Problem 10 of 2001 Euclid

“Points P and Q are located inside the square ABCD such that DP is parallel to QB and DP = QB = PQ. Determine the minimum possible value of $\angle ADP$ .”

The CEMC gave 4 solutions. Here I want to discuss a more general method using analytic geometry and trignometry. Let’s place the diagram on a coordinate plane as shown below.

Let $k$ be the slope of the line DP. The problem is actually asking for the maximum possible value of $k$ such that the conditions “P and Q are inside ABCD”, “DP is parallel to QB” and “DP = QB = PQ” are satisfied.

The equation of DP is $y=kx$ . Because DP//QB, the equation of QB will be $y=kx+b$ . By plugging in B(1, 1), we have $b=1-k$ and $y=kx+1-k$ . Because $DP//QB$ and $DP=QB$ , we should notice the connections between the coordinates of P and Q, that is, If we let the coordinates of P be $(x, kx)$ , then the coordinates of Q will be $(1-x, 1-kx)$ .

The condition $DP=PQ$ gives us the following equation:

$x^2+(kx)^2=(1-x-x)^2+(1-kx-kx)^2$

After calculations, we have the following quadratic equation about $x$ :

$(3k^2+3)x^2-(4k+4)x+2=0$

If the above equation has real solutions, the discriminant must be greater than or equal to 0. That is,

$(4k+4)^2-8(3k^2+3) \geq 0$ .

Then we have $2-\sqrt{3} \leq k \leq 2+\sqrt{3}$ . Therefore, the maximum possible value of $k$ is $2+\sqrt{3}$ .

Next we need to verify that when $k=2+\sqrt{3}$ , P and Q are really inside the square ABCD. This can be done by solving for $x$ and $y$ when $k=2+\sqrt{3}$ . The result shows it is true. (The calculations are omitted here).

Finally, we can say the maximum possible value of $k$ is $2+\sqrt{3}$ indeed. Notice that $tan(75^{\circ})=2+\sqrt{3}$ , therefore the minimum possible value of $\angle ADP$ is $15^{\circ}$ . Done.

Problem 10 of 2001 Euclid

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