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:
x2+6x+45+x2+16x+100=4x2+44x+121⇒2x2+22x+145=4x2+44x+121⇒2x2+22x–24=0⇒x2+11x–12=0
This has the roots x∈{1,−12}. This yields AB∈{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:
- ΔAGE has sides c, x+a, and AE
- ΔBGE has sides c, x+b, and BE
- ΔAEB has sides AE, BE, and 2x+a+b
which gives us:
- AE2=(x+a)2+c2=x2+2ax+a2+c2
- BE2=(x+b)2+c2=x2+2bx+b2+c2
- AE2+BE2=(2x+a+b)2=4x2+2ax+2bx+2ax+a2+ab+2bx+ab+bx2=4x2+4ax+4bx+a2+2ab+b2
So that: x2+2ax+a2+c2+x2+2bx+b2+c2=4x2+4ax+4bx+a2+2ab+b2⇒2x2+2ax+2bx+2c2+a2+b2=4x2+4ax+4bx+a2+2ab+b2⇒2x2+2ax+2bx+2ab–2c2=0⇒x2+(a+b)x+ab–c2=0
The Quadratic Formula allows us to solve this for x: x=−(a+b)±√(a+b)2–4(ab–c2)2
Simplifying the radicand yields: x=−(a+b)±√(a–b)2+4c22
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=−(a+b)+√(a–b)2+4c22
This will have integer solutions when the radicand ((a–b)2+4c2)) 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 ¯BD creates angles ∠BDA and ∠BDC.
For trapezoid ABCD, placing midpoint M on DC and drawing ¯AM and ¯BM creates angle ∠BMA.
It is always the case that m∠BDA<m∠BMA. There will be thus some point E on DC such that m∠BEA=90∘ if and only if m∠BDA≤90∘≤m∠BMA.
Let’s return to an earlier equation. Rather than solving it for x, we’ll solve it for a: x2+ax+bx+ab–c2=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, b1 and b2, and the altitude h. x=Δb2, while b=b1–a. Since c=h, we can rewrite the equation: (Δb)24+aΔb2+(b1–a)Δb2+a(b1–a)–h2=0⇒a2–ab1–b1Δb2–(Δb)24+h2=0⇒a=b1±√b21–4(h2–b1Δb2–(Δb)24)2
Let’s simplify the radicand. Since Δb=b2–b1, b21–4(h2–b1Δb2–(Δb)24)=b21–4(h2–b1b2–b212–b22–2b1b2+b214)=b21–4h2+2b1b2–2b21+b22–2b1b2+b21=b22–4h2
Hence, a=b1±√b22–4h22
This will be real when b2≥2h. Thus, one constraint on our trapezoid is that the length of its bottom base must be at least twice its altitude. If b2=2h, then the midpoint of b1 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 b1≥√b22–4h2⇒b21≥b22–4h2.
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: b21≥b22–4h2b2≥2h
The distance of this vertex from C or D will be a=b1–√b22–4h22