Thursday, June 13, 2013

Leaving Certificate Mathematics 2013: Paper 2, Question 8 “Solution”

So, the State Examination Commission in Ireland made a giant cock-up. At issue is a particular mathematics question, so — as someone not entirely ignorant of basic mathematics — I thought I'd explain the problem.

Leaving Cert Mathematics 2013, Paper 2, Question 8

The problem is that a triangle is completely specified by three values — either angles or lengths — provided that at least one of them is a length: one length and two angles, two lengths and one angle, or three lengths. Given one of these, you can compute the missing values: two lengths and one angle, one length and two angles, or three angles, respectively.

The question text specifies two lengths: \(|HR|=80~\text{km}\) and \(|RP|=~110\text{km}\) and one angle: \(\angle{r}=124^\circ\) (or \(\angle{HRP}=124^\circ\) if you prefer the more verbose notation). The diagram shows one angle consistent with the text, \(r=124^\circ\), and — in the English version of the exam — another angle, \(h=36^\circ\), which is inconsistent with the values given in the text.

If we take the text as correct, and ignore the diagram, let's see what happens — the problem now is: find \(\angle h\) given (dropping the units and angle symbols for convenience): \[|HR|=80 \\
|RP|=110 \\
We know that the sum of the internal angles of any triangle is \(180^\circ\), so: \[r+h+p=180 \Rightarrow h+p = 180-r = 56\]
If we drop a line vertically from the apex, \(R\), to a point, \(X\), on the base, we then have two right-angle triangles “back-to-back”, i.e. sharing the line-segment \(RX\).

From the right-angle triangle, \(HRX\), on the left, we have: \[|RX|=|HR|\sin(h)\] and from the right-angle triangle on the right, \(PRX\), we have: \[|RX|=|RP|\sin(p)\] Combining these two: \[|HR|\sin(h)=|RP|\sin(p)\] But we know that \(h+p=56\) or \(p=56-h\), so \[|HR|\sin(h) = |RP|\sin(56-h)\qquad\qquad(1)\] Now, a basic trigonometric relation (listed in the “log tables” that every candidate gets in the exam) is \[\sin(A-B)=\sin(A)\cos(B)-\cos(A)\sin(B)\] Applying this to the RHS of (1), we get \[|HR|\sin(h)=|RP|(\sin(56)\cos(h)-\cos(56)\sin(h))\] Rearranging: \[(|HR|+|RP|\cos(56))\sin(h) = |RP|\sin(56)\cos(h)\] or \[\tan(h) = \frac{\sin(h)}{\cos(h)} = \frac{|RP|\sin(56)}{|HR|+|RP|\cos(56)}\] or \[h = \tan^{-1}\left( \frac{|RP|\sin(56)}{|HR|+|RP|\cos(56)}\right)\] Substituting in the values we know:  \[h = \tan^{-1}\left( \frac{110\times 0.8290}{80+110\times0.5592}\right) = \tan^{-1}(0.6444) = 32.80^\circ\] Therefore \[32.80\neq 36 \Rightarrow \text{The SEC are morons}\]

Now, the question that is actually asked is “Find the distance from R to HP”. There are at least two further problems:

  • the distance from \(R\) to the line-segment \(HP\) is not uniquely defined (we must assume that the perpendicular distance, i.e. \(|RX|\) is intended); and
  • even under the simplifying assumption that the Earth is a sphere, latitude makes a significant difference to the answer (consider point \(R\) being at the North pole vs. line-segment \(HP\) lying on the equator), and non-Euclidean geometry is not on the syllabus.
If one uses the given value for \(h\), the problem becomes utterly trivial: \[|RX|=|HR|\sin(36^\circ)\] If we don't ignore the diagram, we have two angles and two lengths and must decide to discard either one of the angles or one of the lengths (in the foregoing, we discarded \(h\) from the diagram and kept the two lengths and angle \(r\) from the text), so there are actually four choices for how to proceed.

All in all, a gargantuan cock-up and staggering incompetence from the SEC. Imagine that nobody there spotted any of this!

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