Art of Problem Solving
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

1975 USAMO Problems/Problem 4: Difference between revisions

← Older edit
Suli (talk | contribs)
editor
445 edits
 
(4 intermediate revisions by 2 users not shown)
Line 12: Line 12:
</asy>
</asy>


by Vo Duc Dien


Let E and F be the centers of the small and big circles, respectively, and r and R be their respective radii.
==Solution==
A maximum <math>AP \cdot PB</math> cannot be attained if <math>AB</math> intersects segment <math>O_1O_2</math> because a larger value can be attained by making one of <math>A</math> or <math>B</math> diametrically opposite <math>P</math>, which (as is easily checked) increases the value of both <math>AP</math> and <math>PB</math>. Thus, assume <math>AB</math> does not intersect <math>O_1O_2</math>.


Let M and N be the feet of E and F to AB, and α = ∠APE and ε = ∠BPF
Let <math>E</math> and <math>F</math> be the centers of the small and big circles, respectively, and <math>r</math> and <math>R</math> be their respective radii.
 
Let <math>M</math> and <math>N</math> be the feet of <math>E</math> and <math>F</math> to <math>AB</math>, and <math>\alpha = \angle APE</math> and <math>\epsilon = \angle BPF</math>


We have:
We have:


AP × PB = 2r cosα × 2R cosε = 4 rR cosα cosε
<cmath>AP \times PB = 2r \cos{\alpha} \times 2R \cos{\epsilon} = 4 rR \cos{\alpha} \cos{\epsilon}</cmath>
 
<math>AP\times PB</math> is maximum when the product <math>\cos{\alpha} \cos{\epsilon}</math> is a maximum.
 
We have <math>\cos{\alpha} \cos{\epsilon}= \frac{1}{2} [\cos(\alpha +\epsilon) + \cos(\alpha -\epsilon)]</math>


AP × PB is maximum when the product cosα cosε is a maximum.
But <math>\alpha +\epsilon = 180^{\circ} - \angle EPF</math> and is fixed, so is <math>\cos(\alpha +\epsilon)</math>.


We have cosα cosε = ½ [cos(α + ε) + cos(α - ε)]
So its maximum depends on <math>cos(\alpha -\epsilon)</math> which occurs when <math>\alpha=\epsilon</math>. To draw the line <math>AB</math>:


But α + ε = 180° - ∠EPF and is fixed, so is cos(α + ε)
Draw a circle with center <math>P</math> and radius <math>PE</math> to cut the radius <math>PF</math> at <math>H</math>. Draw the line parallel to <math>EH</math> passing through <math>P</math>. This line meets the small and big circles at <math>A</math> and <math>B</math>, respectively.


So its maximum depends on cos(α - ε) which occurs when α = ε. To draw the line AB:


Draw a circle with center P and radius PE to cut the radius PF at H. Draw the line parallel to EH passing through P. This line meets the small and big circles at A and B, respectively.


Solution with graph posted at


http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.USA1975Problem4
{{alternate solutions}}


==See also==
==See Also==
[http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.USA1975Problem4 Solution with graph at Cut the Knot]


{{USAMO box|year=1975|num-b=3|num-a=5}}
{{USAMO box|year=1975|num-b=3|num-a=5}}
{{MAA Notice}}


[[Category:Olympiad Geometry Problems]]
[[Category:Olympiad Geometry Problems]]
[[Category:Olympiad Inequality Problems]]
[[Category:Olympiad Inequality Problems]]

Latest revision as of 14:13, 13 August 2014

Problem

Two given circles intersect in two points $P$ and $Q$. Show how to construct a segment $AB$ passing through $P$ and terminating on the two circles such that $AP\cdot PB$ is a maximum.

[asy] size(150); defaultpen(fontsize(7)); pair A=(0,0), B=(10,0), P=(4,0), Q=(3.7,-2.5); draw(A--B); draw(circumcircle(A,P,Q)); draw(circumcircle(B,P,Q)); label("A",A,(-1,1));label("P",P,(0,1.5));label("B",B,(1,1));label("Q",Q,(-0.5,-1.5)); [/asy]


Solution

A maximum $AP \cdot PB$ cannot be attained if $AB$ intersects segment $O_1O_2$ because a larger value can be attained by making one of $A$ or $B$ diametrically opposite $P$, which (as is easily checked) increases the value of both $AP$ and $PB$. Thus, assume $AB$ does not intersect $O_1O_2$.

Let $E$ and $F$ be the centers of the small and big circles, respectively, and $r$ and $R$ be their respective radii.

Let $M$ and $N$ be the feet of $E$ and $F$ to $AB$, and $\alpha = \angle APE$ and $\epsilon = \angle BPF$

We have:

\[AP \times PB = 2r \cos{\alpha} \times 2R \cos{\epsilon} = 4 rR \cos{\alpha} \cos{\epsilon}\]

$AP\times PB$ is maximum when the product $\cos{\alpha} \cos{\epsilon}$ is a maximum.

We have $\cos{\alpha} \cos{\epsilon}= \frac{1}{2} [\cos(\alpha +\epsilon) + \cos(\alpha -\epsilon)]$

But $\alpha +\epsilon = 180^{\circ} - \angle EPF$ and is fixed, so is $\cos(\alpha +\epsilon)$.

So its maximum depends on $cos(\alpha -\epsilon)$ which occurs when $\alpha=\epsilon$. To draw the line $AB$:

Draw a circle with center $P$ and radius $PE$ to cut the radius $PF$ at $H$. Draw the line parallel to $EH$ passing through $P$. This line meets the small and big circles at $A$ and $B$, respectively.



Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

See Also

Solution with graph at Cut the Knot

1975 USAMO (Problems • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America.

Retrieved from "https://wiki.aopstest.com/wiki/index.php?title=1975_USAMO_Problems/Problem_4&oldid=63043"