + \frac{1}{4} \cos^2 B + \frac{\sqrt{3}}{2} \cdot \sin B \cdot \cos B$$$$= \frac{1}{4} + \frac{1}{2} \cdot \sin^2 B + \frac{\sqrt{3}}{2} \cdot \sin B \cdot \cos B$$$$= \frac{1}{4} + \sin B \left(\frac{1}{2} \cdot \sin B + \frac{\sqrt{3}}{2} \cdot \cos B \right)$$$$= \frac{1}{4} + \sin B \cdot \sin (B + 60^{\circ}) = \frac{1}{4} + \sin B \cdot \sin(120^{\circ} - B)$$as desired. $\square$ Thus, substituting gives $$\frac{s}{t} = \frac{\frac{1}{4} + \sin B \cdot \sin(120^{\circ} - B)}{2 \cdot \sin B \cdot \sin (120^{\circ} - B) - \frac{1}{4}} = \frac{\frac{1}{4} + \sin B \cdot \sin(120^{\circ} - B)}{2 \left(\frac{1}{4} + \sin B \cdot \sin(120^{\circ} - B) \right) - \frac{3}{4}}.$$Now, let $\angle B = 60^{\circ} + K^{\circ}$. Then, Sum-to-Product yields $$\sin B \cdot \sin(120^{\circ} - B) = \sin \left( \frac{120^{\circ} + 2K^{\circ}}{2} \right) \cdot \sin \left( \frac{120^{\circ} - 2K^{\circ}}{2} \right)$$$$= - \frac{1}{2} \left(\cos 120^{\circ} - \cos 2K^{\circ} \right) = \frac{\cos 2K^{\circ}}{2} + \frac{1}{4}.$$Because $\angle B > \angle C$, we know $0^{\circ} < K^{\circ} < 30^{\circ}$, i.e. $0^{\circ} < 2K^{\circ} < 60^{\circ}$. In addition, since $\cos(\theta)$ is strictly decreasing on the open interval $\left(0^{\circ}, 60^{\circ} \right)$, we have $$\frac{\cos 60^{\circ}}{2} + \frac{1}{4} < \frac{\cos 2K^{\circ}}{2} + \frac{1}{4} < \frac{\cos 0^{\circ}}{2} + \frac{1}{4}$$or $$\frac{1}{2} < \frac{\cos 2K^{\circ}}{2} + \frac{1}{4} < \frac{3}{4}$$which implies $$\frac{3}{4} < \frac{1}{4} + \sin B \cdot \sin(120^{\circ} - B) < 1.$$ Claim: The function $f(x) = \frac{x}{2x - \frac{3}{4}}$ is strictly decreasing for all $x > \frac{3}{8}$. Proof. The proof is trivial and left to the reader. $\square$ Now, it's clear that $$f(1) < \frac{s}{t} = \frac{\frac{1}{4} + \sin B \cdot \sin(120^{\circ} - B)}{2 \left(\frac{1}{4} + \sin B \cdot \sin(120^{\circ} - B) \right) - \frac{3}{4}} < f(\frac{3}{4})$$or $$\frac{4}{5} < \frac{s}{t} < 1.$$$\blacksquare$ Remarks: I'm actually very happy with this solution, as it's quite rare for me to solve a Geometric Inequality and/or Trig Bash an Olympiad Geometry problem! I'm still surprised that I realized $$\sin^2 (B + 30^{\circ}) = \frac{1}{4} + \sin B \cdot \sin(120^{\circ} - B).$$Sometimes, abusing nice constants, such as $\frac{1}{2}$ and $\frac{\sqrt{3}}{2}$, is helpful!

Another USAJMO Geom that is trivialised by Cartesian Coordinates. Bash by setting $A = (0,0),B=(k,0),C=(\dfrac{1}{2},\dfrac{\sqrt3}{2})$ We eventually get $\dfrac{1}{2}\leq k \leq 2$ and $$\dfrac{s}{t}=\dfrac{(k+1)^2}{9k-(k+1)^2}$$Which is between $\dfrac{4}{5}$ and $1$ for that range of $k$. Full proof here: https://infinityintegral.substack.com/p/usajmo-2014-contest-review