Problem:
Let ABCD be a quadrilateral circumscribed about a circle, whose interior and exterior angles are at least 60∘. Prove that
31∣∣∣AB3−AD3∣∣∣≤∣∣∣BC3−CD3∣∣∣≤3∣∣∣AB3−AD3∣∣∣
When does equality hold?
Solution:
By symmetry, we only need to prove the first inequality.
Because quadrilateral ABCD has an incircle, we have AB+CD=BC+AD, or AB− AD=BC−CD. It suffices to prove that
31(AB2+AB⋅AD+AD2)≤BC2+BC⋅CD+CD2
By the given condition, 60∘≤∠A,∠C≤120∘, and so 21≥cosA,cosC≥−21. Applying the law of cosines to triangle ABD yields
BD2=AB2−2AB⋅ADcosA+AD2≥AB2−AB⋅AD+AD2≥31(AB2+AB⋅AD+AD2)
The last inequality is equivalent to the inequality 3AB2−3AB⋅AD+3AD2≥AB2+ AB⋅AD+AD2, or AB2−2AB⋅AD+AD2≥0, which is evident. The last equality holds if and only if AB=AD.
On the other hand, applying the Law of Cosines to triangle BCD yields
BD2=BC2−2BC⋅CDcosC+CD2≤BC2+BC⋅CD+CD2
Combining the last two inequalities gives the desired result.
For the given inequalities to hold, we must have AB=AD. This condition is also sufficient, because all the entries in the equalities are 0 . Thus, the given inequalities hold if and only if ABCD is a kite with AB=AD and BC=CD.
Problem originally by Titu Andreescu.
The problems on this page are the property of the MAA's American Mathematics Competitions