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Difference between revisions of "1982 AHSME Problems/Problem 24"

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<asy> defaultpen(fontsize(10)); real r=sqrt(22); pair B=origin, A=16*dir(60), C=(16,0), D=(10-r,0), E=(10+r,0), F=C+1*dir(120), G=C+14*dir(120), H=13*dir(60), J=6*dir(60), O=circumcenter(G,H,J); dot(A^^B^^C^^D^^E^^F^^G^^H^^J); draw(Circle(O, abs(O-D))^^A--B--C--cycle, linewidth(0.7)); label("$A$", A, N); label("$B$", B, dir(210)); label("$C$", C, dir(330)); label("$D$", D, SW); label("$E$", E, SE); label("$F$", F, dir(170)); label("$G$", G, dir(250)); label("$H$", H, SE); label("$J$", J, dir(0)); label("2", A--G, dir(30)); label("13", F--G, dir(180+30)); label("1", F--C, dir(30)); label("7", H--J, dir(-30));</asy>
 
<asy> defaultpen(fontsize(10)); real r=sqrt(22); pair B=origin, A=16*dir(60), C=(16,0), D=(10-r,0), E=(10+r,0), F=C+1*dir(120), G=C+14*dir(120), H=13*dir(60), J=6*dir(60), O=circumcenter(G,H,J); dot(A^^B^^C^^D^^E^^F^^G^^H^^J); draw(Circle(O, abs(O-D))^^A--B--C--cycle, linewidth(0.7)); label("$A$", A, N); label("$B$", B, dir(210)); label("$C$", C, dir(330)); label("$D$", D, SW); label("$E$", E, SE); label("$F$", F, dir(170)); label("$G$", G, dir(250)); label("$H$", H, SE); label("$J$", J, dir(0)); label("2", A--G, dir(30)); label("13", F--G, dir(180+30)); label("1", F--C, dir(30)); label("7", H--J, dir(-30));</asy>
  
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We know that because <math>AG=2, GF=13,</math> and <math>FC=1</math> the side lengths of the triangle are 16 each. And because <math>2+13+1=16,</math> the equilateral triangle will have side length <math>16</math>. Harnessing the power of point <math>C</math>, we have <math>(CF)(CG)=(CE)(CD).</math> We also have on point <math>A</math>: <math>(AH)(AJ)=(AG)(AF).</math> We will simplify this, by inputting values we already know.
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Let <math>AH=y, BD=a, DE=x,</math> and <math>EC=b.</math> Because <math>AG = 2, GF = 13, HJ = 7 and FC = 1</math> and they sum to 16, the length of each side of the equilateral triangle will be 16. By Power of a Point on A, we have <math>y(y + 7) = 2(2 + 13),</math> or expanding and factoring, <math>0 =y^2 + 7y - 30 = (y - 3)(y + 10).</math> Therefore, y must equal 3 as this triangle must have positive side lengths.
  
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What we know:
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Hence, <math>BJ=16-(3+7)=6.</math>
  
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Side length <math>16</math>
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Power of a point on C yields <math>b(b + x) = l(1 + 13) = 14,</math> and power of a point on B yields a(a + x) = 6(6 + 7) = 78. Also, we know that because a, b, and x are non-intersecting segments of the side of the triangle, a + b + x = 16.
  
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<math>AG=2, GF=13,</math> and <math>FC=1</math>
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So, we have a system of equations:
  
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Now we know what <math>CF</math> is, <math>1.</math> We also can derive <math>CG,</math> as it equals <math>CF+FG=13+1=14.</math>  
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<cmath>a^2 + ax = 78,</cmath>
 +
<cmath>b^2 + bx = 14,</cmath>
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<math>a + b + x =16.</math>
  
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<math>(CF)(CG)=(CE)(CD),</math> so by the prior calculations, <math>(1)(13)=(CE)(CB).</math> Therefore <math>13=(CE)(CD).</math>  
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If we subtract the second equation from the first, it appears to be a difference of squares! And we like differences of squares, because we can factor! Subtracting yields <math>a^2 - b^2 + (a - b)x = (a-b)(a+b)(a-b)x=(a - b)(a + b + x).</math> Because we know that <math>a+b+x=16,</math> inputting this we have that <math>(a - b)16 = 78 - 14 = 64.</math> Therefore <math>a-b=4.</math> Then adding this to the third equation, 2a + x = 20, so <math>a = 10 - (x/2).</math> Substituting this into #1, we can now solve for x, and we have a difference of squares, or <math>100-x^2/4=78.</math> This yields <math>x^2/4=22,</math> so <math>x/2=\sqrt {22}</math> and <math>x= \left(A\right)2\sqrt{22}!</math>
  
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We can update our chart:
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Further clarifications: If you are unsure why we didn't solve for a and b, it is because it is more efficient to use clever manipulations like the ones we used in this problem. We were looking for sums of values we already know, so we indirectly implied dividing <math>a^2 - b^2 + (a - b)x</math> by <math>a+b+x</math> to get <math>a-b,</math> conveniently, then everything fell into place with substitution.
  
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What we know:
 
  
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Side length 16
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~*Solution by ab2024
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<math>AG=2, GF=13,</math> and <math>FC=1</math>
 
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<math>13=(CE)(CD).</math>
 
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<math>CG=14</math>
 
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<math>(AH)(AJ)=(AG)(AF),</math> and it's pretty obvious that <math>AG=2.</math> We can also derive <math>AF, 2+GF=15.</math> Therefore <math>(AH)(AJ)=15.</math> However, we know the value of <math>HJ!</math> Because <math>HJ=AJ-AH,</math> we have <math>AJ=7+AH.</math> Therefore <math>(AH)(7+AH)=15.</math> We can distribute: <math>AH^2+7AH-15=0,</math> and <math>AH = \sqrt{109}/2 - 7/2.</math> But, we know that the side lengths must sum to <math>16,</math> so we have <math>BJ=16-(AH+HJ)=25/2 - \sqrt{109}/2.</math> We can now update our 'What we Know'' section.
 
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What we know:
 
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Side length 16
 
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<math>AG=2, GF=13,</math> and <math>FC=1</math>
 
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<math>13=(CE)(CD).</math>
 
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<math>CG=14</math>
 
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<math>AH = \sqrt{109}/2 - 7/2.</math>
 
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<math>BJ=25/2 - \sqrt{109}/2.</math>
 
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Taking the power of point B, we have <math>(BD)(BE)=(BJ)(BH,)</math> and because we already know <math>AH</math>, we know <math>BH=16-AH=39/2 - \sqrt{109}/2.</math> Now we shall multiply <math>BH</math> and <math>BJ</math> to get <math>(25/2 - \sqrt{109}/2) \cdot (39/2 - \sqrt{109}/2)= 271 - 16 \sqrt{109}.</math>
 
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Therefore <math>BD(BE)=271 - 16 \sqrt{109}.</math>
 
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Updating our chart,
 
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What we know:
 
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Side length 16
 
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<math>AG=2, GF=13,</math> and <math>FC=1</math>
 
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<math>13=(CE)(CD).</math>
 
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<math>CG=14</math>
 
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<math>AH = \sqrt{109}/2 - 7/2.</math>
 
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<math>BJ=25/2 - \sqrt{109}/2.</math>
 
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<math>BD(BE)=271 - 16 \sqrt{109}.</math>
 
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<math>HJ=7</math>
 
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<math>BH=39/2 - \sqrt{109}/2.</math>
 
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<math>CE=16-BE</math>
 
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<math>CD=16-BD</math>
 
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We can now substitute and have 2 equations and two variables.<math>CE=16-BE</math> and <math>CD=16-BD</math> are especially useful. Because <math>13=(CE)(CD)=(16-BE)(16-BD),</math> and <math>BD(BE)=271 - 16 \sqrt{109},</math> we have <math>BD= 15, 3, 29,</math> or <math>17,</math> and <math>BE=15, 3, 29, or 17.</math> However, because <math>BD<BE,</math> we have <math>BD=3</math> or <math>17</math> and <math>BE= 15</math> or <math>29.</math> Because the differences <math>29-17</math> and <math>15-3</math> are constantly <math>12,</math> and <math>BD-BE=DE, DE=12!</math>
 

Revision as of 16:54, 11 October 2020

Solution 1

[asy] defaultpen(fontsize(10)); real r=sqrt(22); pair B=origin, A=16*dir(60), C=(16,0), D=(10-r,0), E=(10+r,0), F=C+1*dir(120), G=C+14*dir(120), H=13*dir(60), J=6*dir(60), O=circumcenter(G,H,J); dot(A^^B^^C^^D^^E^^F^^G^^H^^J); draw(Circle(O, abs(O-D))^^A--B--C--cycle, linewidth(0.7)); label("$A$", A, N); label("$B$", B, dir(210)); label("$C$", C, dir(330)); label("$D$", D, SW); label("$E$", E, SE); label("$F$", F, dir(170)); label("$G$", G, dir(250)); label("$H$", H, SE); label("$J$", J, dir(0)); label("2", A--G, dir(30)); label("13", F--G, dir(180+30)); label("1", F--C, dir(30)); label("7", H--J, dir(-30));[/asy]

Let $AH=y, BD=a, DE=x,$ and $EC=b.$ Because $AG = 2, GF = 13, HJ = 7 and FC = 1$ and they sum to 16, the length of each side of the equilateral triangle will be 16. By Power of a Point on A, we have $y(y + 7) = 2(2 + 13),$ or expanding and factoring, $0 =y^2 + 7y - 30 = (y - 3)(y + 10).$ Therefore, y must equal 3 as this triangle must have positive side lengths.

Hence, $BJ=16-(3+7)=6.$

Power of a point on C yields $b(b + x) = l(1 + 13) = 14,$ and power of a point on B yields a(a + x) = 6(6 + 7) = 78. Also, we know that because a, b, and x are non-intersecting segments of the side of the triangle, a + b + x = 16.

So, we have a system of equations:

\[a^2 + ax = 78,\] \[b^2 + bx = 14,\] $a + b + x =16.$

If we subtract the second equation from the first, it appears to be a difference of squares! And we like differences of squares, because we can factor! Subtracting yields $a^2 - b^2 + (a - b)x = (a-b)(a+b)(a-b)x=(a - b)(a + b + x).$ Because we know that $a+b+x=16,$ inputting this we have that $(a - b)16 = 78 - 14 = 64.$ Therefore $a-b=4.$ Then adding this to the third equation, 2a + x = 20, so $a = 10 - (x/2).$ Substituting this into #1, we can now solve for x, and we have a difference of squares, or $100-x^2/4=78.$ This yields $x^2/4=22,$ so $x/2=\sqrt {22}$ and $x= \left(A\right)2\sqrt{22}!$

Further clarifications: If you are unsure why we didn't solve for a and b, it is because it is more efficient to use clever manipulations like the ones we used in this problem. We were looking for sums of values we already know, so we indirectly implied dividing $a^2 - b^2 + (a - b)x$ by $a+b+x$ to get $a-b,$ conveniently, then everything fell into place with substitution.


~*Solution by ab2024

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