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Problem of the Week
Problem E and Solution
Now I Know My ABCs


In triangle \(ABC\), point \(P\) lies on \(AB\), point \(Q\) lies on \(BC\), and point \(R\) lies on \(AC\) such that \(AQ\), \(BR\), and \(CP\) are altitudes with lengths \(21\) cm, \(24\) cm, and \(56\) cm, respectively.

Determine the measure, in degrees, of \(\angle ABC\), and the lengths, in centimetres, of \(AB\), \(BC\), and \(CA\).

Note the diagram is not drawn to scale.


Let \(BC=a\), \(AC=b\), and \(AB=c\). We will present two methods for determining that \(21a=24b=56c\), and then continue on with the rest of the solution.

We now will continue on with the rest of the solution.

From \(21a=24b\) we obtain \(b=\frac{21}{24}a=\frac{7}{8}a\), and from \(21a=56c\) we obtain \(c=\frac{21}{56}a=\frac{3}{8}a\).
The ratio of the sides in \(\triangle ABC\) is therefore \(a:b:c=a:\frac{7}{8}a:\frac{3}{8}a=8:7:3\). Let \(BC=8x\), \(AC=7x\), and \(AB=3x\), where \(x>0\).

Using the cosine law, \[\begin{aligned} AC^2&=AB^2+BC^2-2(AB)(BC)\cos(\angle ABC)\\ (7x)^2&=(3x)^2+(8x)^2-2(3x)(8x)\cos(\angle ABC)\\ 49x^2&=9x^2+64x^2-48x^2\cos(\angle ABC)\end{aligned}\] Since \(x>0\), we know \(x^2\neq 0\). So dividing by \(x^2\), \[49=73-48\cos(\angle ABC)\] Rearranging, \[\begin{aligned} 48\cos(\angle ABC)&=24\\ \cos(\angle ABC)&=\frac{1}{2}\end{aligned}\] Therefore, \(\angle ABC=60^\circ\).

In right \(\triangle BPC\), \[\begin{aligned} \frac{CP}{BC}&=\sin 60^\circ\\ BC&=\frac{CP}{\sin 60^\circ}\\ BC&=\frac{56}{\frac{\sqrt{3}}{2}}\\ BC&=\frac{112}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}\\ BC&=\frac{112\sqrt{3}}{3}\end{aligned}\] However, \(BC=8x\). Therefore, \[\begin{aligned} 8x&=\frac{112\sqrt{3}}{3}\\ x&=\frac{14\sqrt{3}}{3}\\ 3x&=14\sqrt{3}\\ 7x&=\frac{98\sqrt{3}}{3}\end{aligned}\]

Therefore, \(\angle ABC=60^\circ\), and the side lengths of \(\triangle ABC\) are \(AB=3x=14\sqrt{3}\) cm, \(AC=7x=\dfrac{98\sqrt{3}}{3}\) cm, and \(BC=\dfrac{112\sqrt{3}}{3}\) cm.