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Difference between revisions of "2004 USAMO Problems/Problem 6"

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<cmath>BP\cdot\frac{AI}{IP} + BI\cdot\frac{AP}{IP} = AB.\qquad\qquad (2)</cmath>
 
<cmath>BP\cdot\frac{AI}{IP} + BI\cdot\frac{AP}{IP} = AB.\qquad\qquad (2)</cmath>
 
Because <math>PAIB</math> is cyclic, it is not difficult to see that, as indicated in the figure, <math>m\angle IPB = m\angle IAB = a</math>, <math>m\angle API = m\angle ABI = b</math>, <math>m\angle AIP = m\angle ABP = c</math>, and <math>m\angle PIB = m\angle PAB = d</math>. Note that <math>\triangle AIP</math> and <math>\triangle ICB</math> are similar, implying that
 
Because <math>PAIB</math> is cyclic, it is not difficult to see that, as indicated in the figure, <math>m\angle IPB = m\angle IAB = a</math>, <math>m\angle API = m\angle ABI = b</math>, <math>m\angle AIP = m\angle ABP = c</math>, and <math>m\angle PIB = m\angle PAB = d</math>. Note that <math>\triangle AIP</math> and <math>\triangle ICB</math> are similar, implying that
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<cmath>\frac{AI}{IP = \frac{IC}{CB}\text{  and  }\frac{AP}{IP} = \frac{IB}{CB}.</cmath>
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<cmath>\frac{AI}{IP} = \frac{IC}{CB}\text{  and  }\frac{AP}{IP} = \frac{IB}{CB}.</cmath>
 
Substituting the above equalities into the identity <math>(2)</math>, we arrive at
 
Substituting the above equalities into the identity <math>(2)</math>, we arrive at
 
<cmath>BP\cdot\frac{CI}{BC} + \frac{BI^2}{BC} = AB,</cmath>
 
<cmath>BP\cdot\frac{CI}{BC} + \frac{BI^2}{BC} = AB,</cmath>
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Note also that <math>\triangle BIP</math> and <math>\triangle IDA</math> are similar, implying that <math>\frac{BP}{BI} = \frac{IA}{ID}</math>, or
 
Note also that <math>\triangle BIP</math> and <math>\triangle IDA</math> are similar, implying that <math>\frac{BP}{BI} = \frac{IA}{ID}</math>, or
 
<cmath>BP = \frac{AI}{ID}\cdot IB.</cmath>
 
<cmath>BP = \frac{AI}{ID}\cdot IB.</cmath>
−
Substituting the above identity back into <math>(3)</math> gives the desired relation <math>(1)</math>, establishing the Lemma.
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Substituting the above identity back into <math>(3)</math> gives the desired relation <math>(1)</math>, establishing the Lemma. <math>\blacksquare</math>
  
 
Now we prove our main result. By the Lemma and symmetry, we have
 
Now we prove our main result. By the Lemma and symmetry, we have

Latest revision as of 23:18, 28 November 2014

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 \[BI^2 + \frac{AI}{DI}\cdot BI\cdot CI = AB\cdot BC.\qquad\qquad (1)\]

2004usamo6-1.png

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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