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

(Solution to Problem 28)
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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>.
 
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>.
 +
 +
== Solution 2 (Apollonius Theorem) ==
 +
Let <math>C_1</math> be the circle with a centra <math>O</math> and diameter <math>AB</math>. Let <math>P</math> be any random point inside <math>C_1</math>. Connect <math>OP</math>.
 +
We want <math>AP^2 + BP^2 = 3</math>
 +
In <math>\triangle APB, PO</math> is a median on <math>AB</math>. Using Appolonius Theorem we get:
 +
<cmath> AP^2 + BP^2 = 2(AO^2 + OP^2)</cmath>
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<cmath> 2(OP^2 + 1) = 3</cmath>
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<cmath> 2\cdot OP^2 + 2 = 3</cmath>
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<cmath> OP = \frac{1}{\sqrt{2}} </cmath>
 +
 +
We draw a circle <math>C_2</math> centred at <math>O</math> with a radius of <math>\frac{1}{\sqrt{2}}</math> i.e, <math>OP</math>.
 +
Every point on the circle <math>C_2</math> will satisfy the given conditions. Since there are infinite points on a circle hence there are infinite number of points P that can satisfy the condition.
 +
 +
Hence, the answer is <math>\boxed{\textbf{(E)}}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 05:20, 23 August 2025

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 $\boxed{\textbf{(E)}}$.

Solution 2 (Apollonius Theorem)

Let $C_1$ be the circle with a centra $O$ and diameter $AB$. Let $P$ be any random point inside $C_1$. Connect $OP$. We want $AP^2 + BP^2 = 3$ In $\triangle APB, PO$ is a median on $AB$. Using Appolonius Theorem we get: \[AP^2 + BP^2 = 2(AO^2 + OP^2)\] \[2(OP^2 + 1) = 3\] \[2\cdot OP^2 + 2 = 3\] \[OP = \frac{1}{\sqrt{2}}\]

We draw a circle $C_2$ centred at $O$ with a radius of $\frac{1}{\sqrt{2}}$ i.e, $OP$. Every point on the circle $C_2$ will satisfy the given conditions. Since there are infinite points on a circle hence there are infinite number of points P that can satisfy the condition.

Hence, 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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