Problems for Chapter O1
- Builds on
- Sections O1.1 to O1.8
- Reading time
- 1 h
- Exercises
- 0
- Olympiad problems
- 4
The problems in this section come from the observation rounds of past olympiads, where knowing the sky is tested directly. In the exam they are done under the real sky, in a planetarium or on a printed chart, and here the two that need a picture of the sky come with a chart drawn from a star catalogue, while the other two can be solved from what this chapter teaches. They are not labelled with the sections they need, because recognising which pattern or hop a problem calls for is the skill being tested. Each one opens in place, with a hint and a full solution that is revealed only when asked for.
O1.9.1From textbook to olympiad
Missing stars in Perseus and Andromeda, from IOAA 2009, gives a chart of part of the sky from which four bright stars have been removed, and asks which ones. Treated like a textbook exercise it looks hopeless, since there are hundreds of stars on the chart and nothing to calculate. The way through is the method of Section S2, with patterns taking the place of equations.
- We start by finding our bearings. Polaris is named on the chart, so north is known, and the brightest patterns that are complete, such as the W of Cassiopeia and Capella with its small triangle of stars, fix where everything else should be.
- Next we use every piece of information. The missing stars are bright, so each leaves a gap in a pattern we know well. The table of candidates also helps, since a star that is plainly on the chart can be struck off it.
- Then we name the patterns. In Andromeda the pattern is the chain of three bright stars from the corner of the Great Square, and in Perseus it is the curved line of stars with Mirfak in the middle (Section O1.5).
- Now we work back from the gaps. A pattern with a star too few shows exactly where the star should be, and the hops of Chapter O1 give the distance, such as the 12.5° from Mirach to Almach.
- Last, we check the answer. Each star we name must be in the table and must be missing from the chart, and the four must all lie in Perseus or Andromeda.
The step that unlocks the problem is the third. Once the patterns are named, each missing star shows up as a gap in one of them, and the hops of this chapter give the distance from its neighbours to the place where it belongs.
2009Missing stars in Perseus and Andromeda40 points
Adapted from IOAA 2009 (Tehran), observation problem 1. We have restated it in our own words and kept the original data. The original was solved partly by looking at the real sky, and its chart was drawn for the view from Tehran. Ours is redrawn as a map with north at the top. The original table calls ζ Persei by the name Atik, which the IAU has since given to ο Persei, so we list it by its Bayer letter. The solution was last checked on 18 October 2026.
The original chart showed the sky around Perseus and Andromeda, about 100° by 70°, as it was on 22 October 2009 at 21:00 local time in Tehran. Our chart shows the same part of the sky as a map with north at the top. Four bright stars of Perseus and Andromeda have been left off it. Find the four missing stars and give each its number from the table. Polaris is named on the chart.
| Number | Name | Bayer letter |
|---|---|---|
| 1 | Mirfak | α Per |
| 2 | Alpheratz | α And |
| 3 | ε Per | |
| 4 | Menkib | ξ Per |
| 5 | γ Per | |
| 6 | Algol | β Per |
| 7 | Almach | γ And |
| 8 | δ And | |
| 9 | 51 And | |
| 10 | Mirach | β And |
| 11 | ζ Per |
| Star | Tap a star on the chart. |
|---|---|
| Constellation | |
| Magnitude | |
| Angle between two stars | Tapping a second star gives the angle between them. |
- Three corners of the Great Square, Scheat, Markab and Algenib, are where they should be, but its north-eastern corner is empty. That corner is Alpheratz, number 2, 14.2° east of Scheat and 14.0° north of Algenib.
- The chain of Andromeda runs from Alpheratz through Mirach, number 10, which is on the chart 14.5° from the empty corner. The next star of the chain, Almach, number 7, should lie 12.5° beyond Mirach, and that place is empty too.
- The chain continues, bending only a little, to the curved line of Perseus. The line is there, with γ Per, number 5, and δ Per, but the place in its middle, 15.8° beyond Almach, is empty. That is where Mirfak, number 1, belongs.
- Algol makes a nearly right-angled corner with Almach and Mirfak, 12.1° from Almach and 9.3° from Mirfak, almost due south of Mirfak. That corner is empty too, and it is the place of Algol, number 6.
- The other candidates, ε Per, ξ Per, γ Per, δ And, 51 And, Mirach and ζ Per, are all on the chart, so the four missing stars are found.
AnswerThe missing stars are numbers 1, 2, 6 and 7, which are Mirfak, Alpheratz, Algol and Almach.
2007The celestial pointer10 points
Adapted from IOAA 2007 (Chiang Mai), observation problem 1. We have restated it in our own words and kept the original data, adding the latitude of Chiang Mai and the size of the region to search in the last part, which the original left open. The solution was last checked on 18 October 2026.
An observer in Chiang Mai, Thailand, at latitude 18.8° N, has a pointer to show objects in the night sky. The examiner asks for the following.
- Move the pointer along the celestial equator. (1 point)
- Aim the pointer at the vernal equinox. (1 point)
- Aim the pointer at the brightest star of the Great Square of Pegasus. (2 points)
- Aim the pointer at α Ari. (2 points)
- Start at Aldebaran, α Tau, turn the pointer 35° northward and then 6° westward in equatorial coordinates, and aim at the brightest star within about 10° of that point. (4 points)
- The celestial equator is the great circle through the east and west points of the horizon. It crosses the meridian in the south at an altitude of
- The vernal equinox is the March equinox point, where the Sun crosses the equator northward. It lies in Pisces with no bright star to mark it, 15.5° from Algenib, a little west of due south, which is about the length of the eastern side of the Great Square, 14.0°.
- The four corners of the Great Square are Alpheratz, magnitude 2.07, Scheat, 2.44, Markab, 2.49, and Algenib, 2.83. The brightest is Alpheratz, at the north-eastern corner, even though it belongs to Andromeda.
- α Ari is Hamal, magnitude 2.01, the brightest of the short bent line of three stars that makes up Aries.
- Aldebaran is at right ascension 4h 35.9m and declination +16.5°. Turning 35° north brings the pointer to declination +51.5°, and turning 6° west takes 6° off the right ascension, which brings it to 4h 11.9m. The point is in Perseus. Within 10° of it the brightest star is Mirfak, magnitude 1.8, 7.7° away, brighter than δ Per, magnitude 3.0, 6.0° away. If the 6° is measured along the sky instead, the point is at 3h 57m, and Mirfak, 5.5° away, is still the brightest. Capella is brighter, but it is 12° away.
AnswerThe equator runs from east to west, 71.2° high in the south and just north of Mintaka. The vernal equinox is in Pisces, 15.5° from Algenib, a little west of south. The brightest star of the Square is Alpheratz, α Ari is Hamal, and the last star is Mirfak.
2015Three stars missing from the Magelang sky60 points
Adapted from IOAA 2015 (Magelang), planetarium problem 1. We have restated it in our own words and kept the original data, adding the latitude and longitude of Magelang. The original was shown in a planetarium, and our chart redraws that sky as a map of the whole sky above the horizon. The solution was last checked on 18 October 2026.
A planetarium shows the sky of Magelang, Indonesia, at latitude 7.5° S and longitude 110.2° E, for 28 July 2015 at 21:00 local time (UTC+7). Three bright stars are missing from the image. Find them, and name them by their Bayer letters. The stars that may be missing are α Pav, α Lyr, α Aql, α Vir, α CrB, α Gru, α Ser, α Cen, α Cyg, α Oph, α Cru, α Boo, α PsA, α Sco, β Aqr, β Cru, β Oph, β Peg, γ Sco, γ Cyg, δ Pav, ε Peg and ε Boo. The star called γ Sco in this list is the one now named σ Lib.
| Star | Tap a star on the chart. |
|---|---|
| Constellation | |
| Magnitude | |
| Angle between two stars | Tapping a second star gives the angle between them. |
- At 21:00 local time, 14:00 UT, the local sidereal time is about 17h 45m, so the meridian runs through Ophiuchus and Sagittarius, and Antares is 64° up, west of the meridian. All 23 stars of the list are above the horizon at that moment, the lowest being β Peg at 5°, so none can be struck off for being set, and each pattern has to be checked on the chart.
- The Southern Cross is low in the south-south-west, with Acrux 11° and Mimosa 14° up, and Hadar is beside it. The brighter Pointer, α Cen, should be 4.4° from Hadar, on the side away from the Cross, but that place is empty. α Cen is missing, at an altitude of 27° and an azimuth of 203°.
- Corvus is in the west. The line along its northern edge from Gienah through Algorab leads on, almost straight, for 14.5° to an empty patch of sky where Spica should be. α Vir is missing, 26° up at an azimuth of 261°.
- In the north-west the kite of Boötes is on the chart, with ε Boo, δ Boo, β Boo and γ Boo, and Corona Borealis beside it, but the bright star at the southern tip of the kite, 10.4° from ε Boo, is missing. That star is α Boo, Arcturus, 32° up at an azimuth of 298°.
AnswerThe missing stars are α Cen in the south-south-west, α Vir in the west and α Boo in the west-north-west.
2025The Sun and the Moon around the eclipse of 20084 points
Adapted from IOAA 2025 (Mumbai), sky map problem 1. We have restated two parts of it in our own words and kept the original data. The original asks for the Sun and the Moon to be drawn on a printed sky map, and we ask for their positions in words. The list of stars in the first part comes from another part of the original, which asks for them to be marked on the same map. The solution was last checked on 18 October 2026.
A total eclipse of the Sun took place on 1 August 2008, and at one place on Earth totality came at local noon.
- Give the position of the Sun among the stars at that moment. Name its constellation, and say which of these stars it was closest to: β Aur, δ Cep, δ Cnc, δ Cet, δ Gem, β CVn, α Lyn and β Per. (1 point)
- Describe the position of the Moon among the stars at local noon at the same place on 28 July 2008, and its shape. Which side of it was lit? (3 points)
- During a total eclipse the Moon is in front of the Sun, so both are at the same place on the sky, on or very close to the ecliptic. The Sun passes Regulus, at ecliptic longitude 150°, around 23 August (Section O1.7), and it moves about 1° a day along the ecliptic, so on 1 August, 22 days earlier, it was near longitude
- By the table of Section O1.7 the Sun is in Cancer from 21 July to 11 August, so it was in Cancer. Of the eight stars, δ Cnc, at ecliptic longitude 128.7°, lies within 0.1° of the ecliptic, and the Sun was only 0.7° from it.
- On 28 July, four days before the eclipse, the Moon was west of the Sun. It moves eastward among the stars by about 13.2° a day and the Sun by about 1.0°, so it gains on the Sun by about 12.2° a day, and four days earlier it was about 49° west of it. The full calculation gives 55°, because in those days the Moon was near perigee and moving faster than average. That puts it at ecliptic longitude about 71°, in Taurus, 4.8° north of the ecliptic and about 10° north of Aldebaran.
- A Moon 55° from the Sun is a crescent, with an illuminated fraction of
AnswerOn 1 August 2008 the Sun was in Cancer, 0.7° from δ Cnc. On 28 July the Moon was in Taurus, about 10° north of Aldebaran, as a waning crescent about one fifth lit, with its bright side towards the east.