# 1989 IMO Problems/Problem 4

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Let $ABCD$ be a convex quadrilateral such that the sides $AB,AD,BC$ satisfy $AB=AD+BC$. There exists a point $P$ inside the quadrilateral at a distance $h$ from the line $CD$ such that $AP=h+AD$ and $BP=h+BC$. Show that:

$\frac{1}{\sqrt{h}}\ge\frac{1}{\sqrt{AD}}+\frac{1}{\sqrt{BC}}$

## Solution

Without loss of generality, assume $AD\ge BC$.

Draw the circle with center $A$ and radius $AD$ as well as the circle with center $B$ and radius $BC$. Since $AB=AD+BC$, the intersection of circle $A$ with the line $AB$ is the same as the intersection of circle $B$ with the line $AB$. Thus, the two circles are externally tangent to each other.

Now draw the circle with center $P$ and radius $h$. Since $h$ is the shortest distance from $P$ to $CD$, we have that the circle $P$ is tangent to $CD$. From the conditions $AP=h+AD$ and $BP=h+BC$, by the same reasoning that we found circles $A,B$ are tangent, we find that circles $P,A$ and $P,B$ are externally tangent also.

Fix the two externally tangent circles with centers $A$ and $B$ and let the points $C$ and $D$ vary about their respective circles (without loss of generality, assume that one can read the letters $ABCD$ in counter clockwise order). Notice that the value $\frac{1}{\sqrt{AD}}+\frac{1}{\sqrt{BC}}$ is fixed. Thus, in order to prove the claim presented in the problem, we must maximize the value of $h$. It is clear that all possible circles $P$ are contained within the region created by circles $A,B$ as well as their common tangent line. Now it is clear that $h$ is maximized when circle $P$ is tangent to the common tangent of circles $A,B$ as well as to the circles $A,B$. We shall prove that the value of $h$ at this point is $\frac{1}{\sqrt{AD}}+\frac{1}{\sqrt{BC}}$.

Fix $C,D$ so that $CD$ is the common tangent to circles $A,B$. Since $\angle ADC,\angle BCD$ are right, the length $CD$ is equal to the length of the segment from $B$ to the projection of $B$ onto $AD$ which by the Pythagorean Theorem is $\sqrt{(BC+AD)^2-(AD-BC)^2}=2\sqrt{BC\cdot AD}$. Let the projection of $P$ onto $CD$ be $x$ units away from $D$. Then, as we found the length of $CD$, we find

$x^2+(AD-h)^2=(AD+h)^2\Rightarrow x^2=4AD\cdot h$

$(2\sqrt{BC\cdot AD}-x)^2+(BC-h)^2=(BC+h)^2\Rightarrow (2\sqrt{BC\cdot AD}-x)^2=4BC\cdot h$

Subtracting the second equation from the first yields

$4x\sqrt{BC\cdot AD}-4BC\cdot AD=4AD\cdot h-4BC\cdot h$

$\Rightarrow x=h\left(\sqrt{\frac{AD}{BC}}-\sqrt{\frac{BC}{AD}}\right)+\sqrt{BC\cdot AD}$

Since $x^2=2AD\cdot h$, we have

$0=h\left(\sqrt{\frac{AD}{BC}}-\sqrt{\frac{BC}{AD}}\right)-2\sqrt{AD\cdot h}+\sqrt{BC\cdot AD}$

Solving for $\sqrt{h}$ yields

$\sqrt{h}=\frac{2\sqrt{AD}\pm\sqrt{4AD-4(AD-BC)}}{2\left(\sqrt{\frac{AD}{BC}}-\sqrt{\frac{BC}{AD}}\right)}$

$\Rightarrow \sqrt{h}=\frac{\sqrt{AD}\pm\sqrt{BC}}{\sqrt{\frac{AD}{BC}}-\sqrt{\frac{BC}{AD}}}$

Since $\sqrt{h}\ne \frac{\sqrt{AD}+\sqrt{BC}}{\sqrt{\frac{AD}{BC}}-\sqrt{\frac{BC}{AD}}}$, we have

$\sqrt{h}=\frac{\sqrt{AD}-\sqrt{BC}}{\sqrt{\frac{AD}{BC}}-\sqrt{\frac{BC}{AD}}}$

$=\frac{\sqrt{AD\cdot BC}}{\sqrt{AD}+\sqrt{BC}}$

And since this is the maximimal value of $\sqrt{h}$, the minimal value of $\frac{1}{\sqrt{h}}$ is $\frac{1}{\sqrt{AD}}+\frac{1}{\sqrt{BC}}$ as desired.