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

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== Solution 1 ==  
 
== Solution 1 ==  
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<asy>
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path circ, hyp;
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draw((0,16)--(0,0)--(8,0)--(0,16));
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dot((0,16));
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label("A", (0,16), NW);
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dot((0,0));
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label("C", (0,0), SW);
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dot((8,0));
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label("B", (8,0), SE);
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dot((0,8));
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label("F",(0,8),W);
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markscalefactor=0.1;
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draw(rightanglemark((0,16),(0,0),(8,0)));
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circ=circle((4,0),4);
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draw(circ);
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hyp=(0,16)--(8,0);
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pair [] x=intersectionpoints(circ,hyp);
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dot(x[1]);
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label("D",x[1],NE);
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draw(x[1]--(0,8));
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draw(x[1]--(0,0));
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</asy>
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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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{{AHSME 40p box|year=1965|num-b=29|num-a=31}}
 
{{AHSME 40p box|year=1965|num-b=29|num-a=31}}
 
{{MAA Notice}}
 
{{MAA Notice}}
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[[Category:Introductory Geometry Problems]]

Latest revision as of 08:29, 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

[asy] path circ, hyp; draw((0,16)--(0,0)--(8,0)--(0,16)); dot((0,16)); label("A", (0,16), NW); dot((0,0)); label("C", (0,0), SW); dot((8,0)); label("B", (8,0), SE); dot((0,8)); label("F",(0,8),W); markscalefactor=0.1; draw(rightanglemark((0,16),(0,0),(8,0))); circ=circle((4,0),4); draw(circ); hyp=(0,16)--(8,0); pair [] x=intersectionpoints(circ,hyp); dot(x[1]); label("D",x[1],NE); draw(x[1]--(0,8)); draw(x[1]--(0,0)); [/asy]

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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