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Difference between revisions of "1976 AHSME Problems/Problem 16"

(Created page with "Let <math>AC=BC=DF=EF=x</math> and let the altitude from <math>F</math> to <math>\overline{DE}</math> have length <math>y</math>. This gives that <math>AB=2y</math>. Since <ma...")
 
(No difference)

Latest revision as of 17:04, 20 July 2022

Let $AC=BC=DF=EF=x$ and let the altitude from $F$ to $\overline{DE}$ have length $y$. This gives that $AB=2y$. Since $\triangle ABC$ and $\triangle DEF$ are isosceles, we have that $\angle ACB = 2\sin ^{-1} \left(\frac{y}{x} \right)$ and $\angle DFE = 2\cos ^{-1} \left(\frac{y}{x} \right)$. Since the sum of the inverse sine and cosine of any possible sine/cosine fraction is $90^\circ$, we have that $m \angle ACB + m \angle DFE = 180^\circ$. Thus, $\angle ACB$ and $\angle DFE$ are supplementary, so $\mathbf{II.}$ is true while $\mathbf{I.}$ is not.


Applying the Pythagorean Theorem gives that the length of the altitude from $C$ to $\overline{AB}$ is $\sqrt{x^2-y^2}$ and $DE = 2 \sqrt{x^2 - y^2}$. This means that $[ABC]=[DEF]=y \sqrt{x^2-y^2}.$ So we also have that $\mathbf{III.}$ is true while $\mathbf{IV.}$ is not.


Thus, our answer is $\textbf{(E) }\textbf{II. }\text{and }\textbf{III. }\text{only.}$

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