# You and your "Princess" [closed]

You are an Archer and participating in a tournament. You need to hit the Bull's Eye to win the tournament and the Prize (a big fat pig called "Princess". Wodehouse Anyone?). You love big fat pigs and have to win the Tournament or else loose your sweet heart to one of the other big fat pig lovers.

Anyhow, the task is not easy as the Target is mobile. The target follows a horizontal path of an isosceles triangle with the unequal side being furthest side from you. The trajectory of the path is parallel to the ground. You have taken the measurements and made the calculations beforehand. They are as follows:

• The two equal sides of the trajectory are 5 meters each and the third side being 6 meters.
• You are 80 meters away from the closest vertex of the triangle and are directly in front of it and in a line perpendicular to the furthest unequal side.
• The target is moving at a speed of 28 miles an hour.
• You know that when you release the arrow from a confident and relaxed posture, it always goes at a speed of 98 miles an hour (Yeah always...and you know that).
• Looking from above the target trajectory the target is moving in the counter-clockwise direction.
• You are given special arrows that do not follow the rules of gravity, and will fly parallel to the ground as long as they are within the arena.

After the calculations are you confident that you can get your beloved "Princess"? Why? When and at what angle are to going to shoot your arrow?

If not, are you just going to let another big fat pig lover (!!!RESPECT!!!) going to take your "Princess" away?

• Is this a trick question? Can't you just shoot the arrow into any point of the triangle, timing it so the target will be at that point when the arrow hits based on the travel time?
– xnor
May 19, 2015 at 5:45
• @xnor: well...wasn't that the question in the first place? to find when and at what angle are you going to shoot your arrow? I was assuming that the part "timing" the arrow was an obvious part. May 19, 2015 at 5:54
• In that case, I don't think this is a puzzle, but a textbook algebra/geometry problem, which isn't on-topic here.
– xnor
May 19, 2015 at 5:57
• @xnor: read the tags on the question, and with a little thought i guess even you'll realize that it is a puzzle. May 19, 2015 at 5:59
• In SI units, 28 mph=12.52 m/s and 98 mph=43.81 m/s May 19, 2015 at 6:13

You are 84 meters from the midpoint of the base of the triangle (52 - 32 = 42); assume you shoot at this point.

It will take your arrow 84m / (98mph) = 84m / (43.81m/s) = 1.92s to get to the base of the triangle.

The target is moving at 12.52m/s; therefore, it travels 12.52m/s * 1.92s = 24.04m from the time you shoot your arrow to the time your arrow arrives at the center of the base. This is one full cycle of the triangle, plus 8.04m.

That means you should release the arrow when the target is 0.04m from the apex of the triangle. The target will arrive at the midpoint of the trajectory's base just as your arrow strikes that point. Furthermore, because the target will be traveling along the base during the interval that your arrow goes from 80 to 84m, you do not need to worry about accidentally nipping the side of the target as it moves along one of the 5m sides.

• nice ... will give you the check mark and +1. May 19, 2015 at 7:33
• I did make the error with the units...but i guess it's ok :) May 19, 2015 at 7:35

Looking at this from a logical point of view, two answers spring to mind:

As the bullseye is over six metres wide, just aim for the middle.

If the above is a bit too ridiculous, then:

are you confident that you can get your beloved "Princess"?: No. The bullseye is two metres off the ground, but I am short and can only launch an arrow from 1.75 metres off the ground, and since the arrows will only fly parallel to the ground, I cannot hit regardless of how good my geometry is!