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

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==Problem==
 
==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.  
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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.
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==Hint==
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This problem involves areas.
  
 
==Solution==
 
==Solution==
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==See also==
 
==See also==
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{{IMO box|year=1984|num-b=3|num-a=5}}
  
 
[[Category:Olympiad Geometry Problems]]
 
[[Category:Olympiad Geometry Problems]]

Latest revision as of 23:49, 29 January 2021

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.

Hint

This problem involves areas.

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.

Solution 2

Let $M$ and $N$ be midpoints of $AB$ and $CD$, respectively. Let $[X]$ denote the area of figure $X$.

Suppose that $AD // BC$ and the circle centered at $M$ with diameter $AB$ touches $CD$ at $T$. We know that midsegment $MN // AD$. It follows that $[AMN] = [DMN]$, so $\frac{1}{2} AM \cdot d = \frac{1}{2} MT \cdot DN$, where $d$ is the distance from $N$ to $AB$. But we have $MT = MA$, so $DN = d$. It follows that the circle centered at $N$ with diameter $CD$ touches $AB$.

If on the other hand we have the circle with diameter $CD$ touching $AB$ at $U$(as well as the circle with diameter $AB$ touching $CD$ at $T$), we must have $\frac{1}{2} DN \cdot MT = \frac{1}{2} AM \cdot UN$ because of the equivalence of radii in circles. Hence $[ANM] = [DMN]$, so $A$ and $D$ are equidistant from $MN$ (as $MN$ is the common base). Hence $AB // MN$. Similarly $CD // MN$, and so $AB // CD$, as desired.

See also

1984 IMO (Problems) • Resources
Preceded by
Problem 3
1 2 3 4 5 6 Followed by
Problem 5
All IMO Problems and Solutions
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