Problem:
A convex polygon P in the plane is dissected into smaller convex polygons by drawing all of its diagonals. The lengths of all sides and all diagonals of the polygon P are rational numbers. Prove that the lengths of all sides of all polygons in the dissection are also rational numbers.
Solution:
Let P=A1A2…An, where n is an integer with n≥3. The problem is trivial for n=3 because there are no diagonals and thus no dissections. We assume that n≥4. Our proof is based on the following Lemma.
Lemma 1. Let ABCD be a convex quadrilateral such that all its sides and diagonals have rational length. If segments AC and BD meet at P, then segments AP,BP,CP,DP all have rational lengths.
It is clear by Lemma 1 that the desired result holds when P is a convex quadrilateral. Let AiAj(1≤i<j≤n) be a diagonal of P. Assume that C1,C2,…,Cm are the consecutive division points on diagonal AiAj (where point C1 is the closest to vertex Ai and Cm is the closest to Aj ). Then the segments CℓCℓ+1,1≤ℓ≤m−1, are the sides of all polygons in the dissection. Let Cℓ be the point where diagonal AiAj meets diagonal AsAt. Then quadrilateral AiAsAjAt satisfies the conditions of Lemma 1. Consequently, segments AiCℓ and CℓAj have rational lengths. Therefore, segments AiC1,AiC2,…,AjCm all have rational lengths. Thus, CℓCℓ+1=ACℓ+1−ACℓ is rational. Because i,j,ℓ are arbitrarily chosen, we proved that all sides of all polygons in the dissection are also rational numbers.
Now we present four proofs of Lemma 1 to finish our proof.
First approach: We show only that segment AP is rational, the others being similar. Introduce Cartesian coordinates with A=(0,0) and C=(c,0). Put B=(a,b) and D=(d,e). Then by hypothesis, the numbers
is rational. Because c=0,a is rational. Likewise d is rational.
Now we have that b2=AB2−a2,e2=AD2−d2,(b−e)2=BD2−(a−d)2 are rational, and so that 2be=b2+e2−(b−e)2 is rational. Because quadrilateral ABCD is convex, b and e are nonzero and have opposite sign. Hence eb=2b22be is rational.
We now calculate
P=(b−ebd−ae,0)
so
AP=eb−1eb⋅d−a
is rational.
Second approach
Note that, for an angle α, if cosα is rational, then sinα=rαmα for some rational r and square-free positive integer m (and this expression is unique when r is written in the lowest term). We say two angles α and β with rational cosine are equivalent if mα=mβ, that is, if sinα/sinβ is rational. We establish the following lemma.
Lemma 2. Let α and β be two angles.
(a) If α, β and α+β all have rational cosines, then all three are equivalent.
(b) If α and β have rational cosine values and are equivalent, then α+β has rational cosine value (and is equivalent to the other two).
(c) If α,β and γ are the angles of a triangle with rational sides, then all three have rational cosine values and are equivalent.
Proof: Assume that cosα=s and cosβ=t.
(a) Assume that s and t are rational. By the Addition formula, we have
cos(α+β)=cosαcosβ−sinαsinβ(*)
or, sinαsinβ=st−cos(α+β), which is rational by the given conditions. Hence α and β are equivalent. Thus sinα=ram and sinβ=rbm for some rational numbers ra and rb and some positive square free integer m. By the Addition formula, we have
sin(α+β)=sinαcosβ+cosαsinβ=(tra+srb)m
implying that α+β is equivalent to both α and β.
(b) By (∗),cos(α+β) is rational if s,t are rational and α and β are equivalent. Then by (a), α,β,α+β are equivalent.
(c) Applying the Law of Cosine to triangle ABC shows that cosα,cosβ and cosγ are all rational. Note that cosγ=cos(180∘−α−β)=−cos(α+β). The desired conclusions follow from (a).
We say a triangle rational if all its sides are rational. By Lemma 2 (c), all the angles in a rational triangle have rational cosine values and are equivalent to each other. To prove Lemma 1, we set ∠DAC=A1,∠CAB=A2,∠ABD=B1,∠DBC=B2,∠BCA=C1, ∠ACD=C2,∠CDB=D1,∠BDA=D2. Because triangles ABC,ABD,ADC are rational, angles A2,A1+A2,A1 all have rational cosine values. By Lemma 2 (a), A1 and A2 are equivalent. Similarly, we can show that B1 and B2,C1 and C2,D1 and D2 are equivalent. Because triangle ABC is rational, angles A2 and C1 are equivalent. There all angles A1,A2,B1,…,D2 have rational cosine values and are equivalent.
Because angles A2 and B1 are equivalent, angle A2+B1 has rational values and is equivalent to A2 and B1. Thus, ∠APB=180∘−(A2+B1) has rational cosine value and is equivalent to A2 and B1. Apply the Law of Sine to triangle ABP gives
sin∠APBAB=sin∠B1AP=sin∠A2BP,
implying that both AP and BP have rational length. Similarly, we can show that both CP and DP has rational length, proving Lemma 1.
Third approach: This approach applies the techniques used in the first approach into the second approach. To prove Lemma 1, we set ∠DAP=A1 and ∠BAP=A2. Applying the Law of Cosine to triangle ABC,ABC,ADC shows that angles A1,A2,A1+A2 all has rational cosine values. By the Addition formula, we have
implying that BPPD is rational. Because BP+PD=BD is rational, both BP and PD are rational. Similarly, AP and PC are rational, proving Lemma 1.
Fourth approach: This approach is based on the following lemma.
Lemma 3. Let ABC be a triangle, D be a point on side AC,ϕ1=∠DAB,ϕ2=∠DBA, ϕ3=∠DBC,ϕ4=∠DCB,AB=c,BC=a,AD=x, and DC=y. If the numbers a,c, and cosϕi(1≤i≤4) are all rational, then numbers x and y are also rational.
Proof: Note that x+y=AC=ccosϕ1+acosϕ4 is rational. Hence x is rational if and only if y is rational. Let BD=z. Projecting point D onto the lines AB and BC yields
{xcosϕ1+zcosϕ2=cycosϕ4+zcosϕ3=a
or, denoting ci=cosϕi for i=1,2,3,4,
{c1x+c2z=cc4y+c3z=a
Eliminating z, we get (c1c3)x−(c2c4)y=c3c−c2a, which is rational. Hence there exist rational numbers, r1 and r2, such that
{(c1c3)x−(c2c4)y=r1x+y=r2
We consider two cases.
In this case, we assume that the determinant of the above system, c1c3+c2c4, is not equal to 0 , then this system has a unique solution (x,y) in rational numbers.
In this case, we assume that the determinant c1c3+c2c4=0, or
cosϕ1cosϕ3=−cosϕ2cosϕ4
Let's denote θ=∠BDC, then ϕ2=θ−ϕ1 and ϕ3=180∘−(θ+ϕ4). Then the above equation becomes
It is possible only if [θ+ϕ1+ϕ4]±[θ−ϕ1−ϕ4]=360∘, that is, either θ=180∘ or ϕ1+ϕ4=180∘, which is impossible because they are angles of triangles.
Thus, the determinant c1c3+c2c4 is not equal to 0 and x and y are both rational numbers. Now we are ready to prove Lemma 1. Applying the Law of Cosine to triangles ABC,ACD,ABD shows that cos∠BAC,cos∠CAD,cos∠ABD,cos∠ADB are all rational. Applying Lemma 1 to triangle ABD shows that both of the segments BP and DP have rational lengths. In exactly the same way, we can show that both of the segments AP and CP have rational lengths.
Note: It's interesting how easy it is to get a gap in the proof of the Lemma 1 by using the core idea of the proof of Lemma 3. Here is an example.
Let us project the intersection point of the diagonals, O, onto the four lines of all sides of the quadrilateral. We get the following 4×4 system of linear equations:
Using the Kramer's Rule, we conclude that all x,y,z, and t must be rational numbers, for all the corresponding determinants are rational. However, this logic works only if the determinant of the system is different from 0 .
Unfortunately, there are many geometric configurations for which the determinant of the system vanishes (for example, this occurs for rectangles), and you cannot make a conclusion of rationality of the segments x,y,z, and t. That's why Lemma 2 plays the central role in the solution to this problem.
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