Triangle $ABC$ is inscribed in equilateral triangle $PQR$, as shown.
If $PC = 3$, $BP = CQ = 2$, and $\angle ACB = 90^\circ$, then compute $AQ$.
Let u = BC
Let v = CA
Let w = AB
Let x = AQ
Let PQ = PR = RQ = 5
Let BR = PR - BP \(= 5-2=3\)
Let AR = RQ - AQ \(= 5 - x\)
Using the Law of Cosines:
\(\begin{array}{|rcll|} \hline u^2 &=& BP^2 + PC^2 - 2\cdot BP \cdot PC\cdot \cos(60^{\circ}) \quad & | \quad 2\cdot \cos(60^{\circ}) = 2\cdot \frac12 = 1 \\ u^2 &=& BP^2 + PC^2 - BP \cdot PC \quad & | \quad BP =2 \quad PC = 3 \\ u^2 &=& 2^2 + 3^2 - 2 \cdot 3 \\ \mathbf{u^2} & \mathbf{=} & \mathbf{2^2 + 3^2 - 6} \\ \hline \end{array}\)
Using the Law of Cosines again:
\(\begin{array}{|rcll|} \hline v^2 &=& CQ^2 + AQ^2 - 2\cdot CQ \cdot AQ\cdot \cos(60^{\circ}) \quad & | \quad 2\cdot \cos(60^{\circ}) = 2\cdot \frac12 = 1 \\ v^2 &=& CQ^2 + AQ^2 - CQ \cdot AQ \quad & | \quad CQ =2 \quad AQ = x \\ \mathbf{v^2} & \mathbf{=} & \mathbf{2^2 + x^2 - 2x} \\ \hline \end{array}\)
Using the Law of Cosines again:
\(\begin{array}{|rcll|} \hline w^2 &=& BR^2 + AR^2 - 2\cdot BR \cdot AR\cdot \cos(60^{\circ}) \quad & | \quad 2\cdot \cos(60^{\circ}) = 2\cdot \frac12 = 1 \\ w^2 &=& BR^2 + AR^2 - BR \cdot AR \quad & | \quad BR =3 \quad AR = 5-x \\ \mathbf{w^2} & \mathbf{=} & \mathbf{3^2 + (5-x)^2 - 3 \cdot (5-x)} \\ \hline \end{array}\)
Using Pythagoras' theorem:
\(\begin{array}{|rcll|} \hline u^2+v^2&=& w^2 \\ (2^2 + 3^2 - 6) + (2^2 + x^2 - 2x) &=& 3^2 + (5-x)^2 - 3 \cdot (5-x) \\ 2^2 + \not{3^2} - 6 + 2^2 + x^2 - 2x &=& \not{3^2} + (5-x)^2 - 3 \cdot (5-x) \\ 2^2 - 6 + 2^2 + x^2 - 2x &=& (5-x)^2 - 3 \cdot (5-x) \\ 2 + x^2 - 2x &=& (5-x)^2 - 3 \cdot (5-x) \\ 2 + \not{x^2} - 2x &=& 25 - 10x + \not{x^2} -15+3x \\ 2 - 2x &=& 25 - 10x -15+3x \\ 2 - 2x &=& 10 - 7x \\ 5x &=& 8 \\ x &=& \frac85 \\ \mathbf{x} & \mathbf{=} & \mathbf{1.6} \\ \hline \end{array}\)
AQ is 1.6