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Question: $p,q,r$ are reals such that the following system holds: $$(p+q)(q+r)=-1$$ $$(p-q)^2(1+(p+q)^2)=20$$...

p,q,rp,q,r are reals such that the following system holds:

(p+q)(q+r)=−1(p+q)(q+r)=-1 (p−q)2(1+(p+q)2)=20(p-q)^2(1+(p+q)^2)=20 (q−r)2(1+(q+r)2)=21(q-r)^2(1+(q+r)^2)=21

Find (r−p)2(1+(r+p)2)(r-p)^2(1+(r+p)^2).

Answer

41

Explanation

Solution

Let A=p+qA = p+q, B=q+rB = q+r, C=r+pC = r+p, X=p−qX = p-q, Y=q−rY = q-r, Z=r−pZ = r-p. The given equations are: \begin{enumerate} AB=−1AB = -1 X2(1+A2)=20X^2(1+A^2) = 20 Y2(1+B2)=21Y^2(1+B^2) = 21 \end{enumerate}

We need to find Z2(1+C2)Z^2(1+C^2). Also, X+Y+Z=0X+Y+Z = 0.

We have A=p+qA = p+q, B=q+rB = q+r, C=r+pC = r+p. Then A+B+C=2(p+q+r)A+B+C = 2(p+q+r). So p=A−B+C2p = \frac{A-B+C}{2}, q=A+B−C2q = \frac{A+B-C}{2}, r=−A+B+C2r = \frac{-A+B+C}{2}. Thus X=p−q=C−BX = p-q = C-B, Y=q−r=A−CY = q-r = A-C, Z=r−p=B−AZ = r-p = B-A. So we have: \begin{enumerate} AB=−1AB = -1 (C−B)2(1+A2)=20(C-B)^2(1+A^2) = 20 (A−C)2(1+B2)=21(A-C)^2(1+B^2) = 21 \end{enumerate} We want to find (B−A)2(1+C2)(B-A)^2(1+C^2).

Since AB=−1AB = -1, let A=tan⁡θA = \tan \theta, B=tan⁡ϕB = \tan \phi. Then tan⁡θtan⁡ϕ=−1\tan \theta \tan \phi = -1, so tan⁡ϕ=−cot⁡θ=tan⁡(θ+π2)\tan \phi = -\cot \theta = \tan (\theta + \frac{\pi}{2}). Then ϕ=θ+π2\phi = \theta + \frac{\pi}{2}. Also 1+A2=sec⁡2θ1+A^2 = \sec^2 \theta, 1+B2=sec⁡2ϕ1+B^2 = \sec^2 \phi. Then (C−B)2sec⁡2θ=20(C-B)^2 \sec^2 \theta = 20 and (A−C)2sec⁡2ϕ=21(A-C)^2 \sec^2 \phi = 21.

A−C=q−rA-C = q-r and C−B=p−qC-B = p-q and B−A=r−pB-A = r-p. Let p+q=x1p+q = x_1, p−q=y1p-q = y_1. Let q+r=x2q+r = x_2, q−r=y2q-r = y_2. Let r+p=x3r+p = x_3, r−p=y3r-p = y_3. Then x1x2=−1x_1 x_2 = -1, y12(1+x12)=20y_1^2(1+x_1^2) = 20, y22(1+x22)=21y_2^2(1+x_2^2) = 21. We need to find y32(1+x32)y_3^2(1+x_3^2). We have x1=p+qx_1 = p+q, x2=q+rx_2 = q+r, x3=r+px_3 = r+p. Also y1=p−qy_1 = p-q, y2=q−ry_2 = q-r, y3=r−py_3 = r-p. Then x1−x2=p−r=−y3x_1-x_2 = p-r = -y_3, so y3=x2−x1y_3 = x_2-x_1. x2−x3=q−p=−y1x_2-x_3 = q-p = -y_1, so y1=x3−x2y_1 = x_3-x_2. x3−x1=r−q=−y2x_3-x_1 = r-q = -y_2, so y2=x1−x3y_2 = x_1-x_3.

Then x1x2=−1x_1 x_2 = -1, (x3−x2)2(1+x12)=20(x_3-x_2)^2(1+x_1^2) = 20, (x1−x3)2(1+x22)=21(x_1-x_3)^2(1+x_2^2) = 21. We need to find (x2−x1)2(1+x32)(x_2-x_1)^2(1+x_3^2). Since x1x2=−1x_1 x_2 = -1, x2=−1/x1x_2 = -1/x_1. Then (x3+1/x1)2(1+x12)=20(x_3+1/x_1)^2(1+x_1^2) = 20, (x1−x3)2(1+1/x12)=21(x_1-x_3)^2(1+1/x_1^2) = 21. Then (x1x3+1)2x12(1+x12)=20\frac{(x_1 x_3+1)^2}{x_1^2}(1+x_1^2) = 20 and (x1−x3)2x12(1+x12)=21\frac{(x_1-x_3)^2}{x_1^2}(1+x_1^2) = 21. We want to find (1+x12)2x12(1+x32)\frac{(1+x_1^2)^2}{x_1^2}(1+x_3^2). Let K=1+x12x12K = \frac{1+x_1^2}{x_1^2}. Then (x1x3+1)2K=20(x_1 x_3+1)^2 K = 20 and (x1−x3)2K=21(x_1-x_3)^2 K = 21. Then (x1x3+1)2K+(x1−x3)2K=41(x_1 x_3+1)^2 K + (x_1-x_3)^2 K = 41. K[(x1x3+1)2+(x1−x3)2]=41K [(x_1 x_3+1)^2 + (x_1-x_3)^2] = 41. K[x12x32+2x1x3+1+x12−2x1x3+x32]=41K [x_1^2 x_3^2+2x_1 x_3+1+x_1^2-2x_1 x_3+x_3^2] = 41. K[x12x32+1+x12+x32]=41K [x_1^2 x_3^2+1+x_1^2+x_3^2] = 41. K[(x12+1)(x32+1)]=41K [(x_1^2+1)(x_3^2+1)] = 41. 1+x12x12(1+x12)(1+x32)=41\frac{1+x_1^2}{x_1^2} (1+x_1^2)(1+x_3^2) = 41. (1+x12)2x12(1+x32)=41\frac{(1+x_1^2)^2}{x_1^2} (1+x_3^2) = 41. This is the expression we need to find. Therefore the answer is 41.