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>. | ||
− | 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>. | + | 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 08:58, 19 July 2024
Problem 30
Let of right triangle
be the diameter of a circle intersecting hypotenuse
in
.
At
a tangent is drawn cutting leg
in
. This information is not sufficient to prove that
Solution 1
We will prove every result except for .
By Thales' Theorem, and so
.
and
are both tangents to the same circle, and hence equal. Let
. Then
, and so
. We also have
, which implies
. This means that
, so
indeed bisects
. We also know that
, hence
. And
as
.
Since all of the results except for are true, our answer is
.
Solution 2
It's easy to verify that always equals
. Since
changes depending on the sidelengths of the triangle, we cannot be certain that
. Hence our answer is
.