A box in the shape of a triangle with side lengths \(13\), \(14\), and \(15\) is being used to pack a pizza pi. The pizza is in the shape of a circle and just touches each of the three sides of the triangular box. What is the area of the circle?
Note: For this problem, the following known results about circles may be useful:
If a line is tangent to a circle, then the line is perpendicular to the radius drawn to the point of tangency.
If two tangent segments are drawn to a circle from a point outside of the circle, then the lengths of those segments are equal.
We label the triangle \(ABC\), with \(AB = 13\), \(BC = 14\), and \(AC = 15\).
Let \(P\) be the point where the circle touches the triangle on \(AB\), \(Q\) be the point where the circle touches the triangle on \(BC\), and \(R\) be the point where the circle touches the triangle on \(AC\).
Let \(\angle BAC = \theta\), and let the circle be centred at point \(O\) with radius \(r\).
Using the second given result about circles, we know that \(AP=AR\), \(CR = CQ\), and \(BP = BQ\). Let \(AP = AR = x\). Then \(BP = AB - AP = 13-x\). Since \(BP=BQ\), we have \(BQ = BP = 13-x\). Also, \(CR = AC - AR = 15-x\). Since \(CR=CQ\), we have \(CQ = 15-x\).
Since \(BC=14\) and \(BC = BQ + CQ\), we have \(13-x + 15-x = 14\), thus \(28-2x = 14\), and so \(x=7\).
Using the cosine law in \(\triangle ABC\), \[\begin{aligned} BC^2& = AB^2 +AC^2 - 2(AB)(AC)\cos(\angle BAC)\\ 14^2 &= 13^2 + 15^2 - 2(13)(15)\cos(\theta)\\ \cos\theta &= \frac{14^2-13^2-15^2}{-2(13)(15)}\\ &=\frac{-198}{-390}\\ &=\frac{33}{65} \end{aligned}\]
Consider \(\triangle AOR\) and \(\triangle AOP\). Since \(AR=AP\), \(OR = OP = r\), and \(AO=AO\) (same length), \(\triangle AOR\) is congruent to \(\triangle AOP\). Thus, \(\angle OAP = \angle OAR = \dfrac{\theta}{2}\). Further, using the first given result about circles, we know that \(\triangle AOR\) and \(\triangle AOP\) are right-angled.
In \(\triangle AOR\), \(\tan \left(\dfrac{\theta}{2}\right)= \dfrac{r}{x} = \dfrac{r}{7}\). Thus, \(r = 7\tan\left(\dfrac{\theta}{2}\right)\).
Using the identities \(\cos^2\left(\dfrac{\theta}{2}\right) = \dfrac{1+\cos{\theta}}{2}\) and \(\sin^2\left(\dfrac{\theta}{2}\right) = \dfrac{1-\cos{\theta}}{2}\), we obtain \(\cos^2\left(\dfrac{\theta}{2}\right) = \dfrac{1+\cos{\theta}}{2} = \dfrac{49}{65}\), and \(\sin^2\left(\dfrac{\theta}{2}\right) = \dfrac{1-\cos{\theta}}{2} = \dfrac{16}{65}\).
Since \(0<\dfrac{\theta}{2}<90\degree\), \(\cos\left(\dfrac{\theta}{2}\right)>0\) and \(\sin\left(\dfrac{\theta}{2}\right) >0\). Thus, \(\cos\left(\dfrac{\theta}{2}\right) = \dfrac{7}{\sqrt{65}}\) and \(\sin\left(\dfrac{\theta}{2}\right) = \dfrac{4}{\sqrt{65}}\).
Therefore, we have \[\begin{aligned} r &= 7\tan\left(\frac{\theta}{2}\right) \\ &=\frac{7\sin\left(\frac{\theta}{2}\right)}{\cos\left(\frac{\theta}{2}\right)}\\ &= \frac{7\left(\frac{4}{\sqrt{65}}\right)}{\frac{7}{\sqrt{65}}}\\ &=4 \end{aligned}\]
Therefore, the area of the circle is \(\pi (4)^2 = 16\pi\).