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 11:44, 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