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OM20190322001


g(n)=\displaystyle\sum_{k=1}^{n} (k, n), 其中n\in\mathbb{N}^{*}, (k, n)表示kn的最大公约数, 求g(100).

OM20190320


\triangle ABC中, 已知AB=2, AC=1, \angle BAC=90^{\circ}, D, E分别为BC, AD的中点, 过点E的直线交AB于点P, 交AC于点Q, 求\overrightarrow{BQ}\cdot \overrightarrow{CP}的最大值.

20190318002


如图, 四边形ABCD中, \angle ADB=2\angle ABD, \angle ABC=\angle BDC=30^{\circ}. 求证: AC=AD.

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简证: 作C关于AB的对称点O, 则\triangle OBC为等边三角形. 分别延长BA, CD, 交于点E.

\triangle BOC=2\angle BDC, 知点D在以O为圆心, OB为半径的圆上. 故\triangle OCD是等腰三角形.

易知\triangle CBD\backsim \triangle CEB, 得CB^2=CD\cdot CE. 故CO^2=CD\cdot CE. 所以\triangle CDO\backsim \triangle COE.

\angle ABD=\alpha, 则\angle ADB=2\alpha, \angle DAE=3\alpha, \angle OEC=\angle DOC=2\angle CBD=2(30^{\circ}-\alpha).

所以\angle COE=90^{\circ}-(30^{\circ}-\alpha)=60^{\circ}+\alpha.

所以\angle DOE=\angle COE-\angle COD=(60^{\circ}+\alpha)-(60^{\circ}-2\alpha)=3\alpha.

所以\angle DAE=\angle DOE. 所以A, D, E, O四点共圆.

所以\angle AOD=\angle AED=\dfrac{1}{2}\angle CEO=\dfrac{1}{2}\angle COD. 故\triangle AOC\cong \triangle AOD. 所以AC=AD.

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20190318001


如图, 在平面直角坐标系xOy中, \odot O的半径为2, 点A\odot O上, 点B, C分别在x轴和y轴上, AB=BC, \tan A=\dfrac{4}{3}, PAC上, 且PA=3PC, 连OP, 求OP的取值范围.

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密码保护:OM20190219问题


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2018BALKAN MATHEMATICAL OLYMPIAD-4


求所有质数p, q, 使得3p^{q-1}+1\mid 11^p+17^p.

Find all primes p and q such that 3p^{q-1}+1 divides 11^p+17^p.

密码保护:初三综数一模答案


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密码保护:初一数学思维: 质数和合数


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