An inconstructible quadrilateral

Question

A student tried to draw a quadrilateral $STOP$ with $ST=5cm$, $TO=4cm$, $\angle S = 20^{\circ}, \angle T = 30^{\circ}, \angle O = 40^{\circ}$. But he found out that it was impossible to construct one.We have to show why he failed without using trigonometry.

At first,we calculate the remaining angle P, $\angle P = 270^{\circ}$.But after that,I am stuck.Obviously either of the side lengths is messing things up.But I cannot seem to find any way to show that.A small hint at this point will be appreciated.

Answer 1

The non-convex quadrilateral $STOP$ below satisfies the hypotheses of the question.

The question is, can we construct it, using only the Euclidean straightedge and collapsible compass tools? I had to resort to using a protractor to measure out the $20^{\circ}$ and $40^{\circ}$ angles in order to draw the diagram above, which doesn't count.

So it all boils down to whether or not $20^{\circ}$ angles are constructable, or rather, whether or not there is a general way to trisect an angle. It can't be done. The student referred to in your question was naively attempting to accomplish exactly that, and so doomed.

Answer 2

It appears to me that it is possible to construct such a quadrilateral. Perhaps some problem has crept into giving the data?

Given that the angle at $T$ is supposed to be 30 degrees, we can reconstruct triangle $STO$ using only the side lengths $ST$ and $TO$ according to the familiar Pythagorean relationship and facts about the "special triangle" having 30-60-90 degree angles. [If those facts are not allowed as "using trigonometry", it seems doubtful that any approach using the available angle information would be allowed.]

That is, if we drop a perpendicular from shorter side's vertex $O$ onto $ST$ (the longer side) at point $Q$, then $TQO$ will be a right triangle with hypotenuse $TO$ of length 4cm. It follows from facts about this "special triangle" with angle at $T$ being 30 degrees that $QO$ has half the length of the hypotenuse (2cm) and that $TQ$ has length $2\sqrt{3}$ cm.

Accordingly $SQO$ is also a right triangle, sharing the side $QO$ and having side $SQ$ of length $5 - 2\sqrt{3}$ cm. The hypotenuse $SO$ of this second right triangle is therefore of length $\sqrt{4 + (5-2\sqrt{3})}$ cm.

As pointed out in the Question, the internal angle at $P$ must be 270 degrees, so the external angle at $P$ is 90 degrees. Therefore $P$ lies on the semi-circle with diameter $SO$. Drawing the sides $OP$ and $SP$ at angles 40 and 20 degrees respectively to sides $TO$ and $ST$ will therefore produce a quadrilateral with all the required data.