XXXVI OM - I - Problem 6

Calculate the lengths of the parallel sides of a trapezoid given the lengths of the other sides and the diagonals.
Let $ABCD$ be the considered trapezoid, $AB\parallel CD$, $|AB| = a$, $|BC| = b$, $|CD| = c$, $|DA| = d$, $|AC| = p$, $|BD| = q$ (Figure 2). Given $b$, $d$, $p$, $q$, we need to find $a$ and $c$.
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First, note that if $p = q$, then it must be $b = d < p$ and in this case, the problem does not have a unique solution: there are infinitely many isosceles trapezoids with side length $b$ and diagonal length $p$. More precisely, if $a$ is any number satisfying the inequality $p - b < a < p + b$, then there exists a triangle $ABC$ where $|AB| = a$, $|BC| = b$, $|CA| = p$, and the desired trapezoid $ABCD$ is obtained by taking as vertex $D$ the image of point $C$ under the axial symmetry with respect to the perpendicular bisector of segment $AB$. Denoting by $C'$ the projection of point $C$ onto line $AB$ and setting $x = |AC'|$, $y = |BC'|$, $h = |CC'|$, we see (Figure 3) that if $C' < A$, then $a = x - y$, $c = x + y$, and if $C' > A$, then $a = x - y$, $c = x + y$. In both cases, $ac = x^2 - y^2 = (p^2 - h^2) - (b^2 - h^2) = p^2 - b^2$. Thus, when $b = d < p = q$, the solution to the problem is any pair of numbers $a, c$ such that $p - b < a < p + b$, $c = (p^2 - b^2) / a$.
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Let us assume from now on that $p \neq q$ (the trapezoid is not isosceles). Triangles $ABC$ and $ABD$ have a common base and equal heights, and thus equal areas. We express these areas in terms of the side lengths, using Heron's formula:

\[
\text{Area of } \triangle ABC = \sqrt{s(s-a)(s-b)(s-p)}
\]

where $s = \frac{a + b + p}{2}$, and similarly,

\[
\text{Area of } \triangle ABD = \sqrt{t(t-a)(t-d)(t-q)}
\]

where $t = \frac{a + d + q}{2}$.

Equating the right-hand sides of the obtained equalities, we see that

\[
\sqrt{s(s-a)(s-b)(s-p)} = \sqrt{t(t-a)(t-d)(t-q)}
\]

If the expression in the parentheses on the left side were zero, then one of the factors on the right side would have to be zero; we would then have the alternative system of equations:

\[
\begin{cases}
s - a = 0 \\
s - b = 0 \\
s - p = 0
\end{cases}
\quad \text{or} \quad
\begin{cases}
t - a = 0 \\
t - d = 0 \\
t - q = 0
\end{cases}
\]

In the first case, adding and subtracting the equations of the system, we get $b = d$ and $p = q$, contrary to the previous assumption. In the second case, we get $b = q$ and $d = p$, which is not possible (points $C$ and $D$ would have to coincide). Therefore, $b^2 - d^2 + p^2 - q^2 \neq 0$, so

\[
\frac{(s-a)(s-b)(s-p)}{s} = \frac{(t-a)(t-d)(t-q)}{t}
\]

The left side is a positive number, so the right side is also positive, and thus ultimately

\[
a = \frac{2(p^2 - q^2)}{b^2 - d^2 + p^2 - q^2}
\]

We find the value of $c$ by swapping $p$ and $q$ in the obtained formula for $a$ (this corresponds to repeating the conducted reasoning, starting from the equality of the areas of triangles $CDA$ and $CDB$, with sides $c$, $d$, $p$ and $c$, $b$, $q$, respectively). Thus,

\[
c = \frac{2(q^2 - p^2)}{b^2 - d^2 + q^2 - p^2}
\]

Note. From the formulation of the problem, it follows implicitly that the considered trapezoid actually exists. Therefore, there is no need to seek conditions that the given quantities $b$, $d$, $p$, $q$ should satisfy for the found solution to make geometric sense.