|
|
| (48 intermediate revisions by 19 users not shown) |
| Line 2: |
Line 2: |
| | | | |
| | ==Problem== | | ==Problem== |
| | + | In square <math>ABCD</math>, points <math>E</math> and <math>H</math> lie on <math>\overline{AB}</math> and <math>\overline{DA}</math>, respectively, so that <math>AE=AH.</math> Points <math>F</math> and <math>G</math> lie on <math>\overline{BC}</math> and <math>\overline{CD}</math>, respectively, and points <math>I</math> and <math>J</math> lie on <math>\overline{EH}</math> so that <math>\overline{FI} \perp \overline{EH}</math> and <math>\overline{GJ} \perp \overline{EH}</math>. See the figure below. Triangle <math>AEH</math>, quadrilateral <math>BFIE</math>, quadrilateral <math>DHJG</math>, and pentagon <math>FCGJI</math> each has area <math>1.</math> What is <math>FI^2</math>? |
| | | | |
| − | In square <math>ABCD</math>, points <math>E</math> and <math>H</math> lie on <math>\overline{AB}</math> and <math>\overline{DA}</math>, respectively, so that <math>AE=AH.</math> Points <math>F</math> and <math>G</math> lie on <math>\overline{BC}</math> and <math>\overline{CD}</math>, respectively, and points <math>I</math> and <math>J</math> lie on <math>\overline{EH}</math> so that <math>\overline{FI} \perp \overline{EH}</math> and <math>\overline{GJ} \perp \overline{EH}</math>. See the figure below. Triangle <math>AEH</math>, quadrilateral <math>BFIE</math>, quadrilateral <math>DHJG</math>, and pentagon <math>FCGJI</math> each has area <math>1.</math> What is <math>FI^2</math>?
| |
| | <asy> | | <asy> |
| | real x=2sqrt(2); | | real x=2sqrt(2); |
| Line 38: |
Line 38: |
| | | | |
| | </asy> | | </asy> |
| | + | |
| | <math>\textbf{(A) } \frac{7}{3} \qquad \textbf{(B) } 8-4\sqrt2 \qquad \textbf{(C) } 1+\sqrt2 \qquad \textbf{(D) } \frac{7}{4}\sqrt2 \qquad \textbf{(E) } 2\sqrt2</math> | | <math>\textbf{(A) } \frac{7}{3} \qquad \textbf{(B) } 8-4\sqrt2 \qquad \textbf{(C) } 1+\sqrt2 \qquad \textbf{(D) } \frac{7}{4}\sqrt2 \qquad \textbf{(E) } 2\sqrt2</math> |
| | | | |
| − | ==Solution== | + | == Solution 1 == |
| − | Since the total area is <math>4</math>, the side length of square <math>ABCD</math> is <math>2</math>. We see that since triangle <math>HAE</math> is a right isosceles triangle with area 1, we can determine sides <math>HA</math> and <math>AE</math> both to be <math>\sqrt{2}</math>. Now, consider extending <math>FB</math> and <math>IE</math> until they intersect. Let the point of intersection be <math>K</math>. We note that <math>EBK</math> is also a right isosceles triangle with side <math>2-\sqrt{2}</math> and find it's area to be <math>3-2\sqrt{2}</math>. Now, we notice that <math>FIK</math> is also a right isosceles triangle and find it's area to be <math>\frac{1}{2}</math><math>FI^2</math>. This is also equal to <math>1+3-2\sqrt{2}</math> or <math>4-2\sqrt{2}</math>. Since we are looking for <math>FI^2</math>, we want two times this. That gives <math>\boxed{\textbf{(B)}\ 8-4\sqrt{2}}</math>.~TLiu | + | Since the total area is <math>4</math>, the side length of square <math>ABCD</math> is <math>2</math>. We see that since triangle <math>HAE</math> is a right isosceles triangle with area 1, we can determine sides <math>HA</math> and <math>AE</math> both to be <math>\sqrt{2}</math>. Now, consider extending <math>FB</math> and <math>IE</math> until they intersect. Let the point of intersection be <math>K</math>. We note that <math>EBK</math> is also a right isosceles triangle with side <math>2-\sqrt{2}</math> and find its area to be <math>3-2\sqrt{2}</math>. Now, we notice that <math>FIK</math> is also a right isosceles triangle (because <math>\angle EKB=45^\circ</math>) and find it's area to be <math>\frac{1}{2}</math><math>FI^2</math>. This is also equal to <math>1+3-2\sqrt{2}</math> or <math>4-2\sqrt{2}</math>. Since we are looking for <math>FI^2</math>, we want two times this.(This is a misplaced problem) That gives <math>\boxed{\textbf{(B)}\ 8-4\sqrt{2}}</math>.~TLiu |
| | | | |
| − | ==Solution 2== | + | == Solution 2 == |
| − | Since this is a geometry problem involving sides, and we know that <math>HE</math> is <math>2</math>, we can use our ruler and find the ratio between <math>FI</math> and <math>HE</math>. Measuring(on the booklet), we get that <math>HE</math> is about <math>1.8</math> inches and <math>FI</math> is about <math>1.4</math> inches. Thus, we can then multiply the length of <math>HE</math> by the ratio of <math>\frac{1.4}{1.8},</math> of which we then get <math>FI= \frac{14}{9}.</math> We take the square of that and get <math>\frac{196}{81},</math> and the closest answer to that is <math>\boxed{\textbf{(B)}\ 8-4\sqrt{2}}</math>. ~Celloboy (Note that this is just a strategy I happened to use that worked. Do not press your luck with this strategy, for it was a lucky guess)
| |
| − | | |
| − | ==Solution 3==
| |
| | Draw the auxiliary line <math>AC</math>. Denote by <math>M</math> the point it intersects with <math>HE</math>, and by <math>N</math> the point it intersects with <math>GF</math>. Last, denote by <math>x</math> the segment <math>FN</math>, and by <math>y</math> the segment <math>FI</math>. We will find two equations for <math>x</math> and <math>y</math>, and then solve for <math>y^2</math>. | | Draw the auxiliary line <math>AC</math>. Denote by <math>M</math> the point it intersects with <math>HE</math>, and by <math>N</math> the point it intersects with <math>GF</math>. Last, denote by <math>x</math> the segment <math>FN</math>, and by <math>y</math> the segment <math>FI</math>. We will find two equations for <math>x</math> and <math>y</math>, and then solve for <math>y^2</math>. |
| | | | |
| Line 62: |
Line 60: |
| | Solving for <math>y^2</math> gives <math>\boxed{\textbf{(B)}\ 8-4\sqrt{2}}</math> ~DrB | | Solving for <math>y^2</math> gives <math>\boxed{\textbf{(B)}\ 8-4\sqrt{2}}</math> ~DrB |
| | | | |
| | + | == Solution 3 == |
| | + | Plot a point <math>F'</math> such that <math>F'I</math> and <math>AB</math> are parallel and extend line <math>FB</math> to point <math>B'</math> such that <math>FIB'F'</math> forms a square. Extend line <math>AE</math> to meet line <math>F'B'</math> and point <math>E'</math> is the intersection of the two. The area of this square is equivalent to <math>FI^2</math>. We see that the area of square <math>ABCD</math> is <math>4</math>, meaning each side is of length 2. The area of the pentagon <math>EIFF'E'</math> is <math>2</math>. Length <math>AE=\sqrt{2}</math>, thus <math>EB=2-\sqrt{2}</math>. Triangle <math>EB'E'</math> is isosceles, and the area of this triangle is <math>\frac{1}{2}(4-2\sqrt{2})(2-\sqrt{2})=6-4\sqrt{2}</math>. Adding these two areas, we get <cmath>2+6-4\sqrt{2}=8-4\sqrt{2}\rightarrow \boxed{\textbf{(B)}\ 8-4\sqrt{2}}</cmath>. --OGBooger |
| | | | |
| − | ==Solution 4== | + | == Solution 4 (HARD Calculation) == |
| − | | |
| − | Plot a point <math>F'</math> such that <math>F'I</math> and <math>AB</math> are parallel and extend line <math>FB</math> to point <math>B'</math> such that <math>FIB'F'</math> forms a square. Extend line <math>AE</math> to meet line <math>F'B'</math> and point <math>E'</math> is the intersection of the two. The area of this square is equivalent to <math>FI^2</math>. We see that the area of square <math>ABCD</math> is <math>4</math>, meaning each side is of length 2. The area of the pentagon <math>EIFF'E'</math> is <math>2</math>. Length <math>AE=\sqrt{2}</math>, thus <math>EB=2-\sqrt{2}</math>. Triangle <math>EB'E'</math> is isosceles, and the area of this triangle is <math>\frac{1}{2}(4-2\sqrt{2})(2-\sqrt{2})=6-4\sqrt{2}</math>. Adding these two areas, we get <cmath>2+6-4\sqrt{2}=8-4\sqrt{2}\rightarrow \boxed{\mathrm{(B)}}</cmath>. --OGBooger
| |
| − | | |
| − | ==Solution 5 (HARD Calculation)==
| |
| − | | |
| | We can easily observe that the area of square <math>ABCD</math> is 4 and its side length is 2 since all four regions that build up the square has area 1. | | We can easily observe that the area of square <math>ABCD</math> is 4 and its side length is 2 since all four regions that build up the square has area 1. |
| | Extend <math>FI</math> and let the intersection with <math>AB</math> be <math>K</math>. Connect <math>AC</math>, and let the intersection of <math>AC</math> and <math>HE</math> be <math>L</math>. | | Extend <math>FI</math> and let the intersection with <math>AB</math> be <math>K</math>. Connect <math>AC</math>, and let the intersection of <math>AC</math> and <math>HE</math> be <math>L</math>. |
| Line 78: |
Line 73: |
| | We solve the equation and yield <math>m=\frac{8-2\sqrt{2}-\sqrt{64-32\sqrt{2}}}{2}</math>. | | We solve the equation and yield <math>m=\frac{8-2\sqrt{2}-\sqrt{64-32\sqrt{2}}}{2}</math>. |
| | Now notice that | | Now notice that |
| − | <math>FI=AC-AL=2\sqrt{2}-1-\frac{\sqrt{2}}{2}*\frac{8-2\sqrt{2}-\sqrt{64-32\sqrt{2}}}{2}</math> | + | <math>FI=AC-AL-\frac{m}{\sqrt{2}}=2\sqrt{2}-1-\frac{\sqrt{2}}{2} \cdot \frac{8-2\sqrt{2}-\sqrt{64-32\sqrt{2}}}{2}</math> |
| | <math>=2\sqrt{2}-1-\frac{8\sqrt{2}-4-\sqrt{128-64\sqrt2}}{4}</math> | | <math>=2\sqrt{2}-1-\frac{8\sqrt{2}-4-\sqrt{128-64\sqrt2}}{4}</math> |
| | <math>=\frac{\sqrt{128-64\sqrt{2}}}{4}</math>. | | <math>=\frac{\sqrt{128-64\sqrt{2}}}{4}</math>. |
| | Hence <math>FI^2=\frac{128-64\sqrt{2}}{16}=8-4\sqrt{2}</math>. -HarryW | | Hence <math>FI^2=\frac{128-64\sqrt{2}}{16}=8-4\sqrt{2}</math>. -HarryW |
| | | | |
| − | ==See Also== | + | == Solution 5 (Basically Same as Solution 3)== |
| | | | |
| − | {{AMC10 box|year=2020|ab=B|num-b=20|num-a=22}}
| + | <asy> |
| − | {{AMC12 box|year=2020|ab=B|num-b=17|num-a=19}}
| + | real x=2sqrt(2); |
| | + | real y=2sqrt(16-8sqrt(2))-4+2sqrt(2); |
| | + | real z=2sqrt(8-4sqrt(2)); |
| | + | real k= 8-2sqrt(2); |
| | + | real l= 2sqrt(2)-4; |
| | + | pair A, B, C, D, E, F, G, H, I, J, L, M, K; |
| | + | A = (0,0); |
| | + | B = (4,0); |
| | + | C = (4,4); |
| | + | D = (0,4); |
| | + | E = (x,0); |
| | + | F = (4,y); |
| | + | G = (y,4); |
| | + | H = (0,x); |
| | + | I = F + z * dir(225); |
| | + | J = G + z * dir(225); |
| | + | L = (k,0); |
| | + | M = F + z * dir(315); |
| | + | K = (4,l); |
| | | | |
| − | {{MAA Notice}}
| + | draw(A--B--C--D--A); |
| | + | draw(H--E); |
| | + | draw(J--G^^F--I); |
| | + | draw(F--M); |
| | + | draw(M--L); |
| | + | draw(E--K,dashed+linewidth(.5)); |
| | + | draw(K--L,dashed+linewidth(.5)); |
| | + | draw(B--L); |
| | + | draw(rightanglemark(G, J, I), linewidth(.5)); |
| | + | draw(rightanglemark(F, I, E), linewidth(.5)); |
| | + | draw(rightanglemark(F, M, L), linewidth(.5)); |
| | + | fill((4,0)--(k,0)--M--(4,y)--cycle, gray); |
| | + | dot("$A$", A, S); |
| | + | dot("$C$", C, dir(90)); |
| | + | dot("$D$", D, dir(90)); |
| | + | dot("$E$", E, S); |
| | + | dot("$G$", G, N); |
| | + | dot("$H$", H, W); |
| | + | dot("$I$", I, SW); |
| | + | dot("$J$", J, SW); |
| | + | dot("$K$", K, S); |
| | + | dot("$F(G)$", F, E); |
| | + | dot("$J'$", M, dir(90)); |
| | + | dot("$H'$", L, S); |
| | + | dot("$B(D)$", B, S); |
| | | | |
| | | | |
| | + | </asy> |
| | + | Easily, we can find that: quadrilateral <math>BFIE</math> and <math>DHJG</math> are congruent with each other, so we can move <math>DHJG</math> to the shaded area (<math>F</math> and <math>G</math>, <math>B</math> and <math>D</math> overlapping) to form a square <math>FIKJ'</math> (<math>DG</math> = <math>FB</math>, <math>CG</math> = <math>FC</math>, <math>{\angle} CGF</math> = <math>{\angle}CFG</math> = <math>45^{\circ}</math> so <math>{\angle} IFJ'= 90^{\circ}</math>). Then we can solve <math>AH</math> = <math>AE</math> = <math>\sqrt{2}</math>, <math>EB</math> = <math>2-\sqrt{2}</math>, <math>EK</math> = <math>2\sqrt{2}-2</math>. |
| | | | |
| | + | <math>FI^2=\text{area of} \: BFIE+\text{area of} \:FJ'H'B+\text{area of} \:EH'K \\= 1 + 1 + \frac{1}{2}(2\sqrt{2}-2)^2=8-4\sqrt{2}\rightarrow \boxed{\mathrm{(B)}}</math> |
| | | | |
| | + | --Ryan Zhang @BRS |
| | | | |
| | + | == Solution 6 == |
| | | | |
| | + | <asy> |
| | + | real x=2sqrt(2); |
| | + | real y=2sqrt(16-8sqrt(2))-4+2sqrt(2); |
| | + | real z=2sqrt(8-4sqrt(2)); |
| | + | pair A, B, C, D, E, F, G, H, I, J, K, L; |
| | + | A = (0,0); |
| | + | B = (4,0); |
| | + | C = (4,4); |
| | + | D = (0,4); |
| | + | E = (x,0); |
| | + | F = (4,y); |
| | + | G = (y,4); |
| | + | H = (0,x); |
| | + | I = F + z * dir(225); |
| | + | J = G + z * dir(225); |
| | + | K = (4-x,4); |
| | + | L = J + 1.68 * dir(45); |
| | | | |
| − | ==Solution==
| + | draw(A--B--C--D--A); |
| | + | draw(H--E); |
| | + | draw(J--G^^F--I); |
| | + | draw(H--K,dashed+linewidth(.5)); |
| | + | draw(L--K,dashed+linewidth(.5)); |
| | + | draw(rightanglemark(G, J, I), linewidth(.5)); |
| | + | draw(rightanglemark(F, I, E), linewidth(.5)); |
| | + | draw(rightanglemark(H, K, L), linewidth(.5)); |
| | + | draw(rightanglemark(K, L, G), linewidth(.5)); |
| | | | |
| | + | dot("$A$", A, S); |
| | + | dot("$B$", B, S); |
| | + | dot("$C$", C, dir(90)); |
| | + | dot("$D$", D, dir(90)); |
| | + | dot("$E$", E, S); |
| | + | dot("$F$", F, dir(0)); |
| | + | dot("$G$", G, N); |
| | + | dot("$H$", H, W); |
| | + | dot("$I$", I, SW); |
| | + | dot("$J$", J, SW); |
| | + | dot("$K$", K, N); |
| | + | dot("$L$", L, S); |
| | + | </asy> |
| | | | |
| − | Let <math>x_1=r</math>, then <math>x_2=(-1+i√3)/2</math> r, <math>x_3=((-1+i√3)/2)^2</math> <math>r=(-1-i√3)/2</math> <math>r</math>, <math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| + | <math>[ABCD] = 4</math>, <math>AB = 2</math>, <math>[AHE] = 1</math>, <math>AH = AE = \sqrt{2}</math>, <math>DH = 2 - \sqrt{2}</math>, <math>JL = HK = \sqrt{2} \cdot DH = 2 \sqrt{2} - 2</math> |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| | | | |
| − | ~Yelong_Li
| + | Because <math>ABCD</math> is a square and <math>AH = AE</math>, <math>AC</math> is the line of symmetry of pentagon <math>CDHEB</math>. Because <math>[DHJG] = [BFIE]</math>, <math>DHJG</math> is the reflection of <math>BFIE</math> about line <math>AC</math> |
| − | ==Solution==
| |
| | | | |
| | + | Let <math>FI = GJ = x</math>, <math>KL = LG = GJ - LJ = x - 2 \sqrt{2} + 2</math> |
| | | | |
| − | Let </math>x_1=r<math>, then </math>x_2=(-1+i√3)/2<math> r, </math>x_3=((-1+i√3)/2)^2<math> </math>r=(-1-i√3)/2<math> </math>r<math>, </math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| + | <math>[DHK] = \frac{(2 - \sqrt{2})^2}{2} = 3 - 2 \sqrt {2}</math> |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| | | | |
| − | ~Yelong_Li
| + | <math>[GKL] = \frac{(x - 2 \sqrt{2} + 2)^2}{2} = \frac{x^2}{2} + 2x - 2x \sqrt{2} - 4 \sqrt{2} + 6</math> |
| − | ==Solution==
| |
| | | | |
| | + | <math>[HKJL] = (x - 2 \sqrt{2} + 2) \cdot (2 \sqrt{2} - 2) = 2x \sqrt{2} - 2x + 8 \sqrt{2} -12</math> |
| | | | |
| − | Let <math>x_1=r</math>, then <math>x_2=(-1+i√3)/2</math> r, <math>x_3=((-1+i√3)/2)^2</math> <math>r=(-1-i√3)/2</math> <math>r</math>, <math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| + | <cmath>[DHK] + [GKL] + [HKLJ] = [DHJG]</cmath> |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| | | | |
| − | ~Yelong_Li
| + | <cmath>3 - 2 \sqrt {2} + \frac{x^2}{2} + 2x - 2x \sqrt{2} - 4 \sqrt{2} + 6 + 2x \sqrt{2} - 2x + 8 \sqrt{2} -12= 1</cmath> |
| − | ==Solution==
| |
| | | | |
| | + | <cmath>\frac{x^2}{2} + 2 \sqrt{2} - 4 = 0</cmath> |
| | | | |
| − | Let </math>x_1=r<math>, then </math>x_2=(-1+i√3)/2<math> r, </math>x_3=((-1+i√3)/2)^2<math> </math>r=(-1-i√3)/2<math> </math>r<math>, </math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| + | <cmath>x^2 = 8 - 4 \sqrt{2}</cmath> |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| | | | |
| − | ~Yelong_Li
| + | <cmath>FI^2 = \boxed{\textbf{(B)}\ 8-4\sqrt{2}}</cmath> |
| − | ==Solution==
| |
| | | | |
| | + | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] |
| | | | |
| − | Let <math>x_1=r</math>, then <math>x_2=(-1+i√3)/2</math> r, <math>x_3=((-1+i√3)/2)^2</math> <math>r=(-1-i√3)/2</math> <math>r</math>, <math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| + | == Solution 7 (Easy to See) == |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| | | | |
| − | ~Yelong_Li
| + | Note that the side length of <math>ABCD</math> is 2 and thus the diagonal is of length <math>2\sqrt{2}</math>. However, the height to side <math>HE</math> in triangle <math>HAE</math> is 1, implying that <math>CM = 2\sqrt{2}-1</math> where <math>M</math> is the midpoint of <math>JI</math>. From here suppose that <math>N</math> is the midpoint of <math>\overline{FG}</math> and let <math>x = NC</math>, which means <math>FG=2x</math>. The area of the pentagon is then |
| − | ==Solution== | + | <cmath>[FIJG]+[GCF]=GF \cdot FI + x^2 = (2x)(2\sqrt{2}-1-x)+x^2=1</cmath> |
| | + | Solving this quadratic for <math>x</math> yields <math>x=2\sqrt{2}-1 \pm \sqrt{8-4\sqrt{2}}</math> (technically the smaller value is the correct one but it doesn’t matter for our purposes). We can then calculate <math>FI^2 = (2\sqrt{2} -1 -x)^2 = \boxed{\textbf{(B) } 8-4\sqrt{2}}</math>. |
| | | | |
| | + | ~Dhillonr25 |
| | | | |
| − | Let </math>x_1=r<math>, then </math>x_2=(-1+i√3)/2<math> r, </math>x_3=((-1+i√3)/2)^2<math> </math>r=(-1-i√3)/2<math> </math>r<math>, </math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| + | == Solution 8 == |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| | | | |
| − | ~Yelong_Li
| + | We extend <math>\overline{FB}</math> and <math>\overline{IE}</math> to meet at a point <math>X</math>. Since <math>\angle AEI = \angle BEX</math> and <math>FX</math> is parallel to <math>DA</math>, we know that <math>\triangle{BEX} \sim \triangle{AEH}</math>, and because <math>\angle BXE = \angle IXF</math> and <math>\angle XBE = \angle XIF</math>, we can conclude that <math>\triangle{BEX} \sim \triangle{AEH} \sim \triangle{IFX}</math>. |
| − | ==Solution== | |
| | | | |
| | + | Now, because <math>\triangle{AEH}</math> is isosceles, right, and has an area of 1, we can conclude that <math>AE = AH = \sqrt{2}</math> and that <math>BE = 2-\sqrt{2}</math>. Armed with this knowledge, and setting <math>IF = a</math> and the area of <math>\triangle{BEX} = b</math>, we can use similarity to say that |
| | + | <cmath>(\frac{a}{2-\sqrt{2}})^2 = \frac{1+b}{b}</cmath> |
| | + | Since we know the side lengths of <math>\triangle{BEX}</math> due to the fact that it is also an isosceles right triangle, we know that the area is <math>\frac{(2-\sqrt{2})^2}{2}</math>. |
| | + | Simplifying further and plugging in values, we have |
| | + | <cmath>\frac{a^2}{(2-\sqrt{2})^2} = 1 + \frac{2}{(2-\sqrt{2})^2)}</cmath> |
| | + | Multiplying by <math>(2-\sqrt{2})^2</math> on both sides, we get |
| | + | <cmath>a^2 = (2-\sqrt{2})^2 + 2 = \boxed{\textbf{(B)}\ 8-4\sqrt{2}}</cmath> |
| | + | ~yingkai_0_ |
| | + | ==Video Solution (HOW TO THINK CREATIVELY!!!)== |
| | + | https://youtu.be/oRvHHywcw4w |
| | | | |
| − | Let <math>x_1=r</math>, then <math>x_2=(-1+i√3)/2</math> r, <math>x_3=((-1+i√3)/2)^2</math> <math>r=(-1-i√3)/2</math> <math>r</math>, <math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| + | ~Education, the Study of Everything |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| | | | |
| − | ~Yelong_Li
| + | ==Video Solution by MathEx== |
| − | ==Solution== | + | https://www.youtube.com/watch?v=AKJXB07Sat0 |
| | | | |
| | + | ==Video Solution by TheBeautyOfMath== |
| | + | https://youtu.be/VZYe3Hu88OA?t=189 |
| | | | |
| − | Let </math>x_1=r<math>, then </math>x_2=(-1+i√3)/2<math> r, </math>x_3=((-1+i√3)/2)^2<math> </math>r=(-1-i√3)/2<math> </math>r<math>, </math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| + | == Really Good Vid Explanation == |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| + | https://www.youtube.com/watch?v=AUndgrOH8U8&ab_channel=ReachTheStars |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| | | | |
| − | ~Yelong_Li
| + | ==See Also== |
| − | ==Solution== | + | {{AMC10 box|year=2020|ab=B|num-b=20|num-a=22}} |
| | + | {{AMC12 box|year=2020|ab=B|num-b=17|num-a=19}} |
| | | | |
| − | | + | [[Category:Intermediate Geometry Problems]] |
| − | Let <math>x_1=r</math>, then <math>x_2=(-1+i√3)/2</math> r, <math>x_3=((-1+i√3)/2)^2</math> <math>r=(-1-i√3)/2</math> <math>r</math>, <math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| + | {{MAA Notice}} |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let </math>x_1=r<math>, then </math>x_2=(-1+i√3)/2<math> r, </math>x_3=((-1+i√3)/2)^2<math> </math>r=(-1-i√3)/2<math> </math>r<math>, </math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let <math>x_1=r</math>, then <math>x_2=(-1+i√3)/2</math> r, <math>x_3=((-1+i√3)/2)^2</math> <math>r=(-1-i√3)/2</math> <math>r</math>, <math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let </math>x_1=r<math>, then </math>x_2=(-1+i√3)/2<math> r, </math>x_3=((-1+i√3)/2)^2<math> </math>r=(-1-i√3)/2<math> </math>r<math>, </math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let <math>x_1=r</math>, then <math>x_2=(-1+i√3)/2</math> r, <math>x_3=((-1+i√3)/2)^2</math> <math>r=(-1-i√3)/2</math> <math>r</math>, <math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let </math>x_1=r<math>, then </math>x_2=(-1+i√3)/2<math> r, </math>x_3=((-1+i√3)/2)^2<math> </math>r=(-1-i√3)/2<math> </math>r<math>, </math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let <math>x_1=r</math>, then <math>x_2=(-1+i√3)/2</math> r, <math>x_3=((-1+i√3)/2)^2</math> <math>r=(-1-i√3)/2</math> <math>r</math>, <math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let </math>x_1=r<math>, then </math>x_2=(-1+i√3)/2<math> r, </math>x_3=((-1+i√3)/2)^2<math> </math>r=(-1-i√3)/2<math> </math>r<math>, </math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let <math>x_1=r</math>, then <math>x_2=(-1+i√3)/2</math> r, <math>x_3=((-1+i√3)/2)^2</math> <math>r=(-1-i√3)/2</math> <math>r</math>, <math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let </math>x_1=r<math>, then </math>x_2=(-1+i√3)/2<math> r, </math>x_3=((-1+i√3)/2)^2<math> </math>r=(-1-i√3)/2<math> </math>r<math>, </math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let <math>x_1=r</math>, then <math>x_2=(-1+i√3)/2</math> r, <math>x_3=((-1+i√3)/2)^2</math> <math>r=(-1-i√3)/2</math> <math>r</math>, <math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let </math>x_1=r<math>, then </math>x_2=(-1+i√3)/2<math> r, </math>x_3=((-1+i√3)/2)^2<math> </math>r=(-1-i√3)/2<math> </math>r<math>, </math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let <math>x_1=r</math>, then <math>x_2=(-1+i√3)/2</math> r, <math>x_3=((-1+i√3)/2)^2</math> <math>r=(-1-i√3)/2</math> <math>r</math>, <math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let </math>x_1=r<math>, then </math>x_2=(-1+i√3)/2<math> r, </math>x_3=((-1+i√3)/2)^2<math> </math>r=(-1-i√3)/2<math> </math>r<math>, </math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let <math>x_1=r</math>, then <math>x_2=(-1+i√3)/2</math> r, <math>x_3=((-1+i√3)/2)^2</math> <math>r=(-1-i√3)/2</math> <math>r</math>, <math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let </math>x_1=r<math>, then </math>x_2=(-1+i√3)/2<math> r, </math>x_3=((-1+i√3)/2)^2<math> </math>r=(-1-i√3)/2<math> </math>r<math>, </math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let <math>x_1=r</math>, then <math>x_2=(-1+i√3)/2</math> r, <math>x_3=((-1+i√3)/2)^2</math> <math>r=(-1-i√3)/2</math> <math>r</math>, <math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let </math>x_1=r<math>, then </math>x_2=(-1+i√3)/2<math> r, </math>x_3=((-1+i√3)/2)^2<math> </math>r=(-1-i√3)/2<math> </math>r<math>, </math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let <math>x_1=r</math>, then <math>x_2=(-1+i√3)/2</math> r, <math>x_3=((-1+i√3)/2)^2</math> <math>r=(-1-i√3)/2</math> <math>r</math>, <math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let </math>x_1=r<math>, then </math>x_2=(-1+i√3)/2<math> r, </math>x_3=((-1+i√3)/2)^2<math> </math>r=(-1-i√3)/2<math> </math>r<math>, </math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let <math>x_1=r</math>, then <math>x_2=(-1+i√3)/2</math> r, <math>x_3=((-1+i√3)/2)^2</math> <math>r=(-1-i√3)/2</math> <math>r</math>, <math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let </math>x_1=r<math>, then </math>x_2=(-1+i√3)/2<math> r, </math>x_3=((-1+i√3)/2)^2<math> </math>r=(-1-i√3)/2<math> </math>r<math>, </math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let <math>x_1=r</math>, then <math>x_2=(-1+i√3)/2</math> r, <math>x_3=((-1+i√3)/2)^2</math> <math>r=(-1-i√3)/2</math> <math>r</math>, <math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let </math>x_1=r<math>, then </math>x_2=(-1+i√3)/2<math> r, </math>x_3=((-1+i√3)/2)^2<math> </math>r=(-1-i√3)/2<math> </math>r<math>, </math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let <math>x_1=r</math>, then <math>x_2=(-1+i√3)/2</math> r, <math>x_3=((-1+i√3)/2)^2</math> <math>r=(-1-i√3)/2</math> <math>r</math>, <math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let </math>x_1=r<math>, then </math>x_2=(-1+i√3)/2<math> r, </math>x_3=((-1+i√3)/2)^2<math> </math>r=(-1-i√3)/2<math> </math>r<math>, </math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let <math>x_1=r</math>, then <math>x_2=(-1+i√3)/2</math> r, <math>x_3=((-1+i√3)/2)^2</math> <math>r=(-1-i√3)/2</math> <math>r</math>, <math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let </math>x_1=r<math>, then </math>x_2=(-1+i√3)/2<math> r, </math>x_3=((-1+i√3)/2)^2<math> </math>r=(-1-i√3)/2<math> </math>r<math>, </math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let <math>x_1=r</math>, then <math>x_2=(-1+i√3)/2</math> r, <math>x_3=((-1+i√3)/2)^2</math> <math>r=(-1-i√3)/2</math> <math>r</math>, <math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let </math>x_1=r<math>, then </math>x_2=(-1+i√3)/2<math> r, </math>x_3=((-1+i√3)/2)^2<math> </math>r=(-1-i√3)/2<math> </math>r<math>, </math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let <math>x_1=r</math>, then <math>x_2=(-1+i√3)/2</math> r, <math>x_3=((-1+i√3)/2)^2</math> <math>r=(-1-i√3)/2</math> <math>r</math>, <math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let </math>x_1=r<math>, then </math>x_2=(-1+i√3)/2<math> r, </math>x_3=((-1+i√3)/2)^2<math> </math>r=(-1-i√3)/2<math> </math>r<math>, </math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let <math>x_1=r</math>, then <math>x_2=(-1+i√3)/2</math> r, <math>x_3=((-1+i√3)/2)^2</math> <math>r=(-1-i√3)/2</math> <math>r</math>, <math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let </math>x_1=r<math>, then </math>x_2=(-1+i√3)/2<math> r, </math>x_3=((-1+i√3)/2)^2<math> </math>r=(-1-i√3)/2<math> </math>r<math>, </math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let <math>x_1=r</math>, then <math>x_2=(-1+i√3)/2</math> r, <math>x_3=((-1+i√3)/2)^2</math> <math>r=(-1-i√3)/2</math> <math>r</math>, <math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let </math>x_1=r<math>, then </math>x_2=(-1+i√3)/2<math> r, </math>x_3=((-1+i√3)/2)^2<math> </math>r=(-1-i√3)/2<math> </math>r<math>, </math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let <math>x_1=r</math>, then <math>x_2=(-1+i√3)/2</math> r, <math>x_3=((-1+i√3)/2)^2</math> <math>r=(-1-i√3)/2</math> <math>r</math>, <math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let </math>x_1=r<math>, then </math>x_2=(-1+i√3)/2<math> r, </math>x_3=((-1+i√3)/2)^2<math> </math>r=(-1-i√3)/2<math> </math>r<math>, </math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let <math>x_1=r</math>, then <math>x_2=(-1+i√3)/2</math> r, <math>x_3=((-1+i√3)/2)^2</math> <math>r=(-1-i√3)/2</math> <math>r</math>, <math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let </math>x_1=r<math>, then </math>x_2=(-1+i√3)/2<math> r, </math>x_3=((-1+i√3)/2)^2<math> </math>r=(-1-i√3)/2<math> </math>r<math>, </math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let <math>x_1=r</math>, then <math>x_2=(-1+i√3)/2</math> r, <math>x_3=((-1+i√3)/2)^2</math> <math>r=(-1-i√3)/2</math> <math>r</math>, <math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let </math>x_1=r<math>, then </math>x_2=(-1+i√3)/2<math> r, </math>x_3=((-1+i√3)/2)^2<math> </math>r=(-1-i√3)/2<math> </math>r<math>, </math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let <math>x_1=r</math>, then <math>x_2=(-1+i√3)/2</math> r, <math>x_3=((-1+i√3)/2)^2</math> <math>r=(-1-i√3)/2</math> <math>r</math>, <math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let </math>x_1=r<math>, then </math>x_2=(-1+i√3)/2<math> r, </math>x_3=((-1+i√3)/2)^2<math> </math>r=(-1-i√3)/2<math> </math>r<math>, </math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let <math>x_1=r</math>, then <math>x_2=(-1+i√3)/2</math> r, <math>x_3=((-1+i√3)/2)^2</math> <math>r=(-1-i√3)/2</math> <math>r</math>, <math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let </math>x_1=r<math>, then </math>x_2=(-1+i√3)/2<math> r, </math>x_3=((-1+i√3)/2)^2<math> </math>r=(-1-i√3)/2<math> </math>r<math>, </math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let <math>x_1=r</math>, then <math>x_2=(-1+i√3)/2</math> r, <math>x_3=((-1+i√3)/2)^2</math> <math>r=(-1-i√3)/2</math> <math>r</math>, <math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let </math>x_1=r<math>, then </math>x_2=(-1+i√3)/2<math> r, </math>x_3=((-1+i√3)/2)^2<math> </math>r=(-1-i√3)/2<math> </math>r<math>, </math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let <math>x_1=r</math>, then <math>x_2=(-1+i√3)/2</math> r, <math>x_3=((-1+i√3)/2)^2</math> <math>r=(-1-i√3)/2</math> <math>r</math>, <math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let </math>x_1=r<math>, then </math>x_2=(-1+i√3)/2<math> r, </math>x_3=((-1+i√3)/2)^2<math> </math>r=(-1-i√3)/2<math> </math>r<math>, </math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let <math>x_1=r</math>, then <math>x_2=(-1+i√3)/2</math> r, <math>x_3=((-1+i√3)/2)^2</math> <math>r=(-1-i√3)/2</math> <math>r</math>, <math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let </math>x_1=r<math>, then </math>x_2=(-1+i√3)/2<math> r, </math>x_3=((-1+i√3)/2)^2<math> </math>r=(-1-i√3)/2<math> </math>r<math>, </math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let <math>x_1=r</math>, then <math>x_2=(-1+i√3)/2</math> r, <math>x_3=((-1+i√3)/2)^2</math> <math>r=(-1-i√3)/2</math> <math>r</math>, <math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let </math>x_1=r<math>, then </math>x_2=(-1+i√3)/2<math> r, </math>x_3=((-1+i√3)/2)^2<math> </math>r=(-1-i√3)/2<math> </math>r<math>, </math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let <math>x_1=r</math>, then <math>x_2=(-1+i√3)/2</math> r, <math>x_3=((-1+i√3)/2)^2</math> <math>r=(-1-i√3)/2</math> <math>r</math>, <math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let </math>x_1=r<math>, then </math>x_2=(-1+i√3)/2<math> r, </math>x_3=((-1+i√3)/2)^2<math> </math>r=(-1-i√3)/2<math> </math>r<math>, </math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let <math>x_1=r</math>, then <math>x_2=(-1+i√3)/2</math> r, <math>x_3=((-1+i√3)/2)^2</math> <math>r=(-1-i√3)/2</math> <math>r</math>, <math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let </math>x_1=r<math>, then </math>x_2=(-1+i√3)/2<math> r, </math>x_3=((-1+i√3)/2)^2<math> </math>r=(-1-i√3)/2<math> </math>r<math>, </math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let <math>x_1=r</math>, then <math>x_2=(-1+i√3)/2</math> r, <math>x_3=((-1+i√3)/2)^2</math> <math>r=(-1-i√3)/2</math> <math>r</math>, <math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let </math>x_1=r<math>, then </math>x_2=(-1+i√3)/2<math> r, </math>x_3=((-1+i√3)/2)^2<math> </math>r=(-1-i√3)/2<math> </math>r<math>, </math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let <math>x_1=r</math>, then <math>x_2=(-1+i√3)/2</math> r, <math>x_3=((-1+i√3)/2)^2</math> <math>r=(-1-i√3)/2</math> <math>r</math>, <math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let </math>x_1=r<math>, then </math>x_2=(-1+i√3)/2<math> r, </math>x_3=((-1+i√3)/2)^2<math> </math>r=(-1-i√3)/2<math> </math>r<math>, </math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let <math>x_1=r</math>, then <math>x_2=(-1+i√3)/2</math> r, <math>x_3=((-1+i√3)/2)^2</math> <math>r=(-1-i√3)/2</math> <math>r</math>, <math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let </math>x_1=r<math>, then </math>x_2=(-1+i√3)/2<math> r, </math>x_3=((-1+i√3)/2)^2<math> </math>r=(-1-i√3)/2<math> </math>r<math>, </math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let <math>x_1=r</math>, then <math>x_2=(-1+i√3)/2</math> r, <math>x_3=((-1+i√3)/2)^2</math> <math>r=(-1-i√3)/2</math> <math>r</math>, <math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let </math>x_1=r<math>, then </math>x_2=(-1+i√3)/2<math> r, </math>x_3=((-1+i√3)/2)^2<math> </math>r=(-1-i√3)/2<math> </math>r<math>, </math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let <math>x_1=r</math>, then <math>x_2=(-1+i√3)/2</math> r, <math>x_3=((-1+i√3)/2)^2</math> <math>r=(-1-i√3)/2</math> <math>r</math>, <math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let </math>x_1=r<math>, then </math>x_2=(-1+i√3)/2<math> r, </math>x_3=((-1+i√3)/2)^2<math> </math>r=(-1-i√3)/2<math> </math>r<math>, </math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let <math>x_1=r</math>, then <math>x_2=(-1+i√3)/2</math> r, <math>x_3=((-1+i√3)/2)^2</math> <math>r=(-1-i√3)/2</math> <math>r</math>, <math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let </math>x_1=r<math>, then </math>x_2=(-1+i√3)/2<math> r, </math>x_3=((-1+i√3)/2)^2<math> </math>r=(-1-i√3)/2<math> </math>r<math>, </math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let <math>x_1=r</math>, then <math>x_2=(-1+i√3)/2</math> r, <math>x_3=((-1+i√3)/2)^2</math> <math>r=(-1-i√3)/2</math> <math>r</math>, <math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let </math>x_1=r<math>, then </math>x_2=(-1+i√3)/2<math> r, </math>x_3=((-1+i√3)/2)^2<math> </math>r=(-1-i√3)/2<math> </math>r<math>, </math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let <math>x_1=r</math>, then <math>x_2=(-1+i√3)/2</math> r, <math>x_3=((-1+i√3)/2)^2</math> <math>r=(-1-i√3)/2</math> <math>r</math>, <math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let </math>x_1=r<math>, then </math>x_2=(-1+i√3)/2<math> r, </math>x_3=((-1+i√3)/2)^2<math> </math>r=(-1-i√3)/2<math> </math>r<math>, </math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let <math>x_1=r</math>, then <math>x_2=(-1+i√3)/2</math> r, <math>x_3=((-1+i√3)/2)^2</math> <math>r=(-1-i√3)/2</math> <math>r</math>, <math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let </math>x_1=r<math>, then </math>x_2=(-1+i√3)/2<math> r, </math>x_3=((-1+i√3)/2)^2<math> </math>r=(-1-i√3)/2<math> </math>r<math>, </math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let <math>x_1=r</math>, then <math>x_2=(-1+i√3)/2</math> r, <math>x_3=((-1+i√3)/2)^2</math> <math>r=(-1-i√3)/2</math> <math>r</math>, <math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let </math>x_1=r<math>, then </math>x_2=(-1+i√3)/2<math> r, </math>x_3=((-1+i√3)/2)^2<math> </math>r=(-1-i√3)/2<math> </math>r<math>, </math>x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
| − | ==Solution==
| |
| − | | |
| − | | |
| − | Let <math>x_1=r</math>, then <math>x_2=(-1+i√3)/2</math> r, <math>x_3=((-1+i√3)/2)^2</math> <math>r=(-1-i√3)/2</math> <math>r</math>, $x_4=((-1+i√3)/2)^3 <cmath></cmath>r which means x_4 will be back to x_1
| |
| − | Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.
| |
| − | The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
| |
| − | | |
| − | ~Yelong_Li
| |
, points 
and 
lie on 
and 
, respectively, so that 
Points 
and 
lie on 
and 
, respectively, and points 
and 
lie on 
so that 
and 
. See the figure below. Triangle 
, quadrilateral 
, quadrilateral 
, and pentagon 
each has area 
What is 
?
![[asy] real x=2sqrt(2); real y=2sqrt(16-8sqrt(2))-4+2sqrt(2); real z=2sqrt(8-4sqrt(2)); pair A, B, C, D, E, F, G, H, I, J; A = (0,0); B = (4,0); C = (4,4); D = (0,4); E = (x,0); F = (4,y); G = (y,4); H = (0,x); I = F + z * dir(225); J = G + z * dir(225); draw(A--B--C--D--A); draw(H--E); draw(J--G^^F--I); draw(rightanglemark(G, J, I), linewidth(.5)); draw(rightanglemark(F, I, E), linewidth(.5)); dot("$A$", A, S); dot("$B$", B, S); dot("$C$", C, dir(90)); dot("$D$", D, dir(90)); dot("$E$", E, S); dot("$F$", F, dir(0)); dot("$G$", G, N); dot("$H$", H, W); dot("$I$", I, SW); dot("$J$", J, SW); [/asy]](http://latex.artofproblemsolving.com/4/8/9/4892da26cb6334b359f58d4007f89460883779ab.png)
, the side length of square 
. We see that since triangle 
is a right isosceles triangle with area 1, we can determine sides 
and 
both to be 
. Now, consider extending 
and 
until they intersect. Let the point of intersection be 
. We note that 
is also a right isosceles triangle with side 
and find its area to be 
. Now, we notice that 
is also a right isosceles triangle (because 
) and find it's area to be 
or 
. Since we are looking for 
.~TLiu
. Denote by 
the point it intersects with 
, and by 
the point it intersects with 
. Last, denote by 
the segment 
, and by 
the segment 
. We will find two equations for 
.
, and 
. In addition, the area of 
.
Length of 
.
![$\left[\left(2\sqrt{2}-1\right)-y\right]\cdot \left[\left(2\sqrt{2}-1\right)+y\right]=1$](http://latex.artofproblemsolving.com/7/e/7/7e761f9f7638536a7b413b8109e9b9e59f259b63.png)
such that 
and 
are parallel and extend line 
such that 
forms a square. Extend line 
and point 
is the intersection of the two. The area of this square is equivalent to 
is 
, thus 
. Triangle 
is isosceles, and the area of this triangle is 
. Adding these two areas, we get ![\[2+6-4\sqrt{2}=8-4\sqrt{2}\rightarrow \boxed{\textbf{(B)}\ 8-4\sqrt{2}}\]](http://latex.artofproblemsolving.com/4/9/e/49ea066f508d90208431f7d1bbeecdcacc6d6283.png)
. --OGBooger
.
Notice that since the area of triangle 
, 
, therefore 
.
Let 
, then 
.
Also notice that 
, thus 
.
Now use the condition that the area of quadrilateral 
We solve the equation and yield 
.
Now notice that
.
Hence 
. -HarryW
![[asy] real x=2sqrt(2); real y=2sqrt(16-8sqrt(2))-4+2sqrt(2); real z=2sqrt(8-4sqrt(2)); real k= 8-2sqrt(2); real l= 2sqrt(2)-4; pair A, B, C, D, E, F, G, H, I, J, L, M, K; A = (0,0); B = (4,0); C = (4,4); D = (0,4); E = (x,0); F = (4,y); G = (y,4); H = (0,x); I = F + z * dir(225); J = G + z * dir(225); L = (k,0); M = F + z * dir(315); K = (4,l); draw(A--B--C--D--A); draw(H--E); draw(J--G^^F--I); draw(F--M); draw(M--L); draw(E--K,dashed+linewidth(.5)); draw(K--L,dashed+linewidth(.5)); draw(B--L); draw(rightanglemark(G, J, I), linewidth(.5)); draw(rightanglemark(F, I, E), linewidth(.5)); draw(rightanglemark(F, M, L), linewidth(.5)); fill((4,0)--(k,0)--M--(4,y)--cycle, gray); dot("$A$", A, S); dot("$C$", C, dir(90)); dot("$D$", D, dir(90)); dot("$E$", E, S); dot("$G$", G, N); dot("$H$", H, W); dot("$I$", I, SW); dot("$J$", J, SW); dot("$K$", K, S); dot("$F(G)$", F, E); dot("$J'$", M, dir(90)); dot("$H'$", L, S); dot("$B(D)$", B, S); [/asy]](http://latex.artofproblemsolving.com/c/8/9/c896117a516e4dace31abf9728d278c030cf305c.png)
Easily, we can find that: quadrilateral 
and 
overlapping) to form a square 
(
= 
= 
, 
= 
= 
so 
). Then we can solve 
= 
= 
= 
.
![[asy] real x=2sqrt(2); real y=2sqrt(16-8sqrt(2))-4+2sqrt(2); real z=2sqrt(8-4sqrt(2)); pair A, B, C, D, E, F, G, H, I, J, K, L; A = (0,0); B = (4,0); C = (4,4); D = (0,4); E = (x,0); F = (4,y); G = (y,4); H = (0,x); I = F + z * dir(225); J = G + z * dir(225); K = (4-x,4); L = J + 1.68 * dir(45); draw(A--B--C--D--A); draw(H--E); draw(J--G^^F--I); draw(H--K,dashed+linewidth(.5)); draw(L--K,dashed+linewidth(.5)); draw(rightanglemark(G, J, I), linewidth(.5)); draw(rightanglemark(F, I, E), linewidth(.5)); draw(rightanglemark(H, K, L), linewidth(.5)); draw(rightanglemark(K, L, G), linewidth(.5)); dot("$A$", A, S); dot("$B$", B, S); dot("$C$", C, dir(90)); dot("$D$", D, dir(90)); dot("$E$", E, S); dot("$F$", F, dir(0)); dot("$G$", G, N); dot("$H$", H, W); dot("$I$", I, SW); dot("$J$", J, SW); dot("$K$", K, N); dot("$L$", L, S); [/asy]](http://latex.artofproblemsolving.com/c/5/d/c5d426b3c348804400c5e5a587de7621c723e6e6.png)
![$[ABCD] = 4$](http://latex.artofproblemsolving.com/9/4/0/94000a8639c26a49b9ce21d64b94e5c6c21ed4fd.png)
, 
, ![$[AHE] = 1$](http://latex.artofproblemsolving.com/e/8/a/e8a9657a2d02ed6a858251d5146c2b69eec397b3.png)
, 
, 
, 
, 
. Because ![$[DHJG] = [BFIE]$](http://latex.artofproblemsolving.com/5/2/1/52122265c24304cafaaa466e0e158251164f95da.png)
, 
, 
![$[DHK] = \frac{(2 - \sqrt{2})^2}{2} = 3 - 2 \sqrt {2}$](http://latex.artofproblemsolving.com/6/1/a/61ae4066eb6334dcb14c4e939c1f3c92dd47403d.png)
![$[GKL] = \frac{(x - 2 \sqrt{2} + 2)^2}{2} = \frac{x^2}{2} + 2x - 2x \sqrt{2} - 4 \sqrt{2} + 6$](http://latex.artofproblemsolving.com/7/a/2/7a2d75d2f90ed17e193429ad7c14dad33d4842a9.png)
![$[HKJL] = (x - 2 \sqrt{2} + 2) \cdot (2 \sqrt{2} - 2) = 2x \sqrt{2} - 2x + 8 \sqrt{2} -12$](http://latex.artofproblemsolving.com/b/e/5/be55a6a5489f64cf7985c993bf44204e353b59a2.png)
![\[[DHK] + [GKL] + [HKLJ] = [DHJG]\]](http://latex.artofproblemsolving.com/6/f/3/6f38236b9d510343e38c190e758cddb0d3b97819.png)
![\[3 - 2 \sqrt {2} + \frac{x^2}{2} + 2x - 2x \sqrt{2} - 4 \sqrt{2} + 6 + 2x \sqrt{2} - 2x + 8 \sqrt{2} -12= 1\]](http://latex.artofproblemsolving.com/7/f/1/7f12a5c6ad17cfa08eadb87b61de6371d3cd7abd.png)
![\[\frac{x^2}{2} + 2 \sqrt{2} - 4 = 0\]](http://latex.artofproblemsolving.com/2/a/1/2a1ee4d7207c84132bb4e4be8c447220aeb35d74.png)
![\[x^2 = 8 - 4 \sqrt{2}\]](http://latex.artofproblemsolving.com/9/8/8/988fafef0df95057acce6c88588defd2a5310cb6.png)
![\[FI^2 = \boxed{\textbf{(B)}\ 8-4\sqrt{2}}\]](http://latex.artofproblemsolving.com/2/6/0/260ac550ebdcdb1dd5a721f2cb44bfe94cdca2fb.png)
. However, the height to side 
where 
. From here suppose that 
and let 
, which means 
. The area of the pentagon is then
![\[[FIJG]+[GCF]=GF \cdot FI + x^2 = (2x)(2\sqrt{2}-1-x)+x^2=1\]](http://latex.artofproblemsolving.com/2/9/4/2942ddcd36354b73a74fbe7432f81e216f42bede.png)
Solving this quadratic for 
(technically the smaller value is the correct one but it doesn’t matter for our purposes). We can then calculate 
.
and 
to meet at a point 
. Since 
and 
is parallel to 
, we know that 
, and because 
and 
, we can conclude that 
.
is isosceles, right, and has an area of 1, we can conclude that 
and that 
. Armed with this knowledge, and setting 
and the area of 
, we can use similarity to say that
![\[(\frac{a}{2-\sqrt{2}})^2 = \frac{1+b}{b}\]](http://latex.artofproblemsolving.com/1/2/f/12f59a02abfcaa3629c19053438fceeb4596c06e.png)
Since we know the side lengths of 
due to the fact that it is also an isosceles right triangle, we know that the area is 
.
Simplifying further and plugging in values, we have
![\[\frac{a^2}{(2-\sqrt{2})^2} = 1 + \frac{2}{(2-\sqrt{2})^2)}\]](http://latex.artofproblemsolving.com/f/0/8/f0888449562f33766ffb59efa7d3ea9020478598.png)
Multiplying by 
on both sides, we get
![\[a^2 = (2-\sqrt{2})^2 + 2 = \boxed{\textbf{(B)}\ 8-4\sqrt{2}}\]](http://latex.artofproblemsolving.com/1/f/a/1fa7652c0b220c50da611487afa03169bd1dcc70.png)
~yingkai_0_
