2004 USAMO Problems/Problem 6

Problem

(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.

Solution

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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2004 USAMO (Problems • Resources)
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2004 USAMO Problems/Problem 6

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