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Difference between revisions of "1965 AHSME Problems/Problem 30"

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We will prove every result except for <math>\fbox{B}</math>.  
 
We will prove every result except for <math>\fbox{B}</math>.  
  
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By Thales' Theorem, <math>\angle CDB=90^\circ </math> and so <math>\angle CDA= 90^\circ </math>. <math>FC</math> and <math>FD</math> are both tangents to the same circle, and hence equal. Let <math>\angle CFD=\alpha</math>. Then <math>\angle FDC = \frac{180^\circ - \alpha}{2}</math>, and so <math>\angle FDA = \frac{\alpha}{2}</math>. We also have <math>\angle AFD = 180^\circ - \alpha</math>, which implies <math>\angle FAD=\frac{\alpha}{2}</math>. This means that <math>CF=DF=FA</math>, so <math>DF</math> indeed bisects <math>CA</math>. We also know that <math>\angle BCD=90-\frac{180^\circ - \alpha}{2}=\frac{\alpha}{2}</math>, hence <math>\angle A = \angle BCD</math>. And <math>\angle CFD=2\angle A</math> as <math>\alpha = \frac{\alpha}{2}\times 2</math>.  
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By [[Thales' Theorem]], <math>\angle CDB=90^\circ </math> and so <math>\angle CDA= 90^\circ </math>. <math>FC</math> and <math>FD</math> are both tangents to the same circle, and hence equal. Let <math>\angle CFD=\alpha</math>. Then <math>\angle FDC = \frac{180^\circ - \alpha}{2}</math>, and so <math>\angle FDA = \frac{\alpha}{2}</math>. We also have <math>\angle AFD = 180^\circ - \alpha</math>, which implies <math>\angle FAD=\frac{\alpha}{2}</math>. This means that <math>CF=DF=FA</math>, so <math>DF</math> indeed bisects <math>CA</math>. We also know that <math>\angle BCD=90-\frac{180^\circ - \alpha}{2}=\frac{\alpha}{2}</math>, hence <math>\angle A = \angle BCD</math>. And <math>\angle CFD=2\angle A</math> as <math>\alpha = \frac{\alpha}{2}\times 2</math>.  
  
 
Since all of the results except for <math>B</math> are true, our answer is <math>\fbox{B}</math>.
 
Since all of the results except for <math>B</math> are true, our answer is <math>\fbox{B}</math>.

Revision as of 07:58, 19 July 2024

Problem 30

Let $BC$ of right triangle $ABC$ be the diameter of a circle intersecting hypotenuse $AB$ in $D$. At $D$ a tangent is drawn cutting leg $CA$ in $F$. This information is not sufficient to prove that

$\textbf{(A)}\ DF \text{ bisects }CA \qquad \textbf{(B) }\ DF \text{ bisects }\angle CDA \\ \textbf{(C) }\ DF = FA \qquad \textbf{(D) }\ \angle A = \angle BCD \qquad \textbf{(E) }\ \angle CFD = 2\angle A$


Solution 1

We will prove every result except for $\fbox{B}$.

By Thales' Theorem, $\angle CDB=90^\circ$ and so $\angle CDA= 90^\circ$. $FC$ and $FD$ are both tangents to the same circle, and hence equal. Let $\angle CFD=\alpha$. Then $\angle FDC = \frac{180^\circ - \alpha}{2}$, and so $\angle FDA = \frac{\alpha}{2}$. We also have $\angle AFD = 180^\circ - \alpha$, which implies $\angle FAD=\frac{\alpha}{2}$. This means that $CF=DF=FA$, so $DF$ indeed bisects $CA$. We also know that $\angle BCD=90-\frac{180^\circ - \alpha}{2}=\frac{\alpha}{2}$, hence $\angle A = \angle BCD$. And $\angle CFD=2\angle A$ as $\alpha = \frac{\alpha}{2}\times 2$.

Since all of the results except for $B$ are true, our answer is $\fbox{B}$.

Solution 2

It's easy to verify that $\angle CDA$ always equals $90^\circ$. Since $\angle CDF$ changes depending on the sidelengths of the triangle, we cannot be certain that $\angle CDF=45^\circ$. Hence our answer is $\fbox{B}$.

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