Pool Shark #2

As shown above, reflect the table over the diagonal line to achieve a square.

What you now need to do is aim at the reflected midpoint, M.

Since AMP and CPB are similar with ratio AM:CB=1:2, we achieve AP:CP=1:2, leading to AP= 1/3 AC.

Relevant wiki: Similar Triangles Problem Solving - Basic

Since the triangle is isosceles, we know $\angle A \cong \angle B$ . Since the ball is reflected back at the same angle after hitting the hypotenuse, $\angle ADE \cong \angle BDC$ . Therefore, $\triangle ADE \sim \triangle BDC$ . Noting that $BD = \sqrt{2} - ?$ , we set up an equation:

$\frac{?}{\sqrt{2}-?} = \frac{1}{2} \implies 2? = \sqrt{2}-? \implies ? = \frac{\sqrt{2}}{3}$

Therefore, $?$ is $\boxed{\frac{1}{3}}$ of the hypotenuse.

Reflect the goal in the longer side (hypothenuse), and draw a line from the start position to the goal's reflection. Consider the set of equidistant parallel lines as shown in the diagram. Such lines divide any intersecting line in equal parts. It is then clear that the desired segment is \fbox{\$\frac{1}{3}\$} of the longer side.

Assuming no trick-shot spin, the ball must bounce of the hypotenuse of the triangle with equal incident-reflection angles (alpha) in the diagram below.

Simple geometry shows that triangle A and B in diagram have same proportions and B is 2 x size of A (length dimensions). Therefore the hypotenuse of the original triangle must be divided by ratio 1:2 or 1/3 : 2/3.

Applying the same principle of following a line across adjoining reflected tables used in a solution of the 2 bounce center to upper right corner problem.

I got a really easy intuitive solution without maths: Imagine a line parallel to the hypothesis going through the starting point. Then imagine the square with the goal point in its centre. Now complete the path the ball goes in this square so that the path goes through the goal and ends in the corner opposite to the starting corner. The 3 straight segments of this path obviously have the same length. As well as the shadow they throw on the hypothesis. So it's obvious the aim-point is at 2/3 of the side of the square, which is 1/2 of the hypothesis. 2/3 * 1/2 = 1/3

Since D is the center of segment $AB$ , assign $AB=2$ . Therefore, $AD=DB=\frac12$ . $\angle ABC=\angle ACB=45 \circ$ . $\triangle ADK$ and $\triangle CBK$ are similar, as their angle measures are the same. This means that $\frac{AD}{AK}=\frac{CB}{CK}$ . But $\frac{AD}{BC}=\frac12$ , then $\frac{AK}{CK}=\frac12$ , meaning that $\frac{AK}{CK}=\boxed{\frac{1}{3}}$

Go play pool for a while and you will know that it is about 1/3 down.

Let the triangle vertices be A,B,C, where A is at top and B is the right angle. Let D be the mid-point of the left side, P be the point on the hypotenuse where the ball strikes, and <t = <BPC = <APD, by the reflection law. Let S be a side, S sqrt(2) the hypotenuse, and x = AP. We have the following equations by the Law of Sines: (1) (S/2)/sin(t) = x/sin(135 -t), and (2) (S sqrt(2) - x)/sin(135 - t) = S/sin(t). Substituting, 2x = S sqrt(2) - x, and x = (1/3)S sqrt(2) = (1/3)*hypotenuse. Ed Gray

You do not need fancy maths to solve this problem. Let the Goal be point G. Let the bounce point on AC be D. Let the angle BDC be alpha. If no energy is lost in the bounce then the angle ADG will also be alpha. Angles GAD and DCB are both 45 degrees. So, triangles AGD and BDC are similar triangles. AG is opposite angle ADG (alpha). BC is opposite angle ADG (alpha). BC is twice as long as AG. So, DC is twice as long as AD. So AD is one third as long as AC. So you should aim the ball to hit AC one third of the way between A and C. It takes longer to explain than it does to work it out in your head.

X = 1/3