A Point Determined by Ratios

WLOG let $PA=1$ . Let $P'$ be the image of $P$ under a rotation with centre $B$ that sends $C$ to $A$ . Clearly, $BP' = 2$ , $AP' = 3$ and $\angle PBP' = 90 ^ \circ$ . $PP'^2 = 2^2+2^2= 8 = AP'^2-AP^2$ . Hence $\angle P'PA = 90 ^ \circ$ . $\angle APB = 90 ^ \circ+45 ^ \circ = 135 ^ \circ$ .

[LaTex edits - Calvin]

Let $E,F$ be the foot of the perpendiculars from $P$ to $AB,AD$ , respectively. Let $x=AE,y=AF,l=$ side of the square. We may assume that $PA=1,PB=2,PC=3$ without loss of generality.

$x^2+y^2=1^2,(l-x)^2+y^2=2^2,(l-x)^2+(l-y)^2=3^2$ . Subtracting the first 2 and the last 2 equations gives $l^2-2lx=3$ and $l^2-2ly=5$ , respectively.

We solve for $x,y$ in the 2 new equations and substitute them in the very first equation to get $l^4-10l^2+17=0$ . $l^2$ must be positive, so $l^2=5+2\sqrt2$ (*) by quadratic formula.

Using cosine law, $l^2=5-4cos \angle APB$ ; this and * give $cos \angle APB=-\frac{\sqrt2}{2}$ , and since $\angle APB<180^\circ$ , $\angle APB=135^\circ$ .

Firstly, we construct the point Q, where $\bigtriangleup APB \equiv \bigtriangleup BQC$ .

$\angle PBQ = \angle ABC + \angle CBQ - \angle ABP$

By the congruence of $\bigtriangleup APB$ and $\bigtriangleup CQB$ ,

$\angle ABP=\angle CBQ$ ,

Thus $\angle BPQ = \angle ABC = 90^\circ$

Since $\angle PBQ = 90^\circ$ ,

BP=BQ (By $\bigtriangleup APB \equiv \bigtriangleup CQB$ )

We know $\bigtriangleup PBQ$ is a right isosceles triangle, where $\angle PBQ = 90^\circ$ .

Thus $\angle PQB = 45^\circ$ .

Let $PB = QB = 2 \text{ units}$

By the Pythagorean Theorem, $PQ = \sqrt{PB^2+QB^2} = \sqrt{(2 \text{ units})^2+(2 \text{ units})^2} = \sqrt{8} \text{ units}$ .

By the Pythagorean Theorem, since we can easily see that given:

$PC = 3 \text{ units}$ , $QC = PA = 1 \text{ unit}$ (By $\bigtriangleup APB \equiv \bigtriangleup CQB$ )

We know $PC^2 = (3 \text{ units})^2 = (\sqrt{8} \text{ units})^2 + (1 \text{ unit})^2 = PQ^2 + QC^2$ .

Hence $\bigtriangleup PQC$ is a right triangle, with $\angle PQC = 90^\circ$

$\angle APB = \angle CQB$ (By $\bigtriangleup APB \equiv \bigtriangleup CQB$ )

$= \angle PQB + \angle CQP$ $= 45^\circ + 90^\circ$ $= 135^\circ$ .

Let the foot of perpendicular from P to AB be E and the perpendicular from P to BC be F. Let PF=x. EB=PF=x. BF=sqrt(2^2-x^2)=sqrt(4-x^2). CF=sqrt(3^2-x^2)=sqrt(9-x^2). PE=BF=sqrt(4-x^2). Thus, AF=sqrt(1^2-(4-x^2))=sqrt(x^2-3).

Since ABCD is a square, AB=BC. AB=sqrt(a^2+3)+a, BC=sqrt(4-a^2)+sqrt(9-a^2). Solving for a, we get a=sqrt((2/17)(29+2sqrt2)). We then get (sqrt(7+4sqrt2)+sqrt(58+4sqrt2))/(sqrt17) as the side.

Through the cosine rule, cosAPB=-sqrt2/2, so APB=135.

Firstly, construct a triangle BCE outside the square such that CE = AP and BE = BP. Triangle CEB is congruent to triangle APB. Let PA = u. It suffices to find the measure of angle CEB. Join PE. Since angle ABP + angle PBC = 90 degrees, angle CBE + angle PBC = 90 degrees too. BPE will be an isosceles right angled triangle since BP = BE. Notice that PE = (square root of 8)*u and PC is 3u and CE is u. Triangle CEP is a right angle because of Pythagoras Theorem. Therefore 90 degree + 45 degree is 135 degree.

Let $P'$ be the point outside $ABCD$ such that $\triangle{APB}\cong\triangle{CP'B}$ . Let $AP=CP'=x$ , $PB=P'B=2x$ , and $CP=3x$ . Note that $\angle{P'BP}=\angle{P'BC}+\angle{CBP}=\angle{PBA}+\angle{CBP}=90$ , so $\triangle{BPP'}$ is a 45-45-90 triangle and $PP'=2x\sqrt{2}$ . But then $PP'^2+CP'^2=CP^2$ , so $\angle{CP'P}=90$ . As a result, $\angle{APB}=\angle{CP'B}=\angle{CP'P}+\angle{BP'P}=90+45=\boxed{135}$ .