Difference between revisions of "1982 AHSME Problems/Problem 24"
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=== Solution 1=== | === Solution 1=== | ||
| − | We | + | 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. |
What we know: | What we know: | ||
Revision as of 12:44, 11 October 2020
Solution 1
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 
