In this entry, I’m going to start with a concrete problem and develop an abstract generalization.
The starting problem: Given isosceles trapezoid \(ABCD\) with an altitude of 6. Point \(E\) is on \(\overline{DC}\) such that \(DE = 3\), \(EC = 8\), and \(\angle AEB\) is right. Determine \(AB\).
We can solve this by placing points \(G\) and \(H\) on \(\overline{AB}\) such that \(\overline{EG}\perp\overline{AB}\) and \(\overline{CH}\perp\overline{AB}\).
Since \(ABCD\) is isosceles, \(AF = BH = x\). We can identify three right triangles:
- \(\Delta AGE\) has sides 6, \(x + 3\), and \(AE\)
- \(\Delta BGE\) has sides 6, \(x + 8\), and \(BE\)
- \(\Delta AEB\) has sides \(AE\), \(BE\), and \(2x + 11\)
Since we have three unknowns (\(x, AE, BE\)), it is possible for us to solve a system of three equations. These are the sides of right triangles, so we’ll use the Pythagorean Theorem:
- \(AE^2 = (x + 3)^2 + 6^2 = x^2 + 6x + 9 + 36 = x^2 + 6x + 45\)
- \(BE^2 = (x + 8)^2 + 6^2 = x^2 + 16x + 64 + 36 = x^2 + 16x + 100\)
- \(AE^2 + BE^2 = (2x + 11)^2 = 4x^2 + 44x + 121\)
Adding the two equations and using transitivity yields a single equation with a single unknown:
\[x^2 + 6x + 45 + x^2 + 16x + 100 = 4x^2 + 44x + 121 \Rightarrow \\ 2x^2 + 22x + 145 = 4x^2 + 44x + 121 \Rightarrow \\ 2x^2 + 22x – 24 = 0 \Rightarrow \\ x^2 + 11x – 12 = 0\]
This has the roots \(x \in \{1, -12\}\). This yields \(AB \in \{13, -13\}\). \(AB\) has to be positive, so \(AB = 13\).
Generalizing the Solution
The next level of abstraction is to determine the value of \(x\) given some altitude \(c\) and \(E\) placed to create lengths of \(a\) and \(b\). Returning to our three triangles:
- \(\Delta AGE\) has sides \(c\), \(x + a\), and \(AE\)
- \(\Delta BGE\) has sides \(c\), \(x + b\), and \(BE\)
- \(\Delta AEB\) has sides \(AE\), \(BE\), and \(2x + a + b\)
which gives us:
- \(AE^2 = (x + a)^2 + c^2 = x^2 + 2ax + a^2 + c^2\)
- \(BE^2 = (x + b)^2 + c^2 = x^2 + 2bx + b^2 + c^2\)
- \(AE^2 + BE^2 = (2x + a + b)^2 = 4x^2 + 2ax + 2bx + 2ax + a^2 + ab + 2bx + ab + bx^2 \\ = 4x^2 + 4ax + 4bx + a^2 + 2ab + b^2\)
So that: \[x^2 + 2ax + a^2 + c^2 + x^2 + 2bx + b^2 + c^2 = 4x^2 + 4ax + 4bx + a^2 + 2ab + b^2 \Rightarrow \\ 2x^2 + 2ax + 2bx + 2c^2 + a^2 + b^2 = 4x^2 + 4ax + 4bx + a^2 + 2ab + b^2 \Rightarrow \\ 2x^2 + 2ax + 2bx + 2ab – 2c^2 = 0 \Rightarrow \\ x^2 + (a + b)x + ab – c^2 = 0\]
The Quadratic Formula allows us to solve this for \(x\): \[x = \frac{-(a + b) \pm \sqrt{(a + b)^2 – 4(ab – c^2)}}{2}\]
Simplifying the radicand yields: \[x = \frac{-(a + b) \pm \sqrt{(a – b)^2 + 4c^2}}{2}\]
Note that \(x\) can be negative; that would mean that \(AB < DC\), which is acceptable. However, there is the constraint that \(2x > -(a + b)\) (since \(AB\) has to be positive), so we can only take the larger root: \[x = \frac{-(a + b) + \sqrt{(a – b)^2 + 4c^2}}{2}\]
This will have integer solutions when the radicand (\((a – b)^2 + 4c^2)\)) is a perfect square, such as all trapezoids of altitude \(6\) where \(|a – b| = 5\).
Valid Trapezoids
This raises the question: What are the restrictions on an isosceles trapezoid’s base and altitude that allows for a right angle to be drawn on the base as in the original image?
For trapezoid \(ABCD\), drawing a diagonal \(\overline{BD}\) creates angles \(\angle BDA\) and \(\angle BDC\).
For trapezoid \(ABCD\), placing midpoint \(M\) on \(DC\) and drawing \(\overline{AM}\) and \(\overline{BM}\) creates angle \(\angle BMA\).
It is always the case that \(m\angle{BDA} < m\angle{BMA}\). There will be thus some point \(E\) on \(DC\) such that \(m\angle BEA = 90^{\circ}\) if and only if \(m\angle{BDA} \le 90^{\circ} \le m\angle{BMA}\).
Let’s return to an earlier equation. Rather than solving it for \(x\), we’ll solve it for \(a\): \[x^2 + ax + bx + ab – c^2 = 0\]
\(a\) represents the distance between the trapezoid’s vertex \(D\) and the collinear point \(E\). For a given isosceles trapezoid, we have three basic values: the lengths of the two bases, \(b_1\) and \(b_2\), and the altitude \(h\). \(x = \frac{\Delta b}{2}\), while \(b = b_1 – a\). Since \(c = h\), we can rewrite the equation: \[\frac{(\Delta b)^2}{4} + \frac{a\Delta b}{2} + \frac{(b_1 – a)\Delta b}{2} + a(b_1 – a) – h^2 = 0 \Rightarrow \\ a^2 – ab_1 – \frac{b_1\Delta b}{2} – \frac{(\Delta b)^2}{4} + h^2 = 0 \Rightarrow a = \frac{b_1 \pm \sqrt{b_1^2 – 4\left(h^2 – \frac{b_1\Delta b}{2} – \frac{(\Delta b)^2}{4}\right)}}{2} \]
Let’s simplify the radicand. Since \(\Delta b = b_2 – b_1\), \[b_1^2 – 4\left(h^2 – \frac{b_1\Delta b}{2} – \frac{(\Delta b)^2}{4}\right) = b_1^2 – 4\left(h^2 – \frac{b_1b_2 – b_1^2}{2} – \frac{b_2^2 – 2b_1b_2 + b_1^2}{4}\right) \\ = b_1^2 – 4h^2 + 2b_1b_2 – 2b_1^2 +b_2^2 – 2b_1b_2 + b_1^2 = b_2^2 – 4h^2 \]
Hence, \[a = \frac{b_1 \pm \sqrt{b_2^2 – 4h^2}}{2}\]
This will be real when \(b_2 \ge 2h\). Thus, one constraint on our trapezoid is that the length of its bottom base must be at least twice its altitude. If \(b_2 = 2h\), then the midpoint of \(b_1\) will be the vertex of a right angle with sides that contain \(A\) and \(B\).
The other constraint is that the lower root of \(a\) must be non-negative. If \(a\) is negative, then it is to the left of \(D\) and hence exterior to the trapezoid. The lower root is non-negative when \(b_1 \ge \sqrt{b_2^2 – 4h^2} \Rightarrow b_1^2 \ge b_2^2 – 4h^2\).
Conclusion
Putting these together, it is possible to create a right angle with vertex \(E\) on base \(CD\) and sides containing \(A\) and \(B\) for an isosceles trapezoid ABCD with these constraints: \[b_1^2 \ge b_2^2 – 4h^2 \\ b_2 \ge 2h\]
The distance of this vertex from \(C\) or \(D\) will be \[a = \frac{b_1 – \sqrt{b_2^2 – 4h^2}}{2}\]
