USAAAO 2026, III кръг — Задача 2. Travelers in Outer Space

• (a)(i) 6.8

  Official answer 6.8 ± 0.6; any uncertainty in the printed range is accepted.

• (a)(ii) 590 days

  Official answer 590 ± 30 days (any uncertainty from 10 to 50 days is reasonable).

• (a)(iii) About 0.73 AU — the orbit of Venus.

  The official solution calls the traveller Chert in this part although the problem refers to Esker.

• (b)(i) $0 \leqslant \theta \leqslant \frac{r}{D}$

• (b)(ii) $4 \times 10^{8}$ m

  Official answer (4.0 ± 0.1) × 10⁸ m, from a slope r = 8.7 ± 0.3 mas·ly.

• (b)(iv) On the far side of the Moon (about 4 × 10⁸ m from Earth; the Moon is tidally locked to the Earth).

• (c) Near Capella (all great-circle displacement lines intersect at Capella).

Solution:

(a) (i) (5 points) To get a sense of the motion, we need to make a plot such as the one in Figure 1, and draw a trend line. Note that the line in Figure 1 shows the true value, based on real astronomical data.

From drawing a best fit line, the peak seems to be around $4.1 \pm 0.1$ and the lowest looks to be around $0.6 \pm 0.05$.

This gives a ratio of $\frac{4.1}{0.6} = 6.8$. The relative uncertainty in the quotient is $\sqrt{\left(\frac{0.1}{4.1}\right)^2 + \left(\frac{0.05}{0.6}\right)^2} = 0.09$, meaning that the ratio with uncertainty would be $\boxed{6.8 \pm 0.6}$. This is also the answer one arrives at by reading off the largest and smallest values from the table, which is also acceptable for this part.

Any uncertainty in the range 0.4 to 0.12 would be acceptable, as determining the uncertainty from the graph would be subjective.

(ii) (10 points)

From the plot, we see that the peaks on days 440 and 1030 are close to the true peaks. We can estimate the period to be $1030 - 440 = \boxed{590 \pm 30}$ days (any uncertainty from 10 to 50 days is reasonable).

(iii) (10 points) We first consider the result of part (a)(i). Suppose Chert is orbiting at a distance of $x$ AU from the sun. The farthest distance from the Earth is $1+x$, and the nearest distance from the Earth is either $x-1$ or $1-x$, depending on whether $x > 1$.

If we assume $x < 1$, the equation becomes $\frac{1+x}{1-x} = 6.8 \pm 1.2$, (using the upper bound of the range of uncertainty) which gives $x = 0.74 \pm 0.05$. If $x > 1$, the equation is $\frac{1+x}{x-1} = 6.8$, which gives $x = 1.34 \pm 0.10$. The uncertainties in this part of the solution were computed by solving the equation using the upper and lower bounds of $6.8 \pm 1.2$.

From (a)(ii), the synodic period of Chert's orbit with Earth is $590 \pm 50$ days (using the upper range of acceptable uncertainties). If the sidereal period of Chert's orbit is $T$ days, then either $\frac{1}{T} - \frac{1}{365} = \frac{1}{590}$, or $\frac{1}{365} - \frac{1}{T} = \frac{1}{590}$, depending on whether $T > 365$.

Solving these equations for $T$, we get that $T$ is either $960 \pm 80$ days or $225 \pm 8$ days. By Kepler's 3rd law, these correspond to orbits of radius $\left(\frac{960 \pm 80}{365}\right)^{2/3} = 1.9 \pm 0.2$ AU and $\left(\frac{225 \pm 8}{365}\right)^{2/3} = 0.72 \pm 0.02$ AU.

The only orbital radius consistent with both parts would be about 0.73 AU, which corresponds to the orbit of $\boxed{\text{Venus}}$.

(b) (i) (5 points) Let the location of the Earth be $E$, the location of the star be $S$, and the location of Esker be $K$.

The angular displacement observed by Esker is equal to the measure of $\angle ESK$. Using the law of sines, $\frac{EK}{\sin\angle ESK} = \frac{SK}{\sin\angle SEK}$.

We then use the following:

• $EK = r$ (from the problem statement)
• $\sin\angle ESK = \angle ESK$ (using a small angle approximation)
• $SK = D$ (because $D \gg r$)

Then, we get $\angle ESK = \frac{r}{D}\sin SEK$.

Because $\sin\angle SEK$ ranges from 0 to 1, we get that $\boxed{0 \leqslant \theta \leqslant \dfrac{r}{D}}$.

(ii) (12 points) If we plot the angular displacement $\theta$ vs $1/D$, we have from part (b)(i) that all the points should lie below a line of slope $r$. If we make this plot and draw a line as in Figure 2, the points seem to lie below a line of slope $r = 8.7$ mas/ly$^{-1}$.

Some uncertainty comes from the chance that there aren't any stars with the exact maximum parallax, so any estimate of the uncertainty from 0.1 to 0.5 is reasonable.

Converting mas into radians, we get that

(iii) (3 points) As in part (b)(i), let the location of the Earth be $E$, the location of the star be $S$, and the location of Esker be $K$. Suppose that $\overrightarrow{ES}$ is a random direction on the celestial sphere. $\angle SEK$ is smaller when $\overrightarrow{ES}$ is closer to the poles. Since more of the area of a sphere lies away from the poles, larger values of $\angle SEK$, and thus larger values of $\theta$ relative to the maximum value, are more likely. Figure 3 shows a visualization of this effect.

Any explanation mentioning either the fact that $\angle SEK$ is likely to be close to $90^{\circ}$, or that the distribution of $\sin\theta$ for $\theta \in [0^{\circ}, 180^{\circ}]$ is biased towards larger values, should get full credit.

(iv) (5 points) From part (b)(ii), Esker is around $4 \times 10^8$ m from the Earth, which should be recognizable as being around the Earth-Moon distance.

Since the Moon is tidally locked to the Earth, the reason Esker never sees Earth could be that they are on $\boxed{\text{the far side of the Moon}}$.

(c) (25 points) Let $G$ be the (unknown) apparent location of Gabbro on the celestial sphere as viewed from Earth. For a given star, let $S$ be the apparent location of that star from Earth, and $S'$ be the apparent location of that star from Gabbro's location. The key insight is that $G$, $S$, and $S'$ all lie on a great circle.

One way to prove this assertion is to consider the locations of Earth, Gabbro, and the star in 3D space. Let $O$ be the position of the Earth in 3D space, let $X$ be Gabbros' position, and let $Y$ be the star's location.

Then, the direction of $G$ on the celestial sphere is the direction of the vector $\overrightarrow{OX}$. Similarly, $S$ corresponds to the direction of $\overrightarrow{OY}$, and $S'$ corresponds to $\overrightarrow{XY}$. Note that all three of these vectors lie within the plane passing through $O$, $X$, and $Y$. This means that the direction of those vectors lie on the plane passing through the origin, parallel to this plane. Since the intersection of a sphere and a plane passing through its center is a great circle, this concludes the proof.

Thus, we can plot the position of a few stars from Earth's perspective and from Gabbro's perspective on a gnomonic projection, draw the line connecting them, and see where they all intersect. This is shown in Figure 4.

From the figure, we see that all the lines intersect around $\boxed{\text{Capella}}$, whose location is also listed in the table.