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30

Part of 1987 AMC 12/AHSME

Problems(1)

1987 AMC 12 #30 - Dividing a Triangle

Source:

1/1/2012
In the figure, △ABC\triangle ABC has ∠A=45∘\angle A =45^{\circ} and ∠B=30∘\angle B =30^{\circ}. A line DEDE, with DD on ABAB and ∠ADE=60∘\angle ADE =60^{\circ}, divides △ABC\triangle ABC into two pieces of equal area. (Note: the figure may not be accurate; perhaps EE is on CBCB instead of ACAC.) The ratio ADAB\frac{AD}{AB} is [asy] size((220)); draw((0,0)--(20,0)--(7,6)--cycle); draw((6,6)--(10,-1)); label("A", (0,0), W); label("B", (20,0), E); label("C", (7,6), NE); label("D", (9.5,-1), W); label("E", (5.9, 6.1), SW); label("45∘45^{\circ}", (2.5,.5)); label("60∘60^{\circ}", (7.8,.5)); label("30∘30^{\circ}", (16.5,.5)); [/asy] <spanclass=′latex−bold′>(A)</span> 12<spanclass=′latex−bold′>(B)</span> 22+2<spanclass=′latex−bold′>(C)</span> 13<spanclass=′latex−bold′>(D)</span> 163<spanclass=′latex−bold′>(E)</span> 1124 <span class='latex-bold'>(A)</span>\ \frac{1}{\sqrt{2}} \qquad<span class='latex-bold'>(B)</span>\ \frac{2}{2+\sqrt{2}} \qquad<span class='latex-bold'>(C)</span>\ \frac{1}{\sqrt{3}} \qquad<span class='latex-bold'>(D)</span>\ \frac{1}{\sqrt[3]{6}} \qquad<span class='latex-bold'>(E)</span>\ \frac{1}{\sqrt[4]{12}}
ratiogeometrytrigonometryarea of a triangletrig identitiesLaw of SinesAMC