高中数学联赛模拟试题精选学数学系列第6套几何题
△ A B C \triangle ABC △ABC 的边 B C BC BC 的中点为点 M M M. 点 E E E, F F F 分别是 M M M 关于 A C AC AC, A B AB AB 的对称点, 只需 B F BF BF 和 C E CE CE 交于点 P P P. 点 Q Q Q 满足 Q A = Q M QA=QM QA=QM 且 ∠ Q A P = π 2 \angle QAP = \frac{\pi}{2} ∠QAP=2π. 点 Q Q Q 是 △ P E F \triangle PEF △PEF 的外心. 求证: ∠ A O Q = π 2 \angle AOQ = \frac{\pi}{2} ∠AOQ=2π. (《高中数学联赛模拟试题精选》"学数学"系列第6套)
证明:
∠ F A E = 2 ∠ B A C \angle FAE=2\angle BAC ∠FAE=2∠BAC.
∠ F P E = π − ∠ P B C − ∠ P C B = π − ( π − 2 ∠ A B C ) − ( π − 2 ∠ A C B ) = 2 ∠ A B C + 2 ∠ A C B − π \angle FPE=\pi-\angle PBC-\angle PCB=\pi-(\pi-2\angle ABC)-(\pi-2\angle ACB)=2\angle ABC+2\angle ACB-\pi ∠FPE=π−∠PBC−∠PCB=π−(π−2∠ABC)−(π−2∠ACB)=2∠ABC+2∠ACB−π.
∠ F A E + ∠ F P E = 2 ∠ A B C + 2 ∠ A C B + 2 ∠ B A C − π = π \angle FAE+\angle FPE=2\angle ABC+2\angle ACB+2\angle BAC-\pi=\pi ∠FAE+∠FPE=2∠ABC+2∠ACB+2∠BAC−π=π.
所以 F F F, A A A, E E E, P P P 共圆.
设 A M AM AM 的中点为 K K K, 延长 A O AO AO 交 ( F A E ) (FAE) (FAE) 于点 L L L.
∠ L F B = ∠ L E C \angle LFB=\angle LEC ∠LFB=∠LEC ; F B = B M = M C = E C FB=BM=MC=EC FB=BM=MC=EC ; 由 ∠ A F L = ∠ A E L = π 2 \angle AFL=\angle AEL=\frac{\pi}{2} ∠AFL=∠AEL=2π, A E = A F AE=AF AE=AF 可知 △ A F L ≃ △ A E L \triangle AFL \simeq \triangle AEL △AFL≃△AEL, 所以 L F = L E LF=LE LF=LE, 综上, △ L F B ≃ △ L E C \triangle LFB \simeq \triangle LEC △LFB≃△LEC, 进而 L B = L C LB=LC LB=LC.
L B = L C LB=LC LB=LC, M M M 是 B C BC BC 的中点, 所以 L M ⊥ B C LM \bot BC LM⊥BC.
显然 K O KO KO 是 △ A M L \triangle AML △AML 的中位线, 所以 K O / / L M KO//LM KO//LM, 进而 K O ⊥ B C KO \bot BC KO⊥BC.
∠ K O A = ∠ M L A \angle KOA=\angle MLA ∠KOA=∠MLA.
∠ O A P = ∠ O A F − ∠ F A P = π 2 − ∠ A E F − ∠ F E P = π 2 − ∠ A E P = π 2 − ∠ A M C \angle OAP=\angle OAF-\angle FAP=\frac{\pi}{2}-\angle AEF-\angle FEP=\frac{\pi}{2}-\angle AEP=\frac{\pi}{2}-\angle AMC ∠OAP=∠OAF−∠FAP=2π−∠AEF−∠FEP=2π−∠AEP=2π−∠AMC.
∠ O A P + ∠ A M C = π 2 \angle OAP+\angle AMC=\frac{\pi}{2} ∠OAP+∠AMC=2π.
∠ O A P + ∠ A M L = π \angle OAP+\angle AML=\pi ∠OAP+∠AML=π.
∠ A L M + ∠ M A L = π − ∠ A M L = ∠ O A P = ∠ P A M + ∠ M A L \angle ALM+\angle MAL=\pi-\angle AML=\angle OAP=\angle PAM+\angle MAL ∠ALM+∠MAL=π−∠AML=∠OAP=∠PAM+∠MAL, 所以 ∠ M A P = ∠ M L A \angle MAP=\angle MLA ∠MAP=∠MLA.
∠ K O A = ∠ M L A = ∠ M A P \angle KOA=\angle MLA=\angle MAP ∠KOA=∠MLA=∠MAP.
∠ K Q A = π 2 − ∠ K A Q = ∠ M A P = ∠ K O A \angle KQA=\frac{\pi}{2}-\angle KAQ=\angle MAP=\angle KOA ∠KQA=2π−∠KAQ=∠MAP=∠KOA, 所以 K K K, A A A, Q Q Q, O O O 四点共圆, 进而 ∠ A O Q = ∠ A K Q = π 2 \angle AOQ=\angle AKQ=\frac{\pi}{2} ∠AOQ=∠AKQ=2π.
证毕.
完稿时间: 2025年5月3日.