Quadrilateral $ABCD$ is circumscribed around the circle with centre $I$. Let points $M$ and $N$ be the midpoints of sides $AB$ and $CD$ respectively and let $\frac{IM}{AB} = \frac{IN}{CD}$. Prove that $ABCD$ is either a trapezoid or a parallelogram.
Problem
Source:
Tags: geometry, trapezoid, ratio, geometric transformation, reflection, incenter, analytic geometry
Rofler
09.05.2010 01:46
Outline: Take CID about a spiral similarity about I so that C lies on B. Easy to show that that $A,B/C',D'$ is a triangle. WLOG AC'D'>90 (why?). Now, translate the ratio $\frac{IM}{IN}=\frac{AB}{CD}$ into a fact about this new triangle, and use the fact that AC'D'>90 to show that I is the unique point on segment AD' so that BM/BN'=IM/IN' (the circle of Appolonius helps here). Therefore, you can show that I is the reflection of B about MN'. Thus, BIA is 90, and this implies the required parallel condition. Cheers, Rofler
Batominovski
09.05.2010 02:17
Nice and concise. Thanks a bunch.
Suppose the incenter $I$ lies at the origin (i.e., having complex coordinate $0$) and the inscribed circle $\omega$ of $ABCD$ has radius $1$. Let $\omega$ touch $AB$ at $\alpha$, $BC$ at $\beta$, $CD$ at $\gamma$, and $DA$ at $\delta$. The line $AB$ contains all complex numbers $z$ for which $z-\alpha$ is perpendicular to $\alpha$. In other words, $\frac{z-\alpha}{\alpha}$ is purely imaginary, or
\[\left(\frac{z-\alpha}{\alpha}\right) + \overline{\left(\frac{z-\alpha}{\alpha}\right)}=0\,.\]
Because $|\alpha|=1$, we get $\bar{\alpha}=\frac{1}{\alpha}$. Rewrite the above equation, hence, the line $AB$ has the equation
\[\frac{z}{\alpha^2}+\bar{z}=\frac{2}{\alpha}\,.\]
Likewise, $BC$ has the equation
\[\frac{z}{\beta^2}+\bar{z}=\frac{2}{\beta}\,.\]
So, the point $B$ is the intersection of $AB$ and $BC$. Therefore, the common solution to the previous two equations is the complex coordinate of $B$, which is
\[B = \frac{2\alpha\beta}{\alpha+\beta}\,.\]
Likewise,
\[C = \frac{2\beta\gamma}{\beta+\gamma}\,,\]
\[D = \frac{2\gamma\delta}{\gamma+\delta}\,.\]
and
\[A = \frac{2\delta\alpha}{\delta+\alpha}\,.\]
Consequently,
\[M=\frac{A+B}{2} = \alpha\left(\frac{\beta}{\alpha+\beta}+\frac{\delta}{\alpha+\delta}\right)\]
and
\[N = \frac{C+D}{2} = \gamma\left(\frac{\beta}{\gamma+\beta}+\frac{\delta}{\gamma+\delta}\right)\,.\]
Hence,
\[|MI| = \left|\frac{\beta}{\alpha+\beta}+\frac{\delta}{\alpha+\delta}\right| = \frac{\big|\alpha(\beta+\delta)+2\beta\delta\big|}{|\alpha+\beta||\alpha+\delta|}\]
and
\[|NI| = \frac{\big|\gamma(\beta+\delta)+2\beta\delta\big|}{|\gamma+\beta||\gamma+\delta|}\,.\]
Now, we compute
\[|AB| = \left|\frac{2\alpha\beta}{\alpha+\beta}-\frac{2\delta\alpha}{\delta+\alpha}\right| = \frac{2|\beta-\delta|}{|\alpha+\beta||\alpha+\delta|}\]
and
\[|CD| = \frac{2|\beta-\delta|}{|\gamma+\beta||\gamma+\delta|}\,.\]
That is, $\frac{|MI|}{|AB|}=\frac{|NI|}{|CD|}$ holds if and only if
\[{\big|\alpha(\beta+\delta)+2\beta\delta\big|}= \big|\gamma(\beta+\delta)+2\beta\delta\big|\;\;\;\;\;\text{...(*)}\]
If $\beta+\delta=0$, the above equality holds, and if that happens, then $DA \parallel BC$. ***THIS IS WHAT THE PROBLEM WANTS US TO SHOW.***
However, if $\beta+\delta \neq 0$, let $x$ be the intersection of $DA$ and $BC$. Let $y$ be the image of reflection of $x$ about $I$, that is
\[y=-x=-\frac{2\beta\delta}{\beta+\delta}\,.\]
Hence, Equation (*) is equivalent to
\[|\alpha-y| = |\gamma-y|\,,\]
or $y$ lies on the perpendicular bisector $\ell$ of the segment $\alpha\gamma$. The implication of this result is that $\beta$ and $\delta$ are symmetric with respect to $\ell$ (for the tangents of $\omega$ at $\beta$ and $\gamma$ meet within $\ell$). The consequence is that $ABCD$ has some of its sides crossing one another. Therefore, we are not talking about a convex system (and hence, $\omega$ cannot be said as an inscribed circle). So, the case where $\beta+\delta \neq 0$ is invalid.
PS: Moderators, please move this topic to "solved" category.
Rofler
10.05.2010 08:21
Actually, I would like a solution that is not bash, and has some motivation. The only way I came up with the above is by trying to find a "Russian" solution (elegant, and completely unmotivated). Cheers, Rofler
frenchy
10.05.2010 13:52
it is a very nice problem.taking the relation in square and writing IM^2=(2AI^2+2IB^2-AB^2)\4 so the given relation goes to (AI^2+BI^2)\AB^2=(CI^2+DI^2)\CD^2 but from cosinus law it goes to (AI*BI*cosAIB)\AB^2=(DI*CI*cosCID)\CD^2 but IBA+IAB+ICD+IDC=180 cose point I is the intersection of bisectors so AIB=180-CID so cosAIB=-cosCID that means that the relation obtained gives a contradiction because one of the part will be negative.This is imposible so cosAIB=cosDIC=0 so IBA+IAB=90 so DAB+CBA=180 so AD is parallel to BC so ABCD is a trapezoid at least but trapezoid is a particullar case of parallelogram so we are done
mathophile593
14.05.2010 03:00
This problem is Russian and therefore simpler without cosine law/stewart's theorem. The important lemma is that given triangle ABC with the M the midpoint of BC, AM < BC/2 iff angle BAC > 90. This follows immediately after drawing a circle with diameter BC. Now for the problem, the lemma combined with the condition tells us that AIB and CID are both obtuse, both acute, or both right angles. Since CID + AIB = 180, they're both right angles and the rest follows like in frenchy's proof.