A Triangle With a 45 Degrees Angle in Square

Thinking in terms of paper folding, fold $\Delta ABM\,$ over $AM\,$ and $\Delta ADN\,$ over $AN.\,$ Note that $\angle BAM+\angle DAN=45^{\circ}=\angle MAN.\,$ It follows that the edges $AB\,$ and $AD\,$ of the folded triangles will coincide. If $L\,$ is their common endpoint then $\angle ALM=\angle ALN=90^{\circ}\,$ so that $L\,$ lies on $MN\,$ and $AL\perp MN.\,$ $AL\,$ is an altitude in $\Delta MAN.$

The folding argument shows that $\angle AND=\angle ANL\,$ (and, in passing, $\angle AMB=\angle AML.)\,$ Since $\angle DQN=\angle AQP\,$ and $\angle MAN=45^{\circ}=\angle NDQ,\,$ triangles $DNQ\,$ and $APQ\,$ are similar, implying $\angle APQ=\angle AND=\angle ANM.\,$ Thus triangles $AQP\,$ and $AMN\,$ are similar. Their areas relate as the square of any pair of their corresponding linear elements. Now, assuming $AL=AB=1,\,$ $AH=\displaystyle\frac{\sqrt{2}}{2},\,$ making $[\Delta AMN]=2[\Delta AQP]\,$ and, consequently, $[\Delta AQP]=[MNQP],\,$ as required.

This problem is a converse of the one discussed earlier. The configuration is rich with various properties; some of these could be inferred from the following diagram:

Below, the areas of the blue and red regions are equal.

This is a direct application of Carpets Theorem. It can be easily verified that $\angle PCQ=45^{\circ},\,$ which shows that $[MNQP]=[\Delta AQP]=[\Delta CPQ].\,$ If so, by removing the common part - $IJQP\,$ - from $\Delta CPQ\,$ and the quadrilateral $MNQP\,$ gives the required result.

The problem, due to V. Proizvolov, appeared in Kvant - a popular Russian magazine - (#1, 2004, M1895), with a solution in a later issue (#4, 2004).

The extra properties and the extra problem have been suggested by Bariş Altay (Turkey).