Heavy Machinery, Proved

Study Sheet

Heavy Machinery, Proved

Stewart, Ptolemy, and roots of unity — with their proofs

Stewart's Theorem and the Bisector, Proved

Tip
Proof of Stewart's theorem
ABCD7935d
The example: AB=7AB = 7, AC=9AC = 9, BD=3BD = 3, DC=5DC = 5, and Stewart gives d2=46d^2 = 46.

Let cevian AD=dAD = d hit BCBC with BD=mBD = m, DC=nDC = n. Apply the law of cosines in △ABD\triangle ABD and △ACD\triangle ACD at the angles at DD — which are supplementary, so their cosines are NEGATIVES of each other. Multiply the first equation by nn, the second by mm, and ADD: the cosine terms cancel, leaving b2m+c2n=a(d2+mn)b^2 m + c^2 n = a(d^2 + mn). ■\blacksquare

In plain terms. Two law-of-cosines equations, weighted so the unknown angle disappears. That cancellation IS the theorem.

Example. AB=7AB = 7, AC=9AC = 9, BC=8BC = 8, BD=3BD = 3: 49⋅5+81⋅3−8d2=8⋅3⋅549\cdot5 + 81\cdot3 - 8d^2 = 8\cdot3\cdot5 gives d=46d = \sqrt{46}.

Reminder — Law of Cosines:  c2=a2+b2−2abcos⁡Cc^{2}=a^{2}+b^{2}-2ab\cos C
Tip
The angle bisector length, derived from Stewart

A bisector from AA splits BCBC in ratio c:bc : b (bisector ratio theorem), so m=acb+cm = \tfrac{ac}{b+c}, n=abb+cn = \tfrac{ab}{b+c}. Substituting into Stewart and simplifying collapses to t2=bc−mnt^2 = bc - mn — the bisector length needs only the two sides and the two pieces. ■\blacksquare

In plain terms. Stewart plus the ratio the bisector forces = a two-term formula.

Example. Sides 4,64, 6 around the bisected angle, pieces 2,32, 3: t2=24−6=18t^2 = 24 - 6 = 18, t=32t = 3\sqrt2.

Ptolemy's Theorem, Proved

Tip
Proof of Ptolemy's theorem
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The example: PP on arc BCBC of an equilateral triangle, and Ptolemy on ABPCABPC gives PA=PB+PCPA = PB + PC.

In cyclic ABCDABCD, mark EE on diagonal BDBD with ∠BCE=∠ACD\angle BCE = \angle ACD. Equal inscribed angles give △BCE∼△ACD\triangle BCE \sim \triangle ACD and △CDE∼△CAB\triangle CDE \sim \triangle CAB; the two similarity ratios produce BE⋅AC=AB⋅CDBE \cdot AC = AB \cdot CD and DE⋅AC=AD⋅BCDE \cdot AC = AD \cdot BC. Adding, and using BE+DE=BDBE + DE = BD: AC⋅BD=AB⋅CD+AD⋅BCAC \cdot BD = AB \cdot CD + AD \cdot BC. ■\blacksquare

In plain terms. One clever point splits a diagonal into two pieces, each measured by a similar triangle; adding the pieces is the theorem.

Example. Equilateral △ABC\triangle ABC with PP on arc BCBC: Ptolemy on ABPCABPC collapses to PA=PB+PCPA = PB + PC.

Concept
Reading equality: when Ptolemy becomes an inequality

For a NON-cyclic quadrilateral the same construction gives AC⋅BD<AB⋅CD+AD⋅BCAC \cdot BD < AB \cdot CD + AD \cdot BC — so equality is a TEST for being cyclic. Many AIME problems hide the cyclic condition exactly here.

In plain terms. If the products balance exactly, the four points share a circle.

Example. A quadrilateral with AC⋅BD=AB⋅CD+AD⋅BCAC \cdot BD = AB\cdot CD + AD\cdot BC must be cyclic — no angle chasing required.

Roots of Unity, Proved

Tip
Why xn−1x^n - 1 factors over the nnth roots of unity

By De Moivre (proved by induction on the angle-addition formula), the nn numbers ωk=cis2πkn\omega_k = \text{cis}\tfrac{2\pi k}{n} all satisfy xn=1x^n = 1; a degree-nn polynomial has at most nn roots, so these are ALL of them and xn−1=∏k(x−ωk)x^n - 1 = \prod_k (x - \omega_k). ■\blacksquare

In plain terms. Spinning by 1n\tfrac{1}{n} of a turn nn times lands you back at 11 — and a polynomial cannot have more roots than its degree.

Example. x4−1=(x−1)(x+1)(x−i)(x+i)x^4 - 1 = (x-1)(x+1)(x-i)(x+i).

Tip
Proof: the distances from one vertex multiply to nn
1
The four dashed distances from the vertex 1 of the regular pentagon multiply to exactly 5.

Divide xn−1x^n - 1 by x−1x - 1: ∏k=1n−1(x−ωk)=1+x+⋯+xn−1\prod_{k=1}^{n-1}(x - \omega_k) = 1 + x + \cdots + x^{n-1}. Evaluate at x=1x = 1: the right side is nn, and the left side is exactly the product of distances from vertex 11 to every other vertex of the regular nn-gon. ■\blacksquare

In plain terms. A polynomial identity, evaluated at one point, computes a geometric product no ruler could.

Example. A regular pentagon on the unit circle: the four distances from one vertex multiply to 55.

Two More Proofs Worth Carrying

Tip
Proof: the law of cosines

c2=a2+b2−2abcos⁡Cc^2 = a^2 + b^2 - 2ab\cos C.

Drop the altitude from AA to BCBC, meeting it at HH. Then CH=bcos⁡CCH = b\cos C and AH=bsin⁡CAH = b\sin C, so BH=a−bcos⁡CBH = a - b\cos C. Pythagoras in triangle ABHABH gives c2=(a−bcos⁡C)2+(bsin⁡C)2=a2−2abcos⁡C+b2(cos⁡2C+sin⁡2C)=a2+b2−2abcos⁡Cc^2 = (a - b\cos C)^2 + (b\sin C)^2 = a^2 - 2ab\cos C + b^2(\cos^2C + \sin^2C) = a^2 + b^2 - 2ab\cos C. ■\blacksquare The computation is unchanged when CC is obtuse because cos⁡C\cos C turns negative exactly when HH falls outside the segment — which is why the signed form covers every configuration at once.

Tip
Proof: the arithmetic mean–geometric mean inequality for two terms

a+b2≥ab\dfrac{a+b}{2} \ge \sqrt{ab} for a,b≥0a, b \ge 0, with equality iff a=ba = b.

Since (a−b)2≥0\left(\sqrt a - \sqrt b\right)^2 \ge 0, expanding gives a−2ab+b≥0a - 2\sqrt{ab} + b \ge 0, i.e. a+b≥2aba + b \ge 2\sqrt{ab}. Equality needs a=b\sqrt a = \sqrt b. ■\blacksquare Every use of AM–GM on a contest is this square in disguise, and the equality condition is what names the optimiser — which is usually what the problem actually asks for.