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

Difference between revisions of "1984 IMO Problems/Problem 4"

(page/ problem/ solution)
 
(Wikify)
Line 1: Line 1:
 +
==Problem==
 
Let <math>ABCD</math> be a convex quadrilateral with the line <math>CD</math> being tangent to the circle on diameter <math>AB</math>. Prove that the line <math>AB</math> is tangent to the circle on diameter <math>CD</math> if and only if the lines <math>BC</math> and <math>AD</math> are parallel.  
 
Let <math>ABCD</math> be a convex quadrilateral with the line <math>CD</math> being tangent to the circle on diameter <math>AB</math>. Prove that the line <math>AB</math> is tangent to the circle on diameter <math>CD</math> if and only if the lines <math>BC</math> and <math>AD</math> are parallel.  
  
 +
==Solution==
 
First, we prove that if <math>BC</math> and <math>AD</math> are parallel then the claim is true:  Let <math>AB</math> and <math>CD</math> intersect at <math>E</math> (assume <math>E</math> is closer to <math>AD</math>, the other case being analogous).  Let <math>M,N</math> be the midpoints of <math>AB,CD</math> respectively.  Let the length of the perpendicular from <math>N</math> to <math>AB</math> be <math>r</math>.  It is known that the length of the perpendicular from <math>M</math> to <math>CD</math> is <math>\frac{1}{2}AB</math>. Let the foot of the perpendicular from <math>C</math> to <math>AB</math> be <math>H</math>, and similarly define <math>G</math> for side <math>CD</math>.  Then, since triangles <math>MNE</math> and <math>BCE</math> are similar, we have <math>\frac{CH}{r}=\frac{BG}{\frac{1}{2}AB}</math>.  This gives an expression for <math>r</math>:
 
First, we prove that if <math>BC</math> and <math>AD</math> are parallel then the claim is true:  Let <math>AB</math> and <math>CD</math> intersect at <math>E</math> (assume <math>E</math> is closer to <math>AD</math>, the other case being analogous).  Let <math>M,N</math> be the midpoints of <math>AB,CD</math> respectively.  Let the length of the perpendicular from <math>N</math> to <math>AB</math> be <math>r</math>.  It is known that the length of the perpendicular from <math>M</math> to <math>CD</math> is <math>\frac{1}{2}AB</math>. Let the foot of the perpendicular from <math>C</math> to <math>AB</math> be <math>H</math>, and similarly define <math>G</math> for side <math>CD</math>.  Then, since triangles <math>MNE</math> and <math>BCE</math> are similar, we have <math>\frac{CH}{r}=\frac{BG}{\frac{1}{2}AB}</math>.  This gives an expression for <math>r</math>:
  
Line 28: Line 30:
  
 
Since <math>\frac{1}{2}CD\ne EN</math>, we have <math>NT=0</math> as desired.
 
Since <math>\frac{1}{2}CD\ne EN</math>, we have <math>NT=0</math> as desired.
 +
 +
==See also==
 +
 +
[[Category:Olympiad Geometry Problems]]

Revision as of 09:26, 25 August 2008

Problem

Let $ABCD$ be a convex quadrilateral with the line $CD$ being tangent to the circle on diameter $AB$. Prove that the line $AB$ is tangent to the circle on diameter $CD$ if and only if the lines $BC$ and $AD$ are parallel.

Solution

First, we prove that if $BC$ and $AD$ are parallel then the claim is true: Let $AB$ and $CD$ intersect at $E$ (assume $E$ is closer to $AD$, the other case being analogous). Let $M,N$ be the midpoints of $AB,CD$ respectively. Let the length of the perpendicular from $N$ to $AB$ be $r$. It is known that the length of the perpendicular from $M$ to $CD$ is $\frac{1}{2}AB$. Let the foot of the perpendicular from $C$ to $AB$ be $H$, and similarly define $G$ for side $CD$. Then, since triangles $MNE$ and $BCE$ are similar, we have $\frac{CH}{r}=\frac{BG}{\frac{1}{2}AB}$. This gives an expression for $r$:

$r=\frac{1}{2}AB\cdot\frac{CH}{BG}$

Noticing that $CH=BC\sin EBC,BG=BC\sin ECB$ simplifies the expression to

$r=\frac{1}{2}AB\cdot\frac{\sin EBC}{\sin ECB}$

By the Law of Sines, $\frac{\sin EBC}{\sin ECB}=\frac{EC}{EB}$. Since triangles $EDA,ECB$ are similar, we have $\frac{EC}{EB}=\frac{CD}{AB}$ and thus we have

$r=\frac{1}{2}AB\cdot\frac{CD}{AB}=\frac{1}{2}CD$

and we are done.

Now to prove the converse. Suppose we have the quadrilateral with $BC$ parallel to $AD$, and with all conditions satisfied. We shall prove that there exists no point $T$ on $CD$ such that $T$ is a midpoint of a side $CD^\prime$ of a quadrilateral $ABCD^\prime$ which also satisfies the condition. Suppose there was such a $T$. Like before, define the points $E,M,N$ for quadrilateral $ABCD$. Let $t$ be the length of the perpendicular from $T$ to $AB$. Then, using similar triangles, $\frac{ET}{t}=\frac{EN}{\frac{1}{2}CD}$. This gives

$t=\frac{\frac{1}{2}CD\cdot ET}{EN}$

But, we must have $t=DT$. Thus, we have

$DT=\frac{\frac{1}{2}CD\cdot ET}{EN}$

$\Rightarrow \frac{1}{2}CD\cdot EN+NT\cdot EN=\frac{1}{2}CD\cdot (EN+NT)$

$\Rightarrow NT\cdot EN=\frac{1}{2}CD\cdot NT$

Since $\frac{1}{2}CD\ne EN$, we have $NT=0$ as desired.

See also

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1984_IMO_Problems/Problem_4&oldid=27572"