16. Find the number of non-negative integer solutions \((π, π, π, π)\) to the following inequality:
\(π + π + π + π β€ 12\)
(A) 1772
(B) 1800
(C) 1820
(D) 1825
(E) None of the above
Gunakan teorema star bars Karena \(π + π + π + π β€ 12\), maka ada bilangan bilangan bulat tak negative \(π\) sehingga \(π + π + π + π + π = 12\) Banyak solusi adalah \({12 + 5 β 1\choose {5-1}}= {16\choose 4} =\frac{16!}{4!.12!}=\frac{16.15.14.13}{4.3.2.1}=1820\)
17. A triangle has two medians whose lengths are 12 and 15. Find its maximum possible area.
(A) 100
(B) 125
(C) 128
(D) 144
(E) None of the above
Jika \(π, π, π\) adalah garis berat pada segitiga maka luasnya adalah \(\frac{4}{3}\sqrt{π (π β π)(π β π)(π β π)}\), \(π =\frac{π + π + π}{2}\)
Diketahui median garis dari segitiga \(ABC\) adalah \(12\) dan \(15\), kita misalkan median yang satunya adalah \(π₯\) \(πΏπ’ππ = \frac{4}{3}\sqrt{π (π β π)(π β π)(π β π)}\) \(=\frac{4}{3}\sqrt{(\frac{27+π₯}{2})(\frac{27+π₯}{2}β π₯)(\frac{27+π₯}{2}β 12)(\frac{27+π₯}{2}β 15)}\) \(=\frac{4}{3}\sqrt{(\frac{27+π₯}{2})(\frac{27βπ₯}{2})(\frac{3+π₯}{2})(\frac{π₯β3}{2})}\) \(=\frac{1}{3}\sqrt{(27 + π₯)(27 β π₯)(π₯ + 3)(π₯ β 3)}\) \(=\frac{1}{3}\sqrt{(27^2 β π₯^2)(π₯^2 β 9)}\) Luas maksimum dicapai ketika \((27^2 β π₯^2) = (π₯^2 β 9)β2x^2=738βxΒ²=369\)
Jadi luas maksimum \(=\frac{1}{3}\sqrt{(27^2 β 369)(369 β 9)}=\frac{1}{3}(360)=120\)
20. Let \(Ξπ΄π΅πΆ\) be an acute triangle. \(π΅πΈ\) and \(πΆπΉ\) are the height and median of the triangle, respectively. Suppose \(π΅πΆ = 5\) and \(π΅πΈ = πΆπΉ = 4\).
Evaluate the area of triangle \(Ξπ΄π΅πΆ\).
(A) \(5\sqrt2 β 2\)
(B) \(8\sqrt3 β 3\)
(C) \(8\sqrt2 β 6\)
(D) \(8\sqrt3 β 6\)
(E) None of the above
Perhatikan segitiga siku-siku \(BEC\) \(πΆπΈ = \sqrt{π΅πΆ^2 β π΅πΈ^2} = \sqrt{5^2 β 4^2} = 3\) \(Ξπ΄πΉπ· β π΄π΅πΈ\) Karena \(π΄πΉ = πΉπ΅\) maka \(π΄π· = π·πΈ = π₯\) Karena \(\frac{π΄πΉ}{π΄π΅}=\frac{1}{2}\) maka \(\frac{πΉπ·}{π΅πΈ}=\frac{1}{2}\), diperoleh \(πΉπ· = 2\)