383. Trigonometry in geometry (middle or high school) I.

by Virgil Nicula, Aug 9, 2013, 6:56 PM

PP10 (Israel Diaz Acha). Let $\triangle ABC$ with the incircle $w=C(I,r)$ which touches $\triangle ABC$ at $D\in BC$ , $E\in CA$ and $F\in AB$ . Denote the second intersections $M$ , $N$

of the circle $w$ with $BE$ , $CF$ respectively. Denote $R\in BC\cap FM$ and $S\in BC\cap EN$ . Prove that $\frac {RD}{RB}+\frac {SD}{SC}=2$ and $2RS=BC\iff AB=AC$ .


Proof 1. $DEFM$ is harmonically (<== click), i.e. $\boxed{\frac {MD}{MF}=\frac {ED}{EF}}\ (1)$ . Apply an well-known relation to the cevian $FR/\triangle BFD\ :\ \frac {RD}{RB}=$ $\frac {FD}{FB}\cdot\frac {\sin\widehat {RFD}}{\sin\widehat{RFB}}=$ $\frac {FD}{FB}\cdot\frac {\sin\widehat {MED}}{\sin\widehat{MEF}}=$

$\frac {FD}{FB}\cdot\frac {MD}{MF}\stackrel{(1)}{=}$ $\frac {FD}{FB}\cdot\frac {ED}{EF}\implies$ Thus, $\frac {RD}{RB}=2\sin\frac B2\cdot\frac {2(s-c)\sin\frac C2}{2(s-a)\sin\frac A2}=$ $\frac {2(s-c)}{s-a}\cdot\sqrt{\frac {(s-a)(s-c)}{ac}\cdot\frac {(s-a)(s-b)}{ab}\cdot\frac {bc}{(s-b)(s-c)}}\implies$ $\boxed{\frac {RD}{RB}=\frac {2(s-c)}{a}}\ (2)$ .

Prove analogously $\boxed{\frac {SD}{SC}=\frac {2(s-b)}{a}}\ (3)$ . Therefore, $\frac {RD}{RB}+\frac {SD}{SC}=2$ and get easily $\left\{\begin{array}{c}
RB=\frac {a(s-b)}{2a+b-c}\\\\
SC=\frac{a(s-c)}{2a+c-b}\end{array}\right\|$ . In conclusion, $2RS=BC\iff$ $2(RB+SC)=BC\iff$

$\frac {a+c-b}{2a+b-c}+\frac {a+b-c}{2a+c-b}=1\iff$ $3(b-c)^2=0\iff b=c\iff$ $AB=AC$ .

Proof 2. Suppose w.l.o.g. $AB<AC$ and denote $L\in BC\cap EF$ . Is well-known that the division $(B,C;D,L)$ is
harmonically (<== click), i.e. $\frac {DB}{DC}=\frac {LB}{LC}=\frac {s-b}{s-c}\implies$

$\frac {LB}{s-b}=\frac {LC}{s-c}=$ $\frac {a}{b-c}\implies$ $LB=\frac {a(s-b)}{b-c}$ and $LD=LB+BD=\frac {a(s-b)}{b-c}+(s-b)\implies$ $LD=\frac {2(s-b)(s-c)}{b-c}\implies$ $\boxed{\frac {LD}{LB}=\frac {2(s-c)}a}\ (1)$ . Denote

$X\in DF\cap BE$ . The line $DF$ is the polar of $B$ w.r.t. the incircle $w$ and the point $X$ is the harmonical conjugate of $B$ w.r.t. $\{M,E\}$ , i.e the division $(B,X;M,E)$ is harmonically

$\implies$ the pencil $F(B,X;M,E)$ is harmonic $\implies$ the division $(B,D;R,L)=F(B,X;M,E)\cap BC$ is harmonically, i.e. $\frac {RD}{RB}=\frac {LD}{LB}\stackrel{(1)}{\implies}$ $\boxed{\frac {RD}{RB}=\frac {2(s-c)}a}$ . Prove analogously

that $\boxed{\frac {LD}{LC}=\frac {2(s-b)}a}\ (2)$ and the division $(C,D;S,L)$ is harmonically, i.e. $\frac {SD}{SC}=\frac {LD}{LC}\stackrel{(2)}{\implies}$ $\boxed{\frac {SD}{SC}=\frac {2(s-b)}a}$ a.s.o. Prove easily that $\frac {DR}{DS}=\lambda\ \iff\ \frac {b-c}{a}=2\cdot \frac {\lambda -1}{\lambda +1}$ .



PP11. Let a square $ABCD$ with circumcircle $w=C(O,R)$ . Let midpoints $M$ , $N$ of $[OB]$ , $[CD]$ respectively. $AM$ cut again $w$ in $P$ . Find the measures of the angles of $\triangle APN$ .

Proof 1. Suppose w.l.o.g. $R=2$ . Thus, $\left\{\begin{array}{c}
OM=MB=1\\\
AD=DC=2\sqrt 2\\\
ND=NC=\sqrt 2\end{array}\right\|$ and $AM=\sqrt 5$ . Hence $\triangle AOM\sim\triangle APC\iff$ $\frac {AO}{AP}=\frac {OM}{PC}=\frac {AM}{AC}\iff$ $\frac {2}{AP}=\frac {1}{PC}=\frac {\sqrt 5}{4}\iff$

$\boxed{PC=\frac 4{\sqrt 5}}$ and $AP=\frac 8{\sqrt 5}$ . Since $\triangle AMD\sim\triangle BMP$ obtain $\frac {AM}{BM}=\frac {MD}{MP}=\frac {AD}{BP}\iff$ $\frac {\sqrt 5}{1}=\frac {3}{MP}=\frac {2\sqrt 2}{BP}\iff$ $MP=\frac {3}{\sqrt 5}$ and $BP=\frac {2\sqrt 2}{\sqrt 5}\implies$

$PD^2=BD^2-PB^2=$ $16-\left(\frac {2\sqrt 2}{\sqrt 5}\right)^2=$ $16-\frac 85\implies$ $\boxed{PD=\frac {6\sqrt 2}{\sqrt 5}}$ . Denote $\left\{\begin{array}{c}
\phi =m\left(\widehat{DPN}\right)\\\\
\theta =m\left(\widehat{APN}\right)\end{array}\right\|$ , where $\theta =45^{\circ}+\phi$ . Apply $\frac {ND}{NC}=\frac {PD}{PC}\cdot\frac {\sin \widehat{NPD}}{\sin\widehat{NPC}}\iff$

$1=\frac {\frac {6\sqrt 2}{\sqrt 5}}{\frac 4{\sqrt 5}}\cdot\frac {\sin \phi}{\sin \left(45^{\circ}-\phi\right)}=$ $\frac {3\sqrt 2}{2}\cdot\frac {\sin\phi}{\frac {\sqrt 2}{2}\cdot(\cos\phi-\sin\phi)}=$ $\frac {3\tan\phi}{1-\tan\phi}\implies$ $1=\frac {3\tan\phi}{1-\tan\phi}\iff$ $\tan\phi =\frac 14$ . So $\tan\theta =\tan (45^{\circ}+\phi )=$ $\frac {1+\tan\phi}{1-\tan\phi} =$ $\frac {1+\frac 14}{1-\frac 14}\implies$

$\tan\theta =\frac 53\iff$ $\boxed{\tan\widehat {APN}=\frac 53}$ . Since $\left\{\begin{array}{c}
\tan\widehat{DAN}=\frac {DN}{AD}=\frac 12\\\\
\tan\widehat{MAO}=\frac {OM}{OA}=\frac 12\end{array}\right\|\implies$ $\widehat{DAN}\equiv\widehat{MAO}\implies$ $\widehat{PAN}\equiv\widehat{CAD}\implies$ $\boxed{m\left(\widehat{PAN}\right)=45^{\circ}}$ . In $\triangle APN$ apply identity

$\tan A+\tan P+\tan N=\tan A\tan P\tan N\iff$ $1+\frac 53+\tan N=1\cdot\frac 53\cdot\tan N\iff$ $\boxed{\tan\widehat{ANP}=4}$ .

Proof 2. Suppose w.l.o.g. $R=2$ . Since $\left\{\begin{array}{c}
\tan\widehat{DAN}=\frac {DN}{AD}=\frac 12\\\\
\tan\widehat{MAO}=\frac {OM}{OA}=\frac 12\end{array}\right\|\implies$ $\widehat{DAN}\equiv\widehat{MAO}\implies$ $\widehat{PAN}\equiv\widehat{CAD}\implies$ $\boxed{m\left(\widehat{PAN}\right)=45^{\circ}}$ . Denote the midpoint $R$ of $[OC]$ .

Thus, $AN^2=AR^2+RN^2$ $\implies$ $\boxed{AN=\sqrt{10}}$ . Hence $MA^2=AO^2+MO^2\implies$ $AM=\sqrt 5$ and $\cos \widehat {MAO}=\frac {AO}{AM}\implies$ $\cos\widehat{PAC}=\frac {2}{\sqrt 5}\implies$ $PA=AC\cos\widehat{PAC}\implies$ $\boxed{PA=\frac 8{\sqrt 5}}$ . Apply the generalized Pythagoras' theorem to $[PN]$ in $\triangle APN\ :\ PN^2=PA^2+AN^2-2\cdot PA\cdot AN\cdot\cos\widehat{PAN}=$

$\frac {64}{5}+10-2\cdot\frac 8{\sqrt 5}\cdot\sqrt {10}\cdot\frac {\sqrt 2}2\implies$ $\boxed{PN=\sqrt{\frac {34}{5}}}$ . Otherwise, $S=[PAN]=\frac 12\cdot AP\cdot AN\cdot\sin\widehat{PAN}\implies$ $S=\frac 12\cdot \frac 8{\sqrt 5}\cdot\sqrt{10}\cdot\frac {\sqrt 2}2\implies$ $\boxed{[PAN]=4}$ and

$4S=\left(AP^2+AN^2-PN^2\right)\tan\widehat{PAN}\ (*)\implies$ $16=\frac {64}{5}+10-PN^2\implies$ $PN=\sqrt{\frac {34}{5}}$ . Now can find measures of the other angles for $\triangle APN$ with the same

relation $(*)\ :\ \left\{\begin{array}{ccc}
\tan\widehat{ANP}=\frac {4S}{NA^2+NP^2-AP^2}=\frac{16}{10+\frac {34}{5}-\frac {64}{5}}=\frac {80}{20} & \implies & \boxed{\tan\widehat{ANP}=4}\\\\
\tan\widehat{APN}=\frac {4S}{PA^2+PN^2-AN^2}=\frac{16}{\frac {64}{5}+\frac {34}{5}-10}=\frac {80}{48} & \implies & \boxed{\tan\widehat{APN}=\frac 53}\end{array}\right\|$ .



PP12. Prove that $(\forall )\ \triangle ABC$ and $x\in\mathbb R$ exists the identity $\boxed{4R\sin\frac {A+x}{2}\sin\frac {B+x}{2}\sin\frac {C+x}{2}=2R\sin x\sin\frac x2+r\cos\frac x2+s\sin\frac x2}$ .

PP13. Solve the equation $\boxed{s=2R\sin x+r\cot\frac x2}$ , where $x\in (0,\pi )$ .
See here. Solve the geometrical equation $\sum\sqrt{\left(x-\frac {h_b}{b}\right)\left(x-\frac {h_c}{c}\right)}=x$ in $\triangle ABC$ . See here.


PP14. Prove that $(\forall )\ \triangle ABC$ there is the identity $\cos B\cos C\cos\ (B-C)+\cos C\cos A\cos\ (C-A)+$ $\cos A\cos B\cos\ (A-B)=\prod_{\mathrm{cyc}}\cos\ (B-C)\ -\ 2\cdot\prod_{\mathrm{cyc}}\cos A$ .

Proof. Denote $P(B,C)\equiv \cos B\cos C\cos (B-C)$ and $S\equiv\sum\cos B\cos C\cos (B-C)$ . Thus, $2P(B,C)=[\cos (B+C)+\cos (B-C)]\cos (B-C)\implies$

$4P(B,C)=2\cos (B+C)\cos (B-C)+2\cos^2(B-C)\implies$ $4P(B,C)=\cos 2B+\cos 2C+1+\cos 2(B-C)$ . Therefore, $4S=4\sum P(B,C)=$

$\sum[\cos 2B+\cos 2C+1+\cos 2(B-C)]=$ $3+2\sum\cos 2A+\sum\cos (B-C)]\implies$ $\boxed{4S=3+2U+V}$ , where $\left\{\begin{array}{ccc}
U & \equiv & \sum\cos 2A\\\\
V & \equiv & \sum\cos 2(B-C)\end{array}\right\|$ .

$\blacktriangleright\ U=\sum\cos 2A=\cos 2A+(\cos 2B+\cos 2C)=$ $2\cos^2A-1-2\cos A\cos (B-C)=$

$-1-2\cos A[\cos (B-C)+\cos (B+C)]\implies$ $\boxed{U=\sum\cos 2A==-1-4\cos A\cos B\cos C}$ .

$\blacktriangleright$ I"ll ascertain the sum $\cos 2x+\cos 2y+\cos 2z$ , where $x+y+z=0$ . Thus, $\sum \cos 2x=2\cos^2x-2\cos x\cos (y-z)=$ $-1+2\cos x[\cos x+\cos (y-z)]\implies$

$\sum \cos 2x=-1+4\prod\cos x$ . Thus, $\boxed{x+y+z=0\implies \cos x+\cos y+\cos z=-1+4\cos x\cos y\cos z}\ (*)$ . Particularly for $\left\{\begin{array}{c}
B-C=x\\\\
C-A=y\\\\
A-B=z\end{array}\right\|$ obtain that $\boxed{V=\sum 2\cos (B-C)=-1+4\prod\cos (B-C)}$ and $\boxed{S=V-2U}$ .



PP15. Let a square $ABCD$ , $M\in (AD)$ , $N\in (CD)$ so that $BM\perp AN$ and $I\in AN\cap BM$ , where $IB=b$ , $IN=a$ . Prove that $3b\le 4a$ and $\tan\widehat {ABM}=1+\sqrt{\frac {4a-3b}b}$

Proof. Denote $m\left(\widehat {ABM}\right)=x$ and $U\in (AB)$ , $V\in (CD)$ so that $I\in UV$ and $UV\parallel AD$ . Observe that $AB=\frac b{\cos x}$ and $UV=IU+IV=b\sin x+a\cos x$ . Thus,

$AB=UV\implies $ $b\sin x+a\cos x=\frac b{\cos x}\implies$ $b\sin x\cos x+a\cos^2x=b\implies$ $b\sin 2x+a(1+\cos 2x)=2b\implies$ $\boxed{b\sin 2x+a\cos 2x=2b-a}\ (*)$ . Equation $(*)$

has at least a real zero iff $b^2+a^2\ge (2b-a)^2\iff$ $\boxed{3b\le 4a}\ (1)$ and in this case by the substitution $\boxed{\tan x=t}$ , i.e. $\left\{\begin{array}{ccc}
\sin 2x & = & \frac {2t}{1+t^2}\\\\
\cos 2x & = & \frac {1-t^2}{1+t^2}\end{array}\right\|$ the equation $(*)$ becomes

$2bt+a\left(1-t^2\right)=(2b-a)\left(1+t^2\right)\implies$ $bt^2-bt+(b-a)=0\implies$ $t=\frac {b+\sqrt{4ab-3b^2}}b$ , i.e. $\tan x=1+\sqrt{\frac {4a-3b}b}\implies$ $\tan\widehat {ABM}=1+\sqrt{\frac {4a-3b}b}$ .

If $3b=4a$ , then obtain that $m\left(\widehat {ABM}\right)=45^{\circ}$ , i.e. $\left\{\begin{array}{c}
MA=MD\\\\
NC=ND\end{array}\right\|$ .


PP16 Let $\triangle ABC$ with the incircle $w=C(I,r)$ which touches it at $D\in AC$ , $E\in AB$ and $F\in BC$ . Prove that $\left\{\begin{array}{ccc}
EA & = & EF\\\
EB & = & ED\end{array}\right\|\implies$ $CA=CB$ .

Proof 1 (trigonometric). Observe that $\frac {s-a}{\cot\frac A2}=\frac {s-b}{\cot\frac B2}=\frac {s-c}{\cot\frac C2}=r$ . Therefore,

$EA=EF\iff$ $s-a=2(s-b)\sin \frac B2\iff$ $\cot\frac A2=2\cot\frac B2\sin\frac B2\iff $ $\cot\frac A2=2\cos\frac B2\iff$ $\cos\frac A2=2\sin\frac A2\cos\frac B2\iff$ $\cos \frac A2=\cos\frac C2+\sin\frac {A-B}2\ \ (1)\ .$

$EB=ED\iff$ $s-b=2(s-a)\sin \frac A2\iff$ $\cot\frac B2=2\cot\frac A2\sin\frac A2\iff $ $\cot\frac B2=2\cos\frac A2\iff$ $\cos\frac B2=2\sin\frac B2\cos\frac A2\iff$ $\cos \frac B2=\cos\frac C2-\sin\frac {A-B}2\ \ (2)\ .$

Substract $(2)$ from $(1)\ :\ \cos\frac A2-\cos\frac B2=2\sin\frac {A-B}2\iff$ $2\cos\frac C2\sin\frac {B-A}2=2\sin\frac {A-B}2\iff$ $A=B\ \vee\ \ 2\cos\frac C2+1=0\iff$ $A=B\iff CA=CB\ .$.

Proof 2 (metric).

$\blacktriangleright\ EA=EF\iff$ $s-a=2(s-b)\sin \frac B2\iff$ $(s-a)^2=4(s-b)^2\cdot\frac {(s-a)(s-c)}{ac}\iff$ $ac(s-a)=4(s-b)^2(s-c)\ \ (1)\ .$

$\blacktriangleright\ EB=ED\iff$ $s-b=2(s-a)\sin \frac A2\iff$ $(s-b)^2=4(s-a)^2\cdot\frac {(s-b)(s-c)}{bc}\iff$ $bc(s-b)=4(s-a)^2(s-c)\ \ (2)\ .$

Substract $(2)$ from $(1)\ :\ cs(a-b)-c\left(a^2-b^2\right)=4(s-c)c(a-b)\iff$ $(a-b)[(a+b)-s]+4(a-b)(s-c)=0\iff$ $5(s-c)(a-b)=0\iff$ $a=b\ .$



PP17. Let a rectangle $ABCD$ with $AB>AD$ and $\{M,N\}\subset (BD)$ so that $M\in (BN)$ . Let $\left\{\begin{array}{c}
m\left(\widehat{BAM}\right)=\alpha\\\\
 m\left(\widehat{DCN}\right)=\beta\\\\
m\left(\widehat{DAN}\right)=\phi\end{array}\right\|$ .Prove that $\sqrt[3]{\tan\alpha}=\sqrt{\tan\beta}\iff \phi =45^{\circ}$ .

Proof. Let $\left\{\begin{array}{ccc}
A(0,0) & ; & B(a,0)\\\\
C(a,1) & ; & D(a,1)\end{array}\right\|$ where $a>1$ . Equation of $BD$ is $\frac xa+\frac y1=1\iff x+ay=a$ and its slope is $s(BD)=-\frac 1a$ . Hence there are $0<n<m<1$ so that

$\odot\begin{array}{ccccc}
\nearrow & M\left(x_M,m\right) & \mathrm{where} & x_M=a(1-m) & \searrow \\\\
\searrow &  N\left(x_N,n\right) & \mathrm{where} & x_N=a(1-n) & \nearrow\end{array}\odot$ . Observe that $CM\perp BD\iff$ $s(CM)\cdot s(BD)=-1\iff$ $\frac {1-m}{a-a(1-m)}\cdot \left(-\frac 1a\right)=-1\iff$

$\boxed{\frac {1-m}m=a^2}\ (*)$ . Therefore obtain that $\odot\begin{array}{ccccc}
\nearrow & \tan\beta =s(CN)=\frac {1-n}{a-a(1-n)} & \iff & \tan\beta =\frac {1-n}{an} & \searrow\\\\
\rightarrow & \tan\alpha =s(AM)=\frac m{a(1-m)} & \stackrel{(*)}{\iff} & \tan\alpha =\frac 1{a^3} & \rightarrow\\\\
\searrow & \cot\phi =s(AN)=\frac n{a(1-n)} & \iff & \tan\phi =\frac {a(1-n)}{n} & \nearrow\end{array}\odot$ $\boxed{\tan\phi =a^2\tan\beta}\ (1)$ .

Thus, $\sqrt[3]{\tan\alpha}=\sqrt{\tan\beta}\iff$ $\frac 1a=\sqrt{\tan\beta}\iff$ $\tan\beta =$ $\frac 1{a^2}\ \stackrel{(1)}{\iff}\ \tan\phi =1$ $\iff\phi =45^{\circ}$ .
This post has been edited 356 times. Last edited by Virgil Nicula, Nov 26, 2015, 8:39 AM

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