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第一试
一、选择题（本题满分42分，每小题7分）
1. 设 ，则 （ A ）
A.24. B. 25. C. . D. .
2．在△ABC中，最大角∠A是最小角∠C的两倍，且AB＝7，AC＝8，则BC＝ （ C ）
A. . B. . C. . D. .
3．用 表示不大于 的最大整数，则方程 的解的个数为 （ C ）
A.1. B. 2. C. 3. D. 4.
4．设正方形ABCD的中心为点O，在以五个点A、B、C、D、O为顶点所构成的所有三角形中任意取出两个，它们的面积相等的概率为 （ B ）
A. . B. . C. . D. .
5．如图，在矩形ABCD中，AB＝3，BC＝2，以BC为直径在矩形内作半圆，自点A作半圆的切线AE，则 CBE＝ （ D ）
A. . B. . C. . D. .
6．设 是大于1909的正整数，使得 为完全平方数的 的个数是 （ B ）
A.3. B. 4. C. 5. D. 6.
二、填空题（本题满分28分，每小题7分）
1．已知 是实数，若 是关于 的一元二次方程 的两个非负实根，则 的最小值是_____ _______.
2． 设D是△ABC的边AB上的一点，作DE//BC交AC于点E，作DF//AC交BC于点F，已知△ADE、△DBF的面积分别为 和 ，则四边形DECF的面积为___ ___.
3．如果实数 满足条件 ， ，则 __ ____.
4．已知 是正整数，且满足 是整数，则这样的有序数对 共有___7__对.
第二试 （A）
一．（本题满分20分）已知二次函数 的图象与 轴的交点分别为A、B，与 轴的交点为C.设△ABC的外接圆的圆心为点P.
（1）证明：⊙P与 轴的另一个交点为定点.
（2）如果AB恰好为⊙P的直径且 ，求 和 的值.
解 （1）易求得点 的坐标为 ，设 ， ，则 ， .
设⊙P与 轴的另一个交点为D，由于AB、CD是⊙P的两条相交弦，它们的交点为点O，所以OA×OB＝OC×OD，则 .
因为 ，所以点 在 轴的负半轴上，从而点D在 轴的正半轴上，所以点D为定点，它的坐标为(0,1).
（2）因为AB⊥CD，如果AB恰好为⊙P的直径，则C、D关于点O对称，所以点 的坐标为 ，
即 .
又 ，所以
，解得 .
二．（本题满分25分）设CD是直角三角形ABC的斜边AD上的高， 、 分别是△ADC、△BDC的内心，AC＝3，BC＝4，求 .
解 作 E⊥AB于E， F⊥AB于F.
在直角三角形ABC中，AC＝3，BC＝4， .
又CD⊥AB，由射影定理可得 ，故 ，
.
因为 E为直角三角形ACD的内切圆的半径，所以 ＝ .
连接D 、D ，则D 、D 分别是∠ADC和∠BDC的平分线，所以∠ DC＝∠ DA＝∠ DC＝∠ DB＝45°，故∠ D ＝90°，所以 D⊥ D， .
同理，可求得 ， . 所以 ＝ .
三．（本题满分25分）已知 为正数，满足如下两个条件：
①
②
证明：以 为三边长可构成一个直角三角形.
证法1 将①②两式相乘，得 ，
即 ，
即 ，
即 ，
即 ，
即 ，
即 ，即 ，
即 ，
所以 或 或 ，即 或 或 .
因此，以 为三边长可构成一个直角三角形.
证法2 结合①式，由②式可得 ，
变形，得 ③
又由①式得 ，即 ，
代入③式，得 ，即 .
，
所以 或 或 .
结合①式可得 或 或 .
因此，以 为三边长可构成一个直角三角形.
第二试 （B）
一．（本题满分20分）题目和解答与（A）卷第一题相同.
二． （本题满分25分） 已知△ABC中，∠ACB＝90°，AB边上的高线CH与△ABC的两条内角平分线 AM、BN分别交于P、Q两点.PM、QN的中点分别为E、F.求证：EF‖AB.
解 因为BN是∠ABC的平分线，所以 .
又因为CH⊥AB，所以
，
因此 .
又F是QN的中点，所以CF⊥QN，所以 ，因此C、F、H、B四点共圆.
又 ，所以FC＝FH，故点F在CH的中垂线上.
同理可证，点E在CH的中垂线上.
因此EF⊥CH.又AB⊥CH，所以EF‖AB.
三．（本题满分25分）题目和解答与（A）卷第三题相同.
第二试 （C）
一．（本题满分20分）题目和解答与（A）卷第一题相同.
二．（本题满分25分）题目和解答与（B）卷第二题相同.
三．（本题满分25分）已知 为正数，满足如下两个条件：
①
②
是否存在以 为三边长的三角形？如果存在，求出三角形的最大内角.
解法1 将①②两式相乘，得 ，
即 ，
即 ，
即 ，
即 ，
即 ，
即 ，即 ，
即 ，
所以 或 或 ，即 或 或 .
因此，以 为三边长可构成一个直角三角形，它的最大内角为90°.
解法2 结合①式，由②式可得 ，
变形，得 ③
又由①式得 ，即 ，
代入③式，得 ，即 .
，
所以 或 或 .
结合①式可得 或 或 .
因此，以 为三边长可构成一个直角三角形，它的最大内角为90°.
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绛旓細鍦ㄤ笁瑙掑舰ABC涓紝X鏄疉B涓婄殑涓鐐癸紝Y鏄疊C涓婄殑涓鐐癸紝绾挎AY鍜孋X鐩镐氦浜嶼銆傚亣鑻Y=YC鍙夾B=ZC锛屾眰璇侊細B ,X ,Z 鍜孻 鍥涚偣鍏卞渾銆傝瘉鏄 鎴嚎AZY瀵刮擝CX鏉ヨ锛屾伆濂芥弧瓒虫娑呭姵鏂痆Menelaus]瀹氱悊,鎵浠ュ緱:(CY/YB)*(BA/AX)*(XZ/ZC)=1 (1)鍥犱负AB=ZC,鏁呭緱:CY*XZ=AX*BY (2)鍙圓Y=CY,鎵浠ユ湁 AY*XZ...
绛旓細涓銆 璇蜂綘濉竴濉紙姣忕┖3鍒嗭紝鍏39鍒嗭級1. 瑙勫畾锛歛鈥籦= ,閭ｄ箞2鈥5=___銆2. ,鍒 涓綅鏁板瓧涓篲__銆3. 锛屽垯x=___.4. 锛屽垯 x___.5. 鐢ㄤ笉绛夊彿杩炵粨: [锛嶏紙 锛塢2 6. 杩戜技鏁7.8脳104绮剧‘鍒 浣 7. 鎸夎寰嬪～鏁0锛1锛1锛2锛3锛5锛8锛 21.8. 骞虫柟绛変簬鏈韩鐨勬暟鏄痏__,鍊掓暟...
绛旓細2. 绗竴涓暟鏄疉B锛岀浜屼釜鏁版槸BA锛岀涓変釜鏁版槸A0B 閭ｄ箞锛 BA-AB=10*B+A-(A*10+B)=9(B-A)A0B锛岯A=A*100+B-(10B+A)=9(11A-B)姣忓皬鏃舵墍璧拌矾绋嬬浉绛夛紝鎵浠ワ細 9(B-A)=9(11A-B), B=6A A=1, B=6 鎵浠ユ瘡灏忔椂璺▼锛9(B-A)=9(6-1)=45 km 3. (1) 鐢蹭箼涓欏崟鐙畬鎴...
