Problem:
A circle ω is inscribed in a quadrilateral ABCD. Let I be the center of ω. Suppose that
(AI+DI)2+(BI+CI)2=(AB+CD)2
Prove that ABCD is an isosceles trapezoid.
Solution:
A circle ω is inscribed in a quadrilateral ABCD. Let I be the center of ω. 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 ω, centered at I, is inscribed in a quadrilateral ABCD, then
BI2+DIAI​⋅BI⋅CI=AB⋅BC(*)
Proof: Since circle ω is inscribed in ABCD, we get m∠DAI=m∠IAB=a,m∠ABI= m∠IBC=b,m∠BCI=m∠ICD=c,m∠CDI=m∠IDA=d, and a+b+c+d=180∘. Construct a point P outside of the quadrilateral such that △ABP is similar to △DCI. We obtain
m∠PAI+m∠PBI​=m∠PAB+m∠BAI+m∠PBA+m∠ABI=m∠IDC+a+m∠ICD+b=a+b+c+d=180∘​
implying that the quadrilateral PAIB is cyclic. By Ptolemy's Theorem, we have AI. BP+BIâ‹…AP=ABâ‹…IP, or
BP⋅IPAI​+BI⋅IPAP​=AB
Because PAIB is cyclic, it is not difficult to see that, as indicated in the figure, m∠IPB= m∠IAB=a,m∠API=m∠ABI=b,m∠AIP=m∠ABP=c, and m∠PIB= m∠PAB=d. Note that △AIP and △ICB are similar, implying that
IPAI​=CBIC​ and IPAP​=CBIB​
Substituting the above equalities into the identity ( †), we arrive at
BP⋅BCCI​+BCBI2​=AB
or
BPâ‹…CI+BI2=ABâ‹…BC
Note also that △BIP and △IDA are similar, implying that BIBP​=IDIA​, or
BP=IDAI​⋅IB
Substituting the above identity back into (†′) gives the desired relation (∗), establishing the Lemma.
Now we prove our main result. By the Lemma and symmetry, we have
CI2+AIDI​⋅BI⋅CI=CD⋅BC
Adding the two identities (∗) and (∗′) gives
BI2+CI2+(DIAI​+AIDI​)BI⋅CI=BC(AB+CD)
By the AM-GM Inequality, we have DIAI​+AIDI​≥2. Thus,
BC(AB+CD)≥IB2+IC2+2IB⋅IC=(BI+CI)2
where the equality holds if and only if AI=DI. Likewise, we have
AD(AB+CD)≥(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≤(AD+BC)(AB+CD)=(AB+CD)2
because AD+BC=AB+CD. (The latter equality is true because the circle ω 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 ∠DAB+∠ABC=2a+2b=180∘, implying that AD∥BC. It is not difficult to see that △AIB and △DIC are congruent, implying that AB=CD. Thus, ABCD is an isosceles trapezoid.
Problem originally by Zuming Feng.
The problems on this page are the property of the MAA's American Mathematics Competitions
