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

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== Problem 30==
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== Problem ==
  
 
Let <math>BC</math> of right triangle <math>ABC</math> be the diameter of a circle intersecting hypotenuse <math>AB</math> in <math>D</math>.  
 
Let <math>BC</math> of right triangle <math>ABC</math> be the diameter of a circle intersecting hypotenuse <math>AB</math> in <math>D</math>.  
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\textbf{(C) }\ DF = FA \qquad  
 
\textbf{(C) }\ DF = FA \qquad  
 
\textbf{(D) }\ \angle A = \angle BCD \qquad  
 
\textbf{(D) }\ \angle A = \angle BCD \qquad  
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\textbf{(E) }\ \angle CFD = 2\angle A    </math>  
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\textbf{(E) }\ \angle CFD = 2\angle A    </math>
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== Solution 1 ==  
 
== Solution 1 ==  

Revision as of 09:06, 19 July 2024

Problem

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 $\fbox{B}$.

See Also

1965 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 29 		Followed by
Problem 31
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

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