-2
$\begingroup$

I came across this quiz question in a forum today. I would like to ask for your opinion of this notorious mathematics question, and also, to share.

PSLE 2005 Question

This question came out in Singapore's PSLE examination 2005. Basically, PSLE is one of the most important exams of one's life in Singapore. It literally determines your future, to a huge degree. How did such a question slip past the examiners, we would never know.

As for my real question, is that, is there a valid answer to this question ultimately? I mean, there can be more then one answer, but can it be argued in such a way, that one answer can ultimately be proven to be true?

  • 0
    is SXQ a straight line or does it just look very close to one?2012-11-21

2 Answers 2

3

Triangle $B=QVX$ and $XUQ$ are congruent, triangle $A=SWX$ and $XUQ$ are similar and by comparing their areas, the scaling factor is 2. Hence $XU$ is $\frac23$ of $SR$ and $RU$ is $\frac13$ of $RQ$, thus the the area of rectangle $C$ is $\frac29$ of the big rectangle, i.e. $20\text{ cm}^2$. By the same argument, rectangle $TXVP$ also has area $20\text{ cm}^2$. But then for the total area we obtain $80$ instead of $90\text{ cm}^2$, i.e the given information is inconsistant. That means that suitable reasoning might arrive at any other value for the area of $C$ (ex falsum quodlibet).

We conclude that the problem is ill-posed or we made an unwarranted assumtion. However, the assumptions were essentially that $SXQ$ and $WXV$ are straight lines. If we drop these, there are a lot of degrees of freedom, enough to allow a whole range of different areas to be obtainable for $C$.

  • 2
    It's "ex falso quodlibet", if you inflect "falsum" correctly... (The preposition "ex" calls for the ablative.)2012-11-21
0

Assume line segment $SXQ$ bisects the rectangle $PQRS$, and assume that all line segment intersections form right angles except those intersections involving segment $SXQ$. All areas are in square centimeters:

The total area is $90$. Let's label the rectangle $PVXT$ as having area $D$. The desired area $C$ is given by $C = 90 - 2A - 2B - D$. We know $A$ and $B$, so we need to find $D.$ Considering the triangle $PQS$, with area $90/2 = 45$, we can solve for $D$ with $D = 90/2 - A - B = 25$, thus $C = 25$.