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Then s and t may be \+ chosen so that the triangle C(A+tK)(B+sK) is similar to PQR." }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 1 " " }{TEXT 262 6 "Proof:" }{TEXT -1 268 " Let K=[0,0,1]. Then the t heorem asserts that given any other triangle PQR we can choose t and s so that the triangle (A+tK)C(B+sK) is similar to PQR. 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:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :::::::::::::::::::::j;B:;5:" }{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 282 " Refe rring to the diagram place the triangle ABC (or one similar to it) \+ so that C=[0,0,0], B=[1,0,0], and A=[a,b,0]] with a,b positive. Place \+ the triangle PQR (or one similar to it) so that P=A, Q=C, and R=[p,q ,r[ with p,r positive. Then we want to choose s and t so that the " } {XPPEDIT 18 0 "angle(( A+t*K) *C *(B+s*K))" "-%&angleG6#*(,&%\"AG\"\" \"*&%\"tGF(%\"KGF(F(F(%\"CGF(,&%\"BGF(*&%\"sGF(F+F(F(F(" }{TEXT -1 357 " has the same cosine as QPR and the lengths of the vectors A+tK and B+tK are proportional to QP and RP, the legs of QPR by some fact or k. We have arranged for C and P to be the origin so that we can us e that the cosine of the angle between two vectors is their dot pr oduct divided by the products of their lengths. We then have the thre e equations:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 275 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "dotprod(A+tK , B+tK)/(length(A+t*K)*len gth(B+s*K))= dotprod(Q,R)/(length(Q)*length(R))" "/*&-%(dotprodG6$,&% \"AG\"\"\"%#tKGF),&%\"BGF)F*F)F)*&-%'lengthG6#,&F(F)*&%\"tGF)%\"KGF)F) F)-F/6#,&F,F)*&%\"sGF)F4F)F)F)!\"\"*&-F%6$%\"QG%\"RGF)*&-F/6#F>F)-F/6# F?F)F:" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 21 "and the other t wo are" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 267 "" 0 "" {TEXT -1 26 "length(A+tK) = k length(Q)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 268 "" 0 "" {TEXT -1 26 "length(B+sK) = k length(R)" }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 159 "Using the second \+ two equations in the denominators of the first and knowing that the do t product of Q and R is p we have, on substitution the three equations : " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 269 "" 0 "" {TEXT -1 1 " \+ " }{XPPEDIT 18 0 "dotprod(A+tK , B+sK)= k^2*p" "/-%(dotprodG6$,&%\"AG \"\"\"%#tKGF(,&%\"BGF(%#sKGF(*&%\"kG\"\"#%\"pGF(" }{TEXT -1 3 " ," }} {PARA 270 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 " length(A+t*K) = k" " /-%'lengthG6#,&%\"AG\"\"\"*&%\"tGF(%\"KGF(F(%\"kG" }{TEXT -1 3 " , " } }{PARA 271 "" 0 "" {XPPEDIT 18 0 "length(B+s*K) = k^2*(p^2+q^2)" "/-%' lengthG6#,&%\"BG\"\"\"*&%\"sGF(%\"KGF(F(*&%\"kG\"\"#,&*$%\"pG\"\"#F(*$ %\"qG\"\"#F(F(" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 27 "Putting these into Maplese," }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 237 "A: =[a,b,0]; B:=[1,0,0]; C:=[0,0,0];K:=[0,0,1];\n P:=[0,0,0];Q:=[1,0,0];R :=[p,q,0];\neq1 := linalg[dotprod](A + t*K,B + s*K) = k^2*p;\neq2 := l inalg[dotprod](A + t*K,A + t*K) = k^2;\neq3 := linalg[dotprod](B + s*K ,B + s*K) = k^2*(p^2 + q^2); " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"A G7%%\"aG%\"bG\"\"!" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"BG7%\"\"\"\" \"!F'" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"CG7%\"\"!F&F&" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"KG7%\"\"!F&\"\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"PG7%\"\"!F&F&" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>% \"QG7%\"\"\"\"\"!F'" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"RG7%%\"pG% \"qG\"\"!" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$eq1G/,&%\"aG\"\"\"*&% \"tGF(%\"sGF(F(*&%\"kG\"\"#%\"pGF(" }}{PARA 11 "" 1 "" {XPPMATH 20 "6# >%$eq2G/,(*$%\"aG\"\"#\"\"\"*$%\"bGF)F**$%\"tGF)F**$%\"kGF)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$eq3G/,&\"\"\"F'*$%\"sG\"\"#F'*&%\"kGF*,&* $%\"pGF*F'*$%\"qGF*F'F'" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 " " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 60 "Use eq2 to eliminate k^2 from the first and third euqations." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 61 "EQ1:= lhs(eq1)=lhs(eq2 )*p;\nEQ2:=lhs(eq3)=lhs(eq2)*(p^2+q^2); " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$EQ1G/,&%\"aG\"\"\"*&%\"tGF(%\"sGF(F(*&,(*$F'\"\"#F(*$%\"bGF/F (*$F*F/F(F(%\"pGF(" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$EQ2G/,&\"\"\" F'*$%\"sG\"\"#F'*&,(*$%\"aGF*F'*$%\"bGF*F'*$%\"tGF*F'F',&*$%\"pGF*F'*$ %\"qGF*F'F'" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 73 "We can find solve for values of s and t w hich satisfy these equations by:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "solve(\{EQ1,EQ2\},\{s,t\});" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#<$/%\"tG-%'RootOfG6#,2*&%#_ZG\"\"%%\"qG\"\"#\"\"\"*&,.! \"\"F/*&%\"aGF.%\"pGF.F2*&F4F/F5F/F.*&%\"bGF.F5F.F2*&F4F.F-F.F/*&F8F.F -F.F/F/F+F.F/*&F5F.F8F,F2*$F4F.F2*&F5F/F4\"\"$F.*(F4F/F5F/F8F.F.*&F5F. F4F,F2*(F5F.F4F.F8F.!\"#/%\"sG*&,*F4F2*&F5F/F4F.F/*&F5F/F8F.F/*&F5F/F& F.F/F/F&F2" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 214 "This tells us that if at least one of the roots of \"%1\" non -zero and real then there is a solution. We need to analyze \"%1\" to \+ determine whether it has a non-zero root. The following extracts the a ctual polynomial" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "p1:=op(1,%1);" }}}{EXCHG {PARA 12 "" 1 "" {XPPMATH 20 "6#>%#p1G,2*&%#_ZG\"\"%%\"qG\"\"#\"\"\"*&,.!\"\"F+*&%\"aGF *%\"pGF*F.*&F0F+F1F+F**&%\"bGF*F1F*F.*&F0F*F)F*F+*&F4F*F)F*F+F+F'F*F+* &F1F*F4F(F.*$F0F*F.*&F1F+F0\"\"$F**(F0F+F1F+F4F*F**&F1F*F0F(F.*(F1F*F0 F*F4F*!\"#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 93 "By inspection we see that although p1 is \+ a quartic polynomial it is actually a quadratic in " }{XPPEDIT 18 0 " _Z^2" "*$%#_ZG\"\"#" }{TEXT -1 53 ". We might work with the quadratic by substituting " }{XPPEDIT 18 0 "sqrt(X)" "-%%sqrtG6#%\"XG" }{TEXT -1 7 " for _Z" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "p2:=subs(_Z=sqrt(X),p1);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%#p2G,2*&%\"XG\"\"#%\"qGF(\"\"\"*&,.!\"\"F**&%\"aGF(% \"pGF(F-*&F/F*F0F*F(*&%\"bGF(F0F(F-*&F/F(F)F(F**&F3F(F)F(F*F*F'F*F**&F 0F(F3\"\"%F-*$F/F(F-*&F0F*F/\"\"$F(*(F/F*F0F*F3F(F(*&F0F(F/F7F-*(F0F(F /F(F3F(!\"#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 82 "Thus in order for for there to be a solut ion we must find a positive root to p2. " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 58 " Since the lead coefficient of p2 is the positive number \+ " }{XPPEDIT 18 0 "q^2" "*$%\"qG\"\"#" }{TEXT -1 295 " its graph is a \+ upward-opening parabola which crosses the x-axis twice if p2(z) is neg ative for some value of z. (The \"twice\" is important since we need a non-zero root): if there are two roots one of them must be non-zero. In fact \"const\", the constant term of p2 is the negative of a squa re." }}{PARA 0 "" 0 "" {TEXT -1 2 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "const:=subs(_Z=0,p1);\nconst:=factor(const);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%&constG,.*&%\"pG\"\"#%\"bG\"\"%!\"\" *$%\"aGF(F+*&F'\"\"\"F-\"\"$F(*(F-F/F'F/F)F(F(*&F'F(F-F*F+*(F'F(F-F(F) F(!\"#" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&constG,$*$,(%\"aG!\"\"*&% \"pG\"\"\"F(\"\"#F,*&F+F,%\"bGF-F,F-F)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 120 "If conts is nonzero, the polynomi al will have two real roots. We can easily find when the constant is zero by solving " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "solve(-a+p*a^2+p*b^2=0,p);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#*&%\"aG\"\"\",&*$F$\"\"#F%*$%\"bGF(F%! \"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 35 "Thus const is zero if and only if " }{XPPEDIT 18 0 "p = a/(a^2+b^2)" "/%\"pG*&%\"aG\"\"\",&*$F%\"\"#F&*$%\"bG\"\"#F& !\"\"" }{TEXT -1 130 ". and we must deal with this possibility. If it \+ happens then we can incorporate this into the original equations and s olve again. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 93 "EQ1a:=subs(p=a/(a^2+b^2),EQ1);\nEQ2a:=subs(p=a/( a^2+b^2),EQ2);\nSLNS:=solve(\{EQ1a,EQ2a\},\{s,t\});" }}{PARA 11 "" 1 " " {XPPMATH 20 "6#>%%EQ1aG/,&%\"aG\"\"\"*&%\"tGF(%\"sGF(F(*(,(*$F'\"\"# F(*$%\"bGF/F(*$F*F/F(F(F'F(,&F.F(F0F(!\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%EQ2aG/,&\"\"\"F'*$%\"sG\"\"#F'*&,(*$%\"aGF*F'*$%\"bG F*F'*$%\"tGF*F'F',&*&F.F*,&F-F'F/F'!\"#F'*$%\"qGF*F'F'" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%%SLNSG6$<$/%\"tG\"\"!/%\"sG-%'RootOfG6#,,*&,&*$% \"aG\"\"#\"\"\"*$%\"bGF4F5F5%#_ZGF4F5*&%\"qGF4F3\"\"%!\"\"F6F5*(F:F4F3 F4F7F4!\"#*&F:F4F7F;F<<$/F+**-F-6#,,F0F5F9F5F=F4F?F5F6F " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 153 "The problem is to see that one of these posibilities is a pair of real numbers. We will have a solution with t=0 if there is a real root to the quadratic" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "q1:= (a^2+b ^2)*_Z^2-q^2*a^4-2*q^2*b^2*a^2-q^2*b^4+b^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#q1G,,*&,&*$%\"aG\"\"#\"\"\"*$%\"bGF*F+F+%#_ZGF*F+*&% \"qGF*F)\"\"%!\"\"F,F+*(F0F*F)F*F-F*!\"#*&F0F*F-F1F2" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 67 "O r there will be a solution if the other quadratic has real roots. " }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "q2:=(a^2+b^2)*_Z^2+q^2*a^4+2*q^2*b^2*a^2+q^2*b^4-b^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#q2G,,*&,&*$%\"aG\"\"#\"\"\"*$%\"bGF*F+F+% #_ZGF*F+*&%\"qGF*F)\"\"%F+*(F0F*F)F*F-F*F**&F0F*F-F1F+F,!\"\"" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 274 "" 0 " " {TEXT -1 157 " A quadratic has real roots if and only if its discrim inant is not-negative so we must see that the discriminnant of one of \+ q1, q2 is non-negative. In this " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "discrim(q1,_Z);\ndiscrim(q2, _Z);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&,&*$%\"aG\"\"#\"\"\"*$%\"b GF(F)F),**&%\"qGF(F'\"\"%F)*(F.F(F'F(F+F(F(*&F.F(F+F/F)F*!\"\"F)F/" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&,&*$%\"aG\"\"#\"\"\"*$%\"bGF(F)F), **&%\"qGF(F'\"\"%F)*(F.F(F'F(F+F(F(*&F.F(F+F/F)F*!\"\"F)!\"%" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 155 "Since each of these discriminants is the negative of the other, one of these quadratics is guaranteed to have real roots. Hen ce the problem can be solved." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 135 "One value of having an argument l ike this worked out in Maple is that we can use the argument to solve \+ specific problems. For instance:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}} {SECT 0 {PARA 4 "" 0 "" {TEXT -1 64 "A prism with an equilateral base \+ and a 3-4-5 right cross-section" }}{EXCHG {PARA 260 "" 0 "" {TEXT 267 8 "Exercise" }{TEXT -1 114 ": Cut a triangular prism with an equilate ral base so that the cross-section of the cut is a 3-4-5 right triangl e?" }}{PARA 0 "" 0 "" {TEXT -1 113 "Refering to the original diagram w e want ABC to be the equilateral triangle with C=[0,0,0], B=[1,0,0], \+ and A= [" }{XPPEDIT 18 0 "1/2, sqrt(3)/2" "6$*&\"\"\"\"\"\"\"\"#!\"\" *&-%%sqrtG6#\"\"$F%\"\"#F'" }{TEXT -1 82 ",0]. We want PQR to be a 3-4 -5 triangle. One choice is P=[0,0,0], Q=[1,0,0], R=[0," }{XPPEDIT 18 0 "3/4" "*&\"\"$\"\"\"\"\"%!\"\"" }{TEXT -1 95 ",0]. Once these are d etermined the general calculations become specific and we simply speci fy:" }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 263 "" 0 "" {TEXT -1 34 " a:=1/2;b:=sqrt(3)/2; p:=0;q:=3/4;" }}{PARA 0 "" 0 "" {TEXT -1 0 "" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 50 "Now we sim ply follow the calculations in the proof" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 " " }}}{EXCHG {PARA 263 "" 0 "" {TEXT -1 224 " A:=[a,b ,0]; B:=[1,0,0]; C:=[0,0,0];K:=[0,0,1]; \nP:=[0,0,0];Q:=[1,0,0];R:=[p, q,0];\neq1:=linalg[dotprod](A+t*K,B+s*K) = k^2*p;\n eq2:=linalg[dotpro d](A+t*K,A+t*K) = k^2;\n eq3:=linalg[dotprod](B+s*K,B+s*K) = k^2*(p^2+ q^2); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 59 "Use eq2 to eliminate k^2 from the first and third equations" }} }{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 263 "" 0 "" {TEXT -1 65 " EQ1:= lhs(eq1)=lhs(eq2)*p;\nEQ2:=lhs(eq3)=lhs(eq2)*(p^2+q^2); " } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 73 "we can find solve for values of s and t w hich satisfy these equations by:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 263 "" 0 "" {TEXT -1 28 "ans:=solve(\{EQ1,EQ2\},\{s,t\});" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 67 "At this point we are happy with floating point values for s and t." }}{PARA 263 "" 0 "" {TEXT -1 45 "ans:=solve(\{EQ 1,EQ2\},\{s,t\});\nans:=evalf(ans):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 97 "To get the actual triangles we sub stitute for s and t in the vertices A+tK, B+tK, and the origin." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 263 "" 0 "" {TEXT -1 50 "ANS:=su bs(ans ,[A+s*K,C,B+t*K]);\nANS:=expand(ANS);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" } }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 153 "To check that these are actuall y the vertices of a 3-4-5 triangle we can calculate the lengths of the sides of the triangle. We need a distance function" }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 263 "" 0 "" {TEXT -1 81 "dist:=proc(X,Y) \nsq rt((X[1] - Y[1])^2 + (X[2] - Y[2])^2 + (X[3] - Y[3])^2) \nend;" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 263 "" 0 "" {TEXT -1 70 "legs:=[dist(ANS[1],ANS[2]),dist(ANS[2],ANS[3] ), dist(ANS[3],ANS[1])]; " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 115 "If this is proportional to a 3-4-5 triangle then th e first leg must correspond to one of length 3 thus we scale it." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 263 "" 0 "" {TEXT -1 47 "check:= 3/legs[1] *legs;\ncheck:=expand(check); " }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}} {SECT 0 {PARA 4 "" 0 "" {TEXT -1 9 "EXERCISES" }}{EXCHG {PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 260 "" 0 "" {TEXT 263 8 "Exercise" }{TEXT -1 187 ": Make a three dimensional Maple plot of the of the figure sket ched below. Its base is an equilateral triangle the sides are perpendi cular to the base, and the top is a 3-4-5 triangle." }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 260 "" 0 "" {TEXT 264 8 "Exercise" }{TEXT -1 69 ": Do the previous exercise with \+ the top an isosceles right triangle." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 260 "" 0 "" {TEXT 265 8 "Exercise" }{TEXT -1 10 ": Suppose " } {XPPEDIT 18 0 "f(x) = a*x^4+b*x^3+c*x^2+d*x+e" "/-%\"fG6#%\"xG,,*&%\"a G\"\"\"*$F&\"\"%F*F**&%\"bGF**$F&\"\"$F*F**&%\"cGF**$F&\"\"#F*F**&%\"d GF*F&F*F*%\"eGF*" }{TEXT -1 49 " with a positive. Suppose there is a z such that " }{XPPEDIT 18 0 "f(z)<0" "2-%\"fG6#%\"zG\"\"!" }{TEXT -1 48 ". Prove that f has at least two different roots." }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 260 "" 0 "" {TEXT 266 8 "Exercise" }{TEXT -1 10 ": Suppose " }{XPPEDIT 18 0 "f(x) = a*x^2 + b*x+c" "/-%\"fG6#%\"xG, (*&%\"aG\"\"\"*$F&\"\"#F*F**&%\"bGF*F&F*F*%\"cGF*" }{TEXT -1 22 " with a positive. If " }{XPPEDIT 18 0 "c<0" "2%\"cG\"\"!" }{TEXT -1 12 " p rove that " }{XPPEDIT 18 0 "g (x) = a*x^4 + b*x^2 +c" "/-%\"gG6#%\"xG, (*&%\"aG\"\"\"*$F&\"\"%F*F**&%\"bGF**$F&\"\"#F*F*%\"cGF*" }{TEXT -1 17 " has a real root." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 1 " " }{HYPERLNK 17 "Table of contents" 1 "vp stoc.mws" "" }}}}{MARK "0 1 4 0 1" 0 }{VIEWOPTS 1 1 0 1 1 1803 }