2004 USAMO Problems/Problem 6


(Zuming Feng) A circle $\omega$ is inscribed in a quadrilateral $ABCD$. Let $I$ be the center of $\omega$. Suppose that

$(AI + DI)^2 + (BI + CI)^2 = (AB + CD)^2$.

Prove that $ABCD$ is an isosceles trapezoid.


Our proof is based on the following key Lemma.

Lemma: If a circle $\omega$, centered at $I$, is inscribed in a quadrilateral $ABCD$, then \[BI^2 + \frac{AI}{DI}\cdot BI\cdot CI = AB\cdot BC.\qquad\qquad (1)\]


Proof: Since circle $\omega$ is inscribed in $ABCD$, we get $m\angle DAI = m\angle IAB = a$, $m\angle ABI = m\angle IBC = b$, $m\angle BCI = m\angle ICD = c$, $m\angle CDI = m\angle IDA = d$, and $a + b + c + d = 180^\circ$. Construct a point $P$ outside of the quadrilateral such that $\triangle ABP$ is similar to $\triangle DCI$. We obtain \begin{align*} m\angle PAI + m\angle PBI &= m\angle PAB + m\angle BAI + m\angle PBA + m\angle ABI \\ &= m\angle IDC + a + m\angle ICD + b \\ &= a + b + c + d = 180^\circ, \end{align*} implying that the quadrilateral $PAIB$ is cyclic. By Ptolemy's Theorem, we have $AI\cdot BP + BI\cdot AP = AB\cdot IP$, or \[BP\cdot\frac{AI}{IP} + BI\cdot\frac{AP}{IP} = AB.\qquad\qquad (2)\] Because $PAIB$ is cyclic, it is not difficult to see that, as indicated in the figure, $m\angle IPB = m\angle IAB = a$, $m\angle API = m\angle ABI = b$, $m\angle AIP = m\angle ABP = c$, and $m\angle PIB = m\angle PAB = d$. Note that $\triangle AIP$ and $\triangle ICB$ are similar, implying that \[\frac{AI}{IP} = \frac{IC}{CB}\text{  and  }\frac{AP}{IP} = \frac{IB}{CB}.\] Substituting the above equalities into the identity $(2)$, we arrive at \[BP\cdot\frac{CI}{BC} + \frac{BI^2}{BC} = AB,\] or \[BP\cdot CI + BI^2 = AB\cdot BC.\qquad\qquad (3)\] Note also that $\triangle BIP$ and $\triangle IDA$ are similar, implying that $\frac{BP}{BI} = \frac{IA}{ID}$, or \[BP = \frac{AI}{ID}\cdot IB.\] Substituting the above identity back into $(3)$ gives the desired relation $(1)$, establishing the Lemma. $\blacksquare$

Now we prove our main result. By the Lemma and symmetry, we have \[CI^2 + \frac{DI}{AI}\cdot BI\cdot CI = CD\cdot BC.\qquad\qquad (4)\] Adding the two identities $(1)$ and $(4)$ gives \[BI^2 + CI^2 + \left(\frac{AI}{DI} + \frac{DI}{AI}\right)BI\cdot CI = BC(AB + CD).\] By the AM-GM Inequality, we have $\frac{AI}{DI} + \frac{DI}{AI}\geq 2$. Thus, \[BC(AB + CD)\geq IB^2 + IC^2 + 2IB\cdot IC = (BI + CI)^2,\] where the equality holds if and only if $AI = DI$. Likewise, we have \[AD(AB + CD)\geq (AI + DI)^2,\] where the equality holds if and only if $BI = CI$. Adding the last two identities gives \[(AI + DI)^2 + (BI + CI)^2\leq (AD + BC)(AB + CD) = (AB + CD)^2,\] because $AD + BC = AB + CD$. (The latter equality is true because the circle $\omega$ is inscribed in the quadrilateral $ABCD$.)

By the given condition in the problem, all the equalities in the above discussion must hold, that is, $AI = DI$ and $BI = CI$. Consequently, we have $a = d$, $b = c$, and so $\angle DAB + \angle ABC = 2a + 2b = 180^\circ$, implying that $AD\parallel BC$. It is not difficult to see that $\triangle AIB$ and $\triangle DIC$ are congruent, implying that $AB = CD$. Thus, $ABCD$ is an isosceles trapezoid.

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


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