Difference between revisions of "1982 AHSME Problems/Problem 24"
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=== Solution 1=== | === Solution 1=== | ||
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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> | ||
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. | 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. | ||
Revision as of 14:25, 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]](http://latex.artofproblemsolving.com/7/b/9/7b9dc2915732e0d3d54008ee8a47e6b011655ed3.png)
We know that because 
and 
the side lengths of the triangle are 16 each. And because 
the equilateral triangle will have side length 
. Harnessing the power of point 
, we have 
We also have on point 
: 
We will simplify this, by inputting values we already know.
What we know:
Side length
and
Now we know what 
is, 
We also can derive 
as it equals 
so by the prior calculations, 
Therefore 
We can update our chart:
What we know:
Side length 16
and
and it's pretty obvious that 
We can also derive 
Therefore 
However, we know the value of 
Because 
we have 
Therefore 
We can distribute: 
and 
But, we know that the side lengths must sum to 
so we have 
We can now update our 'What we Know section.
What we know:
Side length 16
and
Taking the power of point B, we have 
and because we already know 
, we know 
Now we shall multiply 
and 
to get 
Therefore 
Updating our chart,
What we know:
Side length 16
and
We can now substitute and have 2 equations and two variables. and are especially useful. Because 
and 
we have 
or 
and 
However, because 
we have 
or 
and 
or 
Because the differences 
and 
are constantly 
and 
