Dansc: As we know, a six-point big eye produces 12 liberties. Counting the four white stones inside the eye, black still has 8 liberties - white, on the other hand, has only three. If white plays at a, black can play elsewhere five times and still win the semeai.
BarkOfDelight: Is this a fair problem for a beginner? The answer seems to be "White can't really do much." Is that the point of this problem? Just doesn't seem to fit the spirit.
Berlioz: But black doesn't have to take the ko at all, he can simply kill whites outer group, so white is dead no matter what.
Herman Hiddema: When white plays to start the ko, it is an atari against the black stones, so if black starts filling liberties on the outside, white can capture. Therefore, black has no choice but to play the ko if he wants to capture white.
Doesn't it only take one play before black captures by playing at b?
Because white will make a ko threat and if black answers white captures back, if black ignores the threat and captures the three white stones, white will carry out what the threat entails. This is how a ko battle works. See https://senseis.xmp.net/?KoFight