PX Formula

24 年 8 月 30 日 星期五
457 字
3 分钟

适用问题:

问题图示

ACB\triangle ACB 中,ACB=α\angle ACB=\alphaAB=aAB=aα\alphaaa 均为定值),求 mAC+nBCmAC+nBC 的最大值。

结论: (mAC+nBC)max=n2+2mncosα+m2sinαa(mAC + nBC)_{\text{max}} = \frac{\sqrt{n^2 + 2mn\cos\alpha + m^2}}{\sin\alpha} \cdot a

推导过程: 将原式转化为 m(AC+nmBC)m\left(AC + \frac{n}{m}BC\right)

  • α>90\alpha > 90^{\circ} 时:

(i) nm>cosα\frac{n}{m} > -\cos\alpha

延长 ACACDD,使得 CD=nmBCCD = \frac{n}{m}BC,作 BHADBH \perp AD,如图:

图示1

λ=nm\lambda = \frac{n}{m}, CH=BC(cosα)CH = BC \cdot (-\cos\alpha), DH=CDCH=(λ+cosα)BCDH = CD - CH = (\lambda + \cos\alpha)BC

BH=BCsinαBH = BC \cdot \sin\alpha

tanADB=HBDH=sinαλ+cosα\therefore \tan\angle ADB = \frac{HB}{DH} = \frac{\sin\alpha}{\lambda + \cos\alpha}(ADB=arctansinαλ+cosα)(\Leftrightarrow\angle ADB = \arctan\frac{\sin\alpha}{\lambda + \cos\alpha})

可以求出 sinADB=sinαλ2+2λcosα+1\sin\angle ADB = \frac{\sin\alpha}{\sqrt{\lambda^2 + 2\lambda\cos\alpha + 1}}

ADB\angle ADB 为定角,ABAB 为定弦 \Rightarrow DDADB\triangle ADB 的外接圆上运动。

ADB\triangle ADB 的外接圆半径的 222r=ABsinADB=asinαλ2+2λcosα+12r = \frac{AB}{\sin\angle ADB} = \frac{a}{\sin\alpha} \cdot \sqrt{\lambda^2 + 2\lambda\cos\alpha + 1}

AD2r\because AD \leq 2r

ADmax=λ2+2λcosα+1sinαa\therefore AD_{\text{max}} = \frac{\sqrt{\lambda^2 + 2\lambda\cos\alpha + 1}}{\sin\alpha} \cdot a

(ii) nm<cosα\frac{n}{m} < -\cos\alpha

延长 ACACDD,使得 CD=nmBCCD = \frac{n}{m}BC,作 BHADBH \perp AD,如图:

图示2

λ=nm\lambda = \frac{n}{m}CH=BC(cosα)CH = BC \cdot (-\cos\alpha), DH=CHCD=(λcosα)BCDH = CH - CD = (-\lambda - \cos\alpha)BC

BH=BCsinαBH = BC \cdot \sin\alpha

tanHDB=HBDH=sinαλ+cosα\therefore \tan\angle HDB = \frac{HB}{DH} = -\frac{\sin\alpha}{\lambda + \cos\alpha}

tanADB=tanDHB=sinαλ+cosα\therefore \tan \angle ADB = -\tan \angle DHB = \frac{\sin\alpha}{\lambda + \cos\alpha} (ADB=arctansinαλ+cosα)(\Leftrightarrow \angle ADB = \arctan \frac{\sin\alpha}{\lambda + \cos\alpha})

ADB\angle ADB 为定角,ABAB 为定弦 \Rightarrow DDADB\triangle ADB 的外接圆上运动。

可以求出 sinADB=sinαλ2+2λcosα+1\sin\angle ADB = \frac{\sin\alpha}{\sqrt{\lambda^2 + 2\lambda\cos\alpha + 1}}

ADB\triangle ADB 的外接圆半径的 222r=ABsinADB=asinαλ2+2λcosα+12r = \frac{AB}{\sin\angle ADB} = \frac{a}{\sin\alpha} \cdot \sqrt{\lambda^2 + 2\lambda\cos\alpha + 1}

AD2r\because AD \leq 2r

ADmax=λ2+2λcosα+1sinαa\therefore AD_{\text{max}} = \frac{\sqrt{\lambda^2 + 2\lambda\cos\alpha + 1}}{\sin\alpha} \cdot a

  • α<90\alpha < 90^{\circ} 时:

延长 ACACDD,使得 CD=nmBCCD = \frac{n}{m}BC,作 BHADBH \perp AD,如图:

图示3

λ=nm\lambda = \frac{n}{m}

CH=BCcosαCH = BC \cdot \cos\alphaDH=CD+CH=(λ+cosα)BCDH = CD + CH = (\lambda + \cos\alpha)BC

BH=BCsinαBH = BC \cdot \sin\alpha

tanADB=HBDH=sinαλ+cosα\therefore \tan\angle ADB = \frac{HB}{DH} = \frac{\sin\alpha}{\lambda + \cos\alpha}(ADB=arctansinαλ+cosα)(\Leftrightarrow\angle ADB = \arctan\frac{\sin\alpha}{\lambda + \cos\alpha})

ADB\angle ADB 为定角,ABAB 为定弦 \Rightarrow DDADB\triangle ADB 的外接圆上运动。

可以求出 sinADB=sinαλ2+2λcosα+1\sin\angle ADB = \frac{\sin\alpha}{\sqrt{\lambda^2 + 2\lambda\cos\alpha + 1}}

ADB\triangle ADB 的外接圆半径的 222r=ABsinADB=asinαλ2+2λcosα+12r = \frac{AB}{\sin\angle ADB} = \frac{a}{\sin\alpha} \cdot \sqrt{\lambda^2 + 2\lambda\cos\alpha + 1}

AD2r\because AD \leq 2r

ADmax=λ2+2λcosα+1sinαa\therefore AD_{\text{max}} = \frac{\sqrt{\lambda^2 + 2\lambda \cos\alpha + 1}}{\sin\alpha} \cdot a

综上,(AC+nmBC)max=ADmax=λ2+2λcosα+1sinαa(AC+\frac{n}{m}\cdot BC)_{\max}=AD_{\max}=\frac{\sqrt{\lambda^2+2\lambda\cos\alpha+1}}{\sin\alpha} \cdot a

(mAC+nBC)max=mADmax=m2+2mncosα+n2sinαa\Leftrightarrow(m\cdot AC+n\cdot BC)_{\text{max}}=m\cdot AD_{\text{max}}=\frac{\sqrt{m^2+2mn\cos\alpha+n^2}}{\sin\alpha}\cdot a

文章标题:PX Formula

文章作者:Walter_Fang

文章链接:https://blog.walterfang.us.kg/posts/px-formula[复制]

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