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Difference between revisions of "1969 AHSME Problems/Problem 28"

(Created page with "== Problem == Let <math>n</math> be the number of points <math>P</math> interior to the region bounded by a circle with radius <math>1</math>, such that the sum of squares of th...")
 
(Solution to Problem 28)
 
Line 10: Line 10:
  
 
== Solution ==
 
== Solution ==
−
<math>\fbox{E}</math>
+
 
 +
<asy>
 +
 
 +
draw(circle((0,0),50));
 +
draw((-50,0)--(-10,20)--(50,0)--(-50,0));
 +
draw((-10,20)--(30,40)--(50,0),dotted);
 +
dot((-50,0));
 +
label("$A$",(-50,0),W);
 +
dot((50,0));
 +
label("$B$",(50,0),E);
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dot((-10,20));
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label("$P$",(-10,20),S);
 +
dot((30,40));
 +
label("$C$",(30,40),NE);
 +
 
 +
 
 +
</asy>
 +
 
 +
Let <math>A</math> and <math>B</math> be points on diameter.  Extend <math>AP</math>, and mark intersection with circle as point <math>C</math>.
 +
 
 +
Because <math>AB</math> is a diameter, <math>\angle ACB = 90^\circ</math>.  Also, by Exterior Angle Theorem, <math>\angle ACB + \angle CBP = \angle APB</math>, so <math>\angle APB > \angle ACB</math>, making <math>\angle APB</math> an obtuse angle.
 +
 
 +
By the [[Law of Cosines]], <math>AP^2 + BP^2 - 2 \cdot AP \cdot BP \cdot \cos{\angle APB} = 4</math>.  Since <math>AP^2 + BP^2 = 3</math>, substitute and simplify to get <math>\cos{\angle APB} = -\frac{1}{2 \cdot AP \cdot BP}</math>.  This equation has infinite solutions because for every <math>AP</math> and <math>BP</math>, where <math>AP + BP \ge 2</math> and <math>AP</math> and <math>BP</math> are both less than <math>2</math>, there can be an obtuse angle that satisfies the equation, so the answer is <math>\boxed{\textbf{(E)}}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 18:57, 9 June 2018

Problem

Let $n$ be the number of points $P$ interior to the region bounded by a circle with radius $1$, such that the sum of squares of the distances from $P$ to the endpoints of a given diameter is $3$. Then $n$ is:

$\text{(A) } 0\quad \text{(B) } 1\quad \text{(C) } 2\quad \text{(D) } 4\quad \text{(E) } \infty$

Solution

[asy] draw(circle((0,0),50)); draw((-50,0)--(-10,20)--(50,0)--(-50,0)); draw((-10,20)--(30,40)--(50,0),dotted); dot((-50,0)); label("$A$",(-50,0),W); dot((50,0)); label("$B$",(50,0),E); dot((-10,20)); label("$P$",(-10,20),S); dot((30,40)); label("$C$",(30,40),NE); [/asy]

Let $A$ and $B$ be points on diameter. Extend $AP$, and mark intersection with circle as point $C$.

Because $AB$ is a diameter, $\angle ACB = 90^\circ$. Also, by Exterior Angle Theorem, $\angle ACB + \angle CBP = \angle APB$, so $\angle APB > \angle ACB$, making $\angle APB$ an obtuse angle.

By the Law of Cosines, $AP^2 + BP^2 - 2 \cdot AP \cdot BP \cdot \cos{\angle APB} = 4$. Since $AP^2 + BP^2 = 3$, substitute and simplify to get $\cos{\angle APB} = -\frac{1}{2 \cdot AP \cdot BP}$. This equation has infinite solutions because for every $AP$ and $BP$, where $AP + BP \ge 2$ and $AP$ and $BP$ are both less than $2$, there can be an obtuse angle that satisfies the equation, so the answer is $\boxed{\textbf{(E)}}$.

See also

1969 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 27 		Followed by
Problem 29
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
All AHSME Problems and Solutions

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