geometric-constructiongeometrynonlinear system
Is it possible to construct (and to calculate) a trapezium from it's two non-parallel sides and it's two diagonals, with other words $b,d,e,f$ are given:
I read out the equations system
$\begin{array}{|l l}
(1) & a =c+p+q \\
(2) & h^2 =e^2 -(a-q)^2 \\
(3) & h^2 =f^2 -(a-p)^2 \\
(4) & b^2 =h^2+q^2 \\
(5) & d^2 =h^2+p^2 \\
\end{array}$
How can I solve that for $a,c,h,p,q$ ?
You can parametrize the trapezium via its height $h$. Then the longer side is $$ 2 \sqrt{k^2-h^2} + k $$ and its area is $$ h (\sqrt{k^2-h^2} + k)\ .$$ The derivative for $h$ is $$k - \frac{h^2}{\sqrt{-h^2 + k^2}} + \sqrt{-h^2 + k^2}\ ,$$ which is zero for $h=0$ or $h=\pm \frac{\sqrt{3}}{2}k$. $h=\frac{\sqrt{3}}{2}k$ gives the proposed optimum, which you can be sure about after also checking the border case $h=k$ (which gives an area of $k^2$).
1) Let the line $MO$ intersects the sides of the trapezoid at the points $E$ and $F$. Then $$\triangle BME \sim \triangle AMF \Rightarrow\frac {BE}{AF}=\frac {ME}{MF}$$
2) $$\triangle EMC \sim \triangle FMD \Rightarrow\frac {ME}{MF}=\frac {EC}{FD}\Rightarrow\frac {BE}{AF}=\frac {EC}{FD}$$
3) $$\triangle AOF \sim \triangle COE \Rightarrow\frac {EC}{AF}=\frac {EO}{OF}$$
$$\triangle BEO \sim \triangle DFO \Rightarrow\frac {BE}{FD}=\frac {EO}{OF} \Rightarrow\frac {BE}{FD}=\frac {EC}{AF}$$ So $$\frac {BE}{AF}=\frac {EC}{FD}$$ and $$\frac{BE}{FD}=\frac {EC}{AF}$$
Then the four points $M,E,O,F$ lie on a straight line
Best Answer
Per the cosine rule $$a^2 = b^2+e^2 -2be \cos C= d^2 +f^2 -2df \cos D\tag1 $$ and note that the triangles ADB and ACB are equal in area
$$\frac12 be \sin C =\frac12df \sin D\implies b^2e^2 (1-\cos^2 C )= d^2f^2(1-\cos^2 D)\tag2 $$ Substitute (1) into (2) to eliminate $\cos C$ and $\cos D$, yielding $$a^2= \frac{(b^2-e^2)^2- (d^2-f^2)^2}{2(b^2+e^2 -d^2-f^2)} $$
Similarly
$$c^2= \frac{(b^2-f^2)^2- (d^2-e^2)^2}{2(b^2+f^2 -d^2-e^2)} $$
Depending on the given values of $b$, $d$, $e$ and $f$, the trapezium is either unique or impossible since the RHS’s of above expressions have to be positive. In the special case of $b=d$ and $e=f$, the trapezium is possible yet not unique.