《2025年亮点给力大试卷九年级数学上册苏科版》

27. (10分)如图,在平面直角坐标系中,⊙O的圆心O在坐标原点,直径AB= 6,P是直径AB上的一个动点(点P不与A,B两点重合),过点P的直线PQ的函数表达式为y= x+m.当直线PQ交y轴于点Q,交⊙O于C,D两点时,过点C作CH垂直于x轴,并延长交⊙O于点E,过点E作EG垂直于y轴,垂足为G,过点C作CF垂直于y轴,垂足为F,连接DE.
(1) 在点P的运动过程中,∠CPB= 45°;
(2) 当m= 2时,试求矩形CEGF的面积;
(3) 在点P的运动过程中,探索$PD^2+PC^2$的值是否会发生变化? 如果发生变化,请你说明理由;如果不发生变化,请你求出这个不变的值.

(2)连接OC.当m = 2时,同(1),得点P的坐标为(-2,0),点Q的坐标为(0,2).所以OP = OQ = 2.设FC = x.因为CF⊥OF,CH⊥OH,∠FOH = 90°,所以四边形CFOH是矩形.所以OH = FC = x.所以PH = OP + OH = 2 + x.因为AB⊥CE,所以∠PCH + ∠CPB = 90°,由(1),得∠CPB = 45°,所以∠PCH = 90° - ∠CPB = 45°,即∠PCH = ∠CPB.所以CH = PH = 2 + x.因为AB是⊙O的直径，AB = 6,所以$OC=\frac{1}{2}AB = 3$,$CE = 2CH = 4 + 2x$,$CH = EH$.所以$S_{矩形CEGF}=CF\cdot CE = x(2x + 4)=2x^2 + 4x$.在Rt△OCH中,由勾股定理，得OH² + CH² = OC²,所以$x^2+(x + 2)^2=3^2$,即$2x^2 + 4x = 5$.所以$S_{矩形CEGF}=5$.
(3)PD² + PC²的值不发生变化.连接PE,OD,OE.因为AB = 6,所以$OD = OE=\frac{1}{2}AB = 3$.由(2),得CH = EH,∠PCH = 45°.又AB⊥CE,所以∠PHC = ∠PHE = 90°.又PH = PH,所以△PHC≌△PHE (SAS).所以PC = PE,∠PCH = ∠PEH,即∠PEH = ∠PCH = 45°.又∠DPE = ∠PCH + ∠PEH,所以∠DPE = 90°,在Rt△DPE中,由勾股定理,得PD² + PE² = DE²,所以PD² + PC² = DE².因为∠DOE = 2∠DCE = 2∠PCH = 90°,所以在Rt△DOE中,由勾股定理，得DE² = OD² + OE² = 18.所以PD² + PC² = 18.则PD² + PC²的值不发生变化，且为18.

答案：
(1)45 解析:对于y = x + m,当x = 0时,y = m;当y = 0时,x + m = 0,解得x = -m.因为直线y = x + m交x轴于点P,交y轴于点Q,所以点P的坐标为(-m,0),点Q的坐标为(0,m).所以OQ = OP = m.又∠POQ = 90°,所以△POQ为等腰直角三角形,所以∠CPB = 45°.
(2)连接OC.当m = 2时,同
(1),得点P的坐标为(-2,0),点Q的坐标为(0,2).所以OP = OQ = 2.设FC = x.因为CF⊥OF,CH⊥OH,∠FOH = 90°,所以四边形CFOH是矩形.所以OH = FC = x.所以PH = OP + OH = 2 + x.因为AB⊥CE,所以∠PCH + ∠CPB = 90°,由
(1),得∠CPB = 45°,所以∠PCH = 90° - ∠CPB = 45°,即∠PCH = ∠CPB.所以CH = PH = 2 + x.因为AB是⊙O的直径，AB = 6,所以$OC=\frac{1}{2}AB = 3$,$CE = 2CH = 4 + 2x$,$CH = EH$.所以$S_{矩形CEGF}=CF\cdot CE = x(2x + 4)=2x^2 + 4x$.在Rt△OCH中,由勾股定理，得OH² + CH² = OC²,所以$x^2+(x + 2)^2=3^2$,即$2x^2 + 4x = 5$.所以$S_{矩形CEGF}=5$.
(3)PD² + PC²的值不发生变化.连接PE,OD,OE.因为AB = 6,所以$OD = OE=\frac{1}{2}AB = 3$.由
(2),得CH = EH,∠PCH = 45°.又AB⊥CE,所以∠PHC = ∠PHE = 90°.又PH = PH,所以△PHC≌△PHE (SAS).所以PC = PE,∠PCH = ∠PEH,即∠PEH = ∠PCH = 45°.又∠DPE = ∠PCH + ∠PEH,所以∠DPE = 90°,在Rt△DPE中,由勾股定理,得PD² + PE² = DE²,所以PD² + PC² = DE².因为∠DOE = 2∠DCE = 2∠PCH = 90°,所以在Rt△DOE中,由勾股定理，得DE² = OD² + OE² = 18.所以PD² + PC² = 18.则PD² + PC²的值不发生变化，且为18.

28. (10分)【思考发现】
(1) 如图①,A,B两点均在⊙O上,且∠AOB= 60°,P,Q两点均在射线AM上.若∠APB= 30°,则点P在⊙O上;若∠AQB＞30°,则点Q在⊙O内(填“内”“上”或“外”);
【问题解决】
如图②,在四边形ABCD中,∠B= ∠D= 90°,∠DAB= 135°,且AB= 2,AD= 4$\sqrt{2}$.
(2) 若P是边BC上任意一点,且∠APD= 45°,求BP的长;如图①,过点D作DN⊥BC于点N,过点A作AF⊥DN于点F,则∠DNB = ∠DNC = ∠AFN = ∠AFD = 90°.又∠B = 90°,所以四边形ABNF是矩形.所以∠BAF = 90°,BN = AF,NF = AB.又∠DAB = 135°,所以∠DAF = ∠DAB - ∠BAF = 45°.又∠DAF + ∠ADF = ∠AFN = 90°,所以∠ADF = 90° - ∠DAF = 45°,所以∠ADF = ∠DAF,即AF = DF.在Rt△ADF中,$AD = 4\sqrt{2}$,由勾股定理，得AD² = AF² + DF²,所以$2AF^2=(4\sqrt{2})^2$,解得AF = 4(负值已舍去).所以DF = BN = AF = 4.以点F为圆心,AF的长为半径作⊙F,交BC于P,P'两点.连接PF,P'F,则PF = P'F = AF = 4.又AB = 2,所以NF = 2.在Rt△P'NF中,由勾股定理,得$P'N=\sqrt{P'F^2 - NF^2}=2\sqrt{3}$.同理,得$PN = 2\sqrt{3}$.所以$BP = BN - PN = 4 - 2\sqrt{3}$,$BP' = BN + P'N = 4 + 2\sqrt{3}$.综上,BP的长为$4 + 2\sqrt{3}$或$4 - 2\sqrt{3}$.
(3) 如图③,以点B为圆心,BC的长为半径作弧,交BA的延长线于点E.若Q为弧EC上的动点,过点Q作QH⊥BC于H,连接BQ,设I为△BQH的内心,连接BI,QI,当点Q从点C运动到点E时,则内心I所经过的路径长为$\frac{5\sqrt{2}\pi}{2}$.

答案：
(1)上 内
(2)如图①,过点D作DN⊥BC于点N,过点A作AF⊥DN于点F,则∠DNB = ∠DNC = ∠AFN = ∠AFD = 90°.又∠B = 90°,所以四边形ABNF是矩形.所以∠BAF = 90°,BN = AF,NF = AB.又∠DAB = 135°,所以∠DAF = ∠DAB - ∠BAF = 45°.又∠DAF + ∠ADF = ∠AFN = 90°,所以∠ADF = 90° - ∠DAF = 45°,所以∠ADF = ∠DAF,即AF = DF.在Rt△ADF中,$AD = 4\sqrt{2}$,由勾股定理，得AD² = AF² + DF²,所以$2AF^2=(4\sqrt{2})^2$,解得AF = 4(负值已舍去).所以DF = BN = AF = 4.以点F为圆心,AF的长为半径作⊙F,交BC于P,P'两点.连接PF,P'F,则PF = P'F = AF = 4.又AB = 2,所以NF = 2.在Rt△P'NF中,由勾股定理,得$P'N=\sqrt{P'F^2 - NF^2}=2\sqrt{3}$.同理,得$PN = 2\sqrt{3}$.所以$BP = BN - PN = 4 - 2\sqrt{3}$,$BP' = BN + P'N = 4 + 2\sqrt{3}$.综上,BP的长为$4 + 2\sqrt{3}$或$4 - 2\sqrt{3}$.
(3)$\frac{5\sqrt{2}\pi}{2}$ 解析:因为∠ABC = ∠ADC = 90°,∠DAB = 135°,∠ABC + ∠BCD + ∠ADC + ∠DAB = 360°,所以∠BCD = 360° - ∠ABC - ∠ADC - ∠DAB = 45°.在图①中,DN⊥BC,所以∠BCD + ∠CDN = 90°.所以∠CDN = 90° - ∠BCD = 45°,即∠CDN = ∠BCD.所以CN = DN.由
(2),得DF = 4,NF = 2,BN = 4,所以CN = DN = DF + NF = 6.所以BC = BN + CN = 10.如图②,连接IC,作△BCI的外接圆.设外接圆圆心为T，在优弧BC上任取一点K，连接BT，CT，BK，CK,则BT = CT,∠BTC = 2∠BKC.又四边形BICK为⊙T的内接四边形,所以∠BKC + ∠BIC = 180°.因为I为△BQH的内心,所以BI平分∠QBH,QI平分∠BQH,即$\angle QBI=\angle HBI=\frac{1}{2}\angle QBH$,$\angle BQI=\angle HQI=\frac{1}{2}\angle BQH$.又QH⊥BC,所以∠BQH + ∠QBH = 90°,即$\angle BQI+\angle QBI=\frac{1}{2}(\angle BQH+\angle QBH)=45°$.又∠BQI + ∠QBI + ∠BIQ = 180°,所以∠BIQ = 180° - ∠BQI - ∠QBI = 135°.在$\triangle BIQ$和$\triangle BIC$中,$\begin{cases} BQ = BC \\ \angle QBI = \angle CBI \\ BI = BI \end{cases}$,所以△BIQ≌△BIC (SAS).所以∠BIQ = ∠BIC,即∠BIC = 135°.由题意，得当点Q从点C运动到点E时，点I的运动路径是$\overset{\frown}{BC}$.所以∠BKC = 180° - ∠BIC = 45°.所以∠BTC = 90°.所以△BTC是等腰直角三角形.在Rt△BTC中,由勾股定理,得BT² + CT² = BC²,即$2BT^2 = 10^2$,解得$BT = 5\sqrt{2}$(负值已舍去).则$l_{\overset{\frown}{BC}}=\frac{90\pi×5\sqrt{2}}{180}=\frac{5\sqrt{2}\pi}{2}$,即内心I所经过的路径长为$\frac{5\sqrt{2}\pi}{2}$.