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 ¯DC such that DE=3, EC=8, and ∠AEB is right. Determine AB.
We can solve this by placing points G and H on ¯AB such that ¯EG⊥¯AB and ¯CH⊥¯AB.
Since ABCD is isosceles, AF=BH=x. We can identify three right triangles:
- ΔAGE has sides 6, x+3, and AE
- ΔBGE has sides 6, x+8, and BE
- ΔAEB 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:
- AE2=(x+3)2+62=x2+6x+9+36=x2+6x+45
- BE2=(x+8)2+62=x2+16x+64+36=x2+16x+100
- AE2+BE2=(2x+11)2=4x2+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}