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2008 AMC 12B 第 25 题

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25.

设 ABCDABCD 是梯形，满足 AB∥CDAB \parallel CD、AB=11AB = 11、BC=5BC = 5、CD=19CD = 19、DA=7DA = 7。∠A\angle A 与 ∠D\angle D 的角平分线交于 PP，∠B\angle B 与 ∠C\angle C 的角平分线交于 QQ。六边形 ABQCDPABQCDP 的面积是多少？

Let ABCDABCD be a trapezoid with AB∥CD,AB \parallel CD, AB=11,AB = 11, BC=5,BC = 5, CD=19,CD = 19, and DA=7.DA = 7. Bisectors of ∠A\angle A and ∠D\angle D meet at P,P, and bisectors of ∠B\angle B and ∠C\angle C meet at Q.Q. What is the area of hexagon ABQCDP?ABQCDP?

28328\sqrt{3}

30330\sqrt{3}

32332\sqrt{3}

35335\sqrt{3}

36336\sqrt{3}

答案：B
知识点：梯形角平分线余弦定理
难度评级：2230
小提示：

∠A+∠D=180∘\angle A + \angle D = 180^\circ，所以它们的角平分线垂直相交，且 AD‾\overline{AD} 的中点到 A,D,PA, D, P 等距。

∠A+∠D=180∘,\angle A + \angle D = 180^\circ, so their bisectors meet at right angles, and the midpoint of AD‾\overline{AD} is equidistant from A,D,PA, D, P

大提示：

这使 M,P,Q,NM, P, Q, N 位于中位线上，且 PQ=AB+CD−AD−BC2PQ = \tfrac{AB + CD - AD - BC}{2}；再由 △ADE\triangle ADE（AE∥BCAE \parallel BC，边长为 7,5,87, 5, 8）求高。

This puts M,P,Q,NM, P, Q, N on the midline with PQ=AB+CD−AD−BC2;PQ = \tfrac{AB + CD - AD - BC}{2}; get the height from △ADE\triangle ADE (AE∥BC,AE \parallel BC, sides 7,5,87, 5, 8)

解答：

因为 AB∥CDAB \parallel CD，有 ∠A+∠D=180∘\angle A + \angle D = 180^\circ，所以 ∠A\angle A 与 ∠D\angle D 的角平分线垂直相交，即 ∠APD=90∘\angle APD = 90^\circ。于是 AD‾\overline{AD} 的中点 MM 是直角三角形 APDAPD 的外心，得到 MP=MA=MDMP = MA = MD。所以 ∠MPA=∠PAM=∠PAB\angle MPA = \angle PAM = \angle PAB，最后一个等号使用了 AA 处的角平分线。因此 MP∥ABMP \parallel AB。对 BC‾\overline{BC} 的中点 NN 同理可得 QN∥ABQN \parallel AB。所以 M,P,Q,NM, P, Q, N 共线于中位线上。

中位线长为 AB+CD2=15\tfrac{AB + CD}{2} = 15，而 MP=AD2=72MP = \tfrac{AD}{2} = \tfrac72、QN=BC2=52QN = \tfrac{BC}{2} = \tfrac52。因此 PQ=15−72−52=9PQ = 15 - \tfrac72 - \tfrac52 = 9。

作 AE∥BCAE \parallel BC，其中 EE 位于 CD‾\overline{CD} 上，则 AE=5AE = 5，且 DE=CD−AB=8DE = CD - AB = 8。在 △ADE\triangle ADE 中，cos⁡(∠AED)=82+52−722⋅8⋅5=12\cos(\angle AED) = \tfrac{8^2 + 5^2 - 7^2}{2 \cdot 8 \cdot 5} = \tfrac12，所以 ∠AED=60∘\angle AED = 60^\circ，梯形高为 AF=5sin⁡60∘=532AF = 5\sin 60^\circ = \tfrac{5\sqrt3}{2}。

线段 PQPQ 位于半高处，所以六边形分成两个梯形，并且 [ABQCDP]=AF4(AB+CD+2 PQ)=5324(11+19+18)=303。 \begin{aligned} &[ABQCDP] \\ &= \frac{AF}{4}\bigl(AB + CD + 2\,PQ\bigr) \\ &= \frac{\frac{5\sqrt3}{2}}{4}(11 + 19 + 18) \\ &= 30\sqrt3 \end{aligned}\text{。}

所以正确答案是 B。

Because AB∥CD,AB \parallel CD, ∠A+∠D=180∘,\angle A + \angle D = 180^\circ, so the bisectors of ∠A\angle A and ∠D\angle D meet at right angles, ∠APD=90∘.\angle APD = 90^\circ. Then the midpoint MM of AD‾\overline{AD} is the circumcenter of right triangle APD,APD, giving MP=MA=MD.MP = MA = MD. Thus ∠MPA=∠PAM=∠PAB,\angle MPA = \angle PAM = \angle PAB, where the last equality uses the angle bisector at A.A. Therefore MP∥AB.MP \parallel AB. The same argument gives QN∥ABQN \parallel AB for the midpoint NN of BC‾.\overline{BC}. Hence M,P,Q,NM, P, Q, N are collinear on the midline.

The midline has length AB+CD2=15,\tfrac{AB + CD}{2} = 15, while MP=AD2=72MP = \tfrac{AD}{2} = \tfrac72 and QN=BC2=52.QN = \tfrac{BC}{2} = \tfrac52. Hence PQ=15−72−52=9.PQ = 15 - \tfrac72 - \tfrac52 = 9.

Drawing AE∥BCAE \parallel BC with EE on CD‾\overline{CD} gives AE=5AE = 5 and DE=CD−AB=8.DE = CD - AB = 8. In △ADE,\triangle ADE, cos⁡(∠AED)=82+52−722⋅8⋅5=12,\cos(\angle AED) = \tfrac{8^2 + 5^2 - 7^2}{2 \cdot 8 \cdot 5} = \tfrac12, so ∠AED=60∘\angle AED = 60^\circ and the trapezoid’s height is AF=5sin⁡60∘=532.AF = 5\sin 60^\circ = \tfrac{5\sqrt3}{2}.

The segment PQPQ sits at half the height, so the hexagon splits into two trapezoids and [ABQCDP]=AF4(AB+CD+2 PQ)=5324(11+19+18)=303. \begin{aligned} &[ABQCDP] \\ &= \frac{AF}{4}\bigl(AB + CD + 2\,PQ\bigr) \\ &= \frac{\frac{5\sqrt3}{2}}{4}(11 + 19 + 18) \\ &= 30\sqrt3. \end{aligned}

Thus, the correct answer is B.

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