103 Trigonometry Problems

or
5. Solutions to Advanced Problems 177
|BP|·|CI|+|BI| 2 =|AB|·|BC|. (†††)
Note also that triangle BIP and triangle IDAare similar, implying that |BP|
|BI|
=
|IA|
|ID| ,or
|BP|= |AI|
|ID| ·|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
|CI| 2 + |DI| ·|BI|·|CI|=|CD|·|BC|. (‡)
|AI|
Adding the two identities (†) and (‡) gives
( |AI|
|BI| 2 +|CI| 2 +
|DI| + |DI| )
|BI|·|CI|=|BC|(|AB|+|CD|).
|AI|
By the arithmetic–geometric means inequality, wehave |AI|
Thus,
|DI| + |DI|
|AI|
≥ 2.
|BC|(|AB|+|CD|) ≥|IB| 2 +|IC| 2 + 2|IB|·|IC|=(|BI|+|CI|) 2 ,
where equality holds if and only if |AI| =|DI|. Likewise, we have
|AD|(|AB|+|CD|) ≥ (|AI|+|DI|) 2 ,
where equality holds if and only if |BI|=|CI|. Adding the last two identities
gives the desired inequality (∗).
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 triangle AIB and triangle DIC
are congruent, implying that |AB| = |CD|. Thus, ABCD is an isosceles
trapezoid.
44. [USAMO 2001] Let a,b, and c be nonnegative real numbers such that
a 2 + b 2 + c 2 + abc = 4.
Prove that
0 ≤ ab + bc + ca − abc ≤ 2.