351. Problems with areas.

by Virgil Nicula, Jul 23, 2012, 1:31 AM

PP11. Let an acute $\triangle ABC$ and a mobile point $D\in (AB)$ . Let $M$ , $N$ be the feet of the perpendiculars from $D$ to $BC$ , $AC$ respectively.

Let $H_1$ , $H_2$ be the orthocentres of triangles $MNC$ , $MND$ respectively. Prove that the area $[AH_1BH_2]$ is constant.

Proof. Observe that $H_1NDM$ and $H_2NCM$ are parallelograms. Apply an well-known property to them :

$\left\{\begin{array}{ccc} H_1NDM\ : & \implies & \delta_{AB}(H_1)+\delta_{AB}(D)=\delta _{AB}(M)+\delta_{AB}(N)\\\\ H_2NCM\ : & \implies & \delta_{AB}(H_2)+\delta_{AB}(C)=\delta _{AB}(M)+\delta_{AB}(N)\end{array}\right|\implies$ $\delta_{AB}(H_1)+\delta_{AB}(H_2)=\delta_{AB}(C)\implies$ $[AH_1BH_2]=[ABC]$ .

PP12. Let $\triangle ABC$ with $AB=c=13$ , $AC=b=12$ and $[ABC]=S=72$ . Find $a=BC$ .

Proof 1 (without trig.). Appear two cases $B\in\left\{B_1,B_2\right\}$ , where $A$ is midpoint of $[B_1B_2]$ , i.e. $CA$ is median in $\triangle B_1CB_2$ . Suppose w.l.o.g. that $m\left(\widehat{CAB_1}\right)\le 90^{\circ}$ . Thus,

$AB_1=AB_2=c=13$ . Denote $\{X_1,X_2\}\subset AC$ such that $B_1X_1\perp AC$ and $B_2X_2\perp AC$ . Prove easily that $B_1X_1=B_2X_2=\frac {2S}{b}=12$ and $AX_1=AX_2=5\implies$

$CX_1=7\ ,\ CX_2=17\implies$ $\left\{\begin{array}{ccc} a_1=CB_1=\sqrt {B_1X_1^2+CX_1^2}=\sqrt {12^2+7^2}=\sqrt {193}\\\\ a_2=CB_2=\sqrt {B_2X_2^2+CX_2^2}=\sqrt {12^2+17^2}=\sqrt {433}\end{array}\right|$ $\implies a=BC\in\left\{\sqrt {193},\sqrt {433}\right\}$ .

Proof 2 (metric - brute force). Denote the midpoint $M$ of $[AB]$ and the projection $D$ of $C$ on $AB$ . Thus, $CD\perp AB$ $\iff$ $CA^2-CB^2=DA^2-DB^2$ $\iff$ $2c\cdot MD=$ $\left|a^2-b^2\right|$ .

Therefore, $MC^2=CD^2+MD^2$ $\iff$ $m_c^2=h_c^2+MD^2$ $\iff$ $4c^2m_c^2=4c^2h_c^2+$ $\left(a^2-b^2\right)^2$ $\iff$ $c^2\left[2\left(a^2+b^2\right)-c^2\right]=$ $16S^2+\left(a^2-b^2\right)^2$ $\iff$

$\boxed{16S^2=2\left(a^2b^2+b^2c^2+c^2a^2\right)-\left(a^4+b^4+c^4\right)}$ $\iff$ $a^4-2a^2\left(b^2+c^2\right)+$ $16S^2+\left(b^2-c^2\right)^2=0$ . Observe that

$\Delta '=\left(b^2+c^2\right)^2-$ $16S^2-\left(b^2-c^2\right)^2$ $\iff$ $\Delta '=4b^2c^2-16S^2$ . In conclusion, $a^2\in\left\{b^2+c^2\pm 2\sqrt {(bc-2S)(bc+2s)}\right\}$ .

In our particular case obtain $a^2\in \left\{144+169\pm 2\sqrt {12\cdot 300}\right\}=$ $\left\{313\pm 120\right\}=$ $\{433,193\}$ , what means $a\in\{\sqrt {193},\sqrt{433}\}$ .

Proof 3 (trigonometric). Since $2S=bc\sin A$ obtain that $\sin A=\frac {2\cdot 72}{12\cdot 13}=\frac {12}{13}\iff$ $\cos A\in \left\{\pm\frac {5}{13}\right\}$ . Since $a^2=b^2+c^2-2bc\cos A$

obtain that $a^2=144+169\pm 2\cdot 12\cdot 13\cdot\frac {5}{13}\iff$ $a^2\in\{313\pm 120\}\iff$ $a\in\left\{\sqrt {193},\sqrt {433}\right\}$ .

Remark. $\boxed{\tan\frac A2=\frac {\sin A}{1+\cos A}}=$ $\frac {\frac {12}{13}}{1\pm\frac {5}{13}}\implies$ $\tan\frac A2\in\left\{\frac 23,\frac 32\right\}$ . I"ll use an well-known relation $\boxed{S=s(s-a)\tan\frac A2}$ .

In conclusion, $b+c=25$ and $\left\{\begin{array}{ccccccc} \tan\frac A2=\frac 23 & \implies & s(s-a)=108 & \implies & (b+c)^2-a^2=432 & \implies & a=\sqrt {193}\\\\ \tan\frac A2=\frac 32 & \implies & s(s-a)=48 & \implies & (b+c)^2-a^2=192 & \implies & a=\sqrt {433}\end{array}\right|$ .

PP13. Let acute $\triangle ABC$ with circumcentre $O$ , centroid $G$ and orthocenter $H$ . Let $\left\{\begin{array}{ccc} D\in BC & ; & OD\perp BC\\\ E\in CA & ; & HE\perp CA\\\ F\in (AB) & ; & FA=FB\end{array}\right|$ . Suppose that $[ODC]=[HEA]=[GFB]$ . Find $m\left(\widehat{ACB}\right)$ .

Proof. Show easily $4\cdot [ODC]=R^2\sin 2A\ ,\ [HEA]=$ $R^2\cos^2A\sin 2C\ ,\ 3\cdot [GFB]=$ $R^2\sin A\sin B\sin C$ . Let $\left\{\begin{array}{c} \tan A=x\\\ \tan B=y\\\ \tan C=z\end{array}\right|$ , where $x+y+z=xyz$ .

From $\mathrm{area} (ODC)=\mathrm{area}(HEA)=\mathrm{area}(GFB)$ obtain that $\mathrm{area}(ODC)=\mathrm{area}(GFB)\implies$ $3\cos A=2\sin B\sin C\implies$ $3(1-yz)+2yz=0\implies \boxed{yz=3}\ (1)\ .$

$[ODC]=[HEA]\implies$ $\sin A=2\cos A\sin 2C\implies$ $x=\frac {4z}{1+z^2}\implies$ $\boxed{4z=x(1+z^2)}\ (2)\ .$ Thus, $x+y+z=xyz\stackrel{(1)}{\implies}\ y$ $+z=2x\stackrel{(1\wedge 2)}{\implies}$

$\frac 3z$ $+z=\frac {8z}{1+z^2}\implies$ $(z^2+1)(z^2+3)=8z^2\implies$ $z^4-4z^2+3=0\implies$ $z^2\in\{1,3\}\implies$ $z\in\left\{1,\sqrt 3\right\}\implies$ $\boxed{C\in\left\{\frac {\pi}{4},\frac {\pi}{3}\right\}}$ . Nice problem !

PP14. Let $P$ be a point on the circumcircle $w=C(O,R)$ of the given $\triangle ABC$ . Denote $Q\in AP\cap BC$ . From $Q$ perpendiculars are drawn

to $AB$ and $AC$ with feet $K$ and $M$ respectively. Find all points $P$ such that quadrilateral $AKPM$ and triangle $ABC$ have equal areas.

Proof. Denote $m\left(\widehat{BAP}\right)=x$ . Thus, $m\left(\widehat{CAP}\right)=A-x$ , $\boxed{AP=2R\sin(C+x)}$ and $a\cdot AQ\sin (B+x)=2S$ $\implies$ $\boxed{AQ=\frac {2S}{a\sin (B+x)}}$ . Observe that

$\left\{\begin{array}{ccc} AK & = & AQ\cos x\\\ AM & = & AQ\cos (A-x)\end{array}\right|$ . Therefore, $[AKPM]=[ABC]\iff$ $[AKP]+[AMP]=S\iff$ $AP\cdot AK\cdot \sin x+AP\cdot AM\cdot\sin (A-x)=2S\iff$

$AP\cdot AQ\cdot \left[\cos x\sin x+\cos (A-x)\sin (A-x)\right]=2S\iff$ $2R\sin (C+x)\cdot \frac {2S}{a\sin (B+x)}\cdot\frac 12\left[\sin 2x+\sin 2(A-x)\right]=2S\iff$

$\sin (C+x)\sin A\cos (2x-A)=\sin A\sin(B+x)\iff$ $\sin (C+x)\cos (2x-A)=\sin(B+x)\iff$ $\sin (C-A+3x)+\sin (A+C-x)=$

$2\sin (B+x)\iff$ $\sin (C-A+3x)=\sin (B+x)\iff$ $C-A+3x=B+x\ \ \vee\ \ (C-A+3x)+(B+x)=180^{\circ}\iff$

$2x=A+B-C\ \ \vee\ \ 4x=180^{\circ}+A-B-C\iff$ $x=90^{\circ}-C\ \ \vee\ \ x=\frac A2$ . Thus, $P\in \{P_1,P_2\}$ , where $[AP_1]$ is diameter of $w$ and ray $[AP_2$ is bisector of $\widehat{BAC}$ .

Verification.

$\blacktriangleright\ P:=P_1$ . In this case $BKQP$ , $CMQP$ are trapezoids and $\left\{\begin{array}{ccccc} PB\perp AB & \iff & QK\parallel PB & \iff & [ABQ]=[APK]\\\\ PC\perp AC & \iff & QM\parallel PC & \iff & [ACQ]=[APM]\end{array}\right|$ $\bigoplus\implies [AKPM]=[ABC]$ .

$\blacktriangleright\ P:=P_2$ . In this case $AK=AM=AQ\cos\frac A2$ and (an well-known property) $AQ\cdot AP_2=bc$ . Therefore, $2\cdot [AKP_2M]=$

$2\cdot \left([AKP_2]+[AMP_2]\right)=$ $AP_2\cdot AK\sin \frac A2+AP_2\cdot AM\cdot\sin\frac A2=$ $2\cdot AP_2\cdot AQ\cos\frac A2\sin\frac A2=bc\sin A=2S$ .

PP15. Let $D$ , $E$ , $F$ be the points on the sides of triangle $\Delta ABC$ respectively such that $DE\perp CA$ , $EF\perp AB$ , $FD\perp BC$ . Denote the intersections

$M=AD\cap BE$ , $N=BE\cap CF$ , $P=CF\cap AD$ . Prove that $\frac{BD}{BC}+\frac{CE}{CA}+\frac{AF}{AB}=1$ and $[BMD]+[CNE]+[APF]=[MNP]$ .

Proof (Sunken Rock). Take $X\in AB$ so that $DX\bot DE$ , $Y\in AB$ so that $EY\parallel BC\implies$ $\frac{BD}{BC}=\frac{BX}{AB}\ (\ 1\ )$ and $\frac{CE}{AC}=\frac{BY}{AB}\ (\ 2\ )$ . So must prove $BY=FX\ (*)$ .

As constructed, $\Delta DEF\sim\Delta CAB\implies \frac{EF}{DE}=\frac{AB}{AC}\ (\ 3\ )$ . Since $DX\bot DE$ and $FD\bot BC$ we get $\angle CDE=\angle FDX$ , but $DEFX$ is cyclic, so $\angle FDX=\angle FEX$ .

Thus $\angle FEX=\angle CDE$ and $\Delta FEX\sim\Delta EDC$ . So $\frac{CE}{FX}=\frac{DE}{EF}\ (\ 4\ )$ . With $(3)$ we got $\frac{FX}{AB}=\frac{CE}{AC}$ . Since $\frac{BX+BY+AF}{AB}=1$ we get the wanted $(*)$ .

PP16. Let $ABCD$ be a rectangle and $E\in (CD)$ , $F\in (BC)$ so that $AEF$ is an equilateral triangle. Prove that $[ABF]+[ADE]=[CEF]$ .

Proof 1 (trigonometric). Suppose w.l.o.g. that $AE=EF=FA=1$ and denote $\left\{\begin{array}{c} m\left(\widehat{BAF}\right)=x\\\ m\left(\widehat{DAE}\right)=y\end{array}\right|$ , where $x+y=30^{\circ}$ . Thus, $[ABF]+[ADE]=[CEF]\iff$

$\cos x\sin x+\cos y\sin y=\cos \left(x+30^{\circ}\right)\cos\left(y+30^{\circ}\right)\iff$ $\sin 2x+\sin 2y=\cos 90^{\circ}+\cos (x-y)\iff$ $2\sin (x+y)\cos (x-y)=\cos (x-y)$ , what is truly.

Proof 2 (complex numbers). Denote $AD=BC=a$ , $AB=CD=b$ and $BF=n$ , $CE=m$ . Thus, $A(bi)$ , $B(0)$ , $C(a)$ , $D(a+bi)$ and $F(n)$ , $E(a+mi)$ . Therefore,

$AEF$ is equilateral $\iff$ $(bi)^2+(a+mi)^2+n^2=n\cdot bi+n\cdot (a+mi)+bi\cdot (a+mi)\iff$ $\left\{\begin{array}{cc} a^2-an+n^2=b^2-bm+m^2 & (1)\\\\ a(b-m)+bn=m(a-n) & (2)\end{array}\right|$ . The first relation is for

the existence of the equilateral triangle $AEF$ (compatibility of this problem) and the second relation is equivalently with the uor conclusion, i.e. $[ADE]+[ABF]=[CEF]$ .

Remark. Construct outside of the rectangle the equilateral triangles $BCX$ and $CDY$ . The relation $(1)$ means $FX=EY\iff$ the hypotesis is substantially.

Proof 3 (metric). Denote the midpoint $P$ of $[EF]$ , the projections $M$ , $N$ of $P$ on $BC$ , $CD$ respectively and $CF=2m$ , $CE=2n$ . Thus, $\left\{\begin{array}{c} PN=FM=MC=m\\\\ PM=NC=EN=n\end{array}\right|$ . Observe

that $ADEP$ , $ABFP$ are cyclical quadrilaterals and $\left\{\begin{array}{ccccccc} m\left(\widehat{PDN}\right)=30^{\circ} & \implies & \cot\widehat{PDN}=\frac {DN}{PN} & \implies & DN=m\sqrt 3 & \implies & DE=m\sqrt 3-n\\\\ m\left(\widehat{PBM}\right)=30^{\circ} & \implies & \cot\widehat{PBM}=\frac {BM}{PM} & \implies & BM=n\sqrt 3 & \implies & BF=n\sqrt 3-m\end{array}\right|$ . Therefore,

$\left\{\begin{array}{c} AD=BC=n\sqrt 3+m\\\\ AB=CD=m\sqrt 3+n\end{array}\right|$ and $[ABF]+[ADE]=[ECF]\iff$ $AB\cdot BF+AD\cdot DE=CE\cdot CF\iff$ $\left(m\sqrt 3+n\right)\left(n\sqrt 3-m\right)+$ $\left(n\sqrt 3+m\right)\left(m\sqrt 3-n\right)=$

$4mn$ , what is true. Let $\left\{\begin{array}{c} AD=BC=a\\\\ AB=CD=b\end{array}\right|$ . Thus, $\left\{\begin{array}{c} m+n\sqrt 3=a\\\\ m\sqrt 3+n=b\end{array}\right|\implies$ $\left\{\begin{array}{c} m=\frac {b\sqrt 3-a}{2}\\\\ n=\frac {a\sqrt 3-b}{2}\end{array}\right|$ . Therefore, $\left\{\begin{array}{c} 0<2m<a\\\\ 0<2n<b\end{array}\right|\iff$ $\left\{\begin{array}{c} 0<b\sqrt 3-a<a\\\\ 0<a\sqrt 3-b<b\end{array}\right|\iff$

$\left\{\begin{array}{c} a<b\sqrt 3<2a\\\\ b<a\sqrt 3<2b\end{array}\right|\iff$ $\frac {1}{\sqrt 3}<\frac {\sqrt 3}{2}<\frac ab<\frac {2}{\sqrt 3}<\sqrt 3\implies$ $\frac {AD}{AB}\in \left(\frac {\sqrt 3}{2},\frac {2}{\sqrt 3}\right)$ . In conclusion, there is an equilateral $\triangle AEF$ with $E\in (CD)$ and $F\in (BC)$ $\iff$

$\frac {AD}{AB}\in\left(\frac {\sqrt 3}{2},\frac {2}{\sqrt 3}\right)\iff$ $\tan\widehat{ABD}\in\left(\frac {\sqrt 3}{2},\frac {2}{\sqrt 3}\right)$ .

An easy extension. Let $ABCD$ be a rectangle and $E\in (CD)$ , $F\in (BC)$ so that $AE=AF$ and $m\left(\widehat{EAF}\right)=2\phi$ . Prove that $[ABF]+[ADE]=[CEF]\cdot\cot\phi\cdot\cot 2\phi$ .

PP17. Let $P$ be the interior point of $\triangle ABC$ so that the interior segments through $P$ and what are parallel to the sides have same length $d$ . Find $d$ .

Proof 1. Let $\left\{\begin{array}{cc} \{E,F\}\subset (BC)\\\\ \{G,H\}\subset (CA)\\\\ \{I,J\}\subset (AB)\end{array}\right|$ so that $\left\{\begin{array}{c} P\in JG\cap IF\cap HE\\\\ JG\parallel BC\ ;\ EF=x\\\\ FI\parallel CA\ ;\ FC=y\\\\ HE\parallel AB\ ;\ BE=z\end{array}\right|$ . , where $\left\{\begin{array}{c} x+y+z=a\\\\ JG=IF=HE=d\\\\ PEF\sim HPG\sim IJP\sim ABC\end{array}\right|\implies$

$y+z=\frac ba\cdot (z+x)=\frac ca\cdot (x+y)=d$ . Therefore, $ad=\frac {y+z}{\frac 1a}=\frac {z+x}{\frac 1b}=\frac {x+y}{\frac 1c}=$ $\frac {2a}{\frac 1a+\frac 1b+\frac 1c}\implies$ $\boxed{d=\frac {2}{\frac 1a+\frac 1b+\frac 1c}}$ .

Proof 2 (nice). Denote $\left\{\begin{array}{cc} \{E,F\}\subset (BC)\\\\ \{G,H\}\subset (CA)\\\\ \{I,J\}\subset (AB)\end{array}\right|$ so that $P\in JG\cap IF\cap HE$ and $\left\{\begin{array}{c} JG\parallel BC\\\\ FI\parallel CA\\\\ HE\parallel AB\end{array}\right|$ . Denote the area $S=[ABC]$ and the distancies $u$ , $v$ , $w$ of $P$

to the sidelines $BC$ , $CA$ , $AB$ respectively. Therefore, $\boxed{au+bv+cw=2S}\ (*)$ and $\frac {GJ}{a}+\frac {FI}{b}+\frac {EH}{c}=$ $\sum \frac {h_a-u}{h_a}=$ $\sum\left(1-\frac {u}{h_a}\right)=$ $3-\sum\frac {au}{ah_a}=$

$3-\frac {au+bv+cw}{2S}\stackrel{(*)}{\implies}$ $\boxed{\frac {GJ}{a}+\frac {FI}{b}+\frac {EH}{c}=2}\ .$ In the particular case when $GJ=FI=EH=d$ obtain that $\boxed{\frac 2d=\frac 1a+\frac 1b+\frac 1c}$ . Remark that $d\le \frac 29\cdot (a+b+c)$ .

PP18. A circle of radius 2 is centered at $O$ . Square $OABC$ has side length 1 . Sides $\overline{AB}$ and $\overline{CB}$ are extended past $b$ to meet the circle at $D$

and $E$ respectively. What is the area of the shaded region in the figure, which is bounded by $\overline{BD}$ , $\overline{BE}$ and the minor arc connecting $D$ and $E$ ?

$[asy] defaultpen(linewidth(0.8)); pair O=origin, A=(1,0), C=(0,1), B=(1,1), D=(1, sqrt(3)), E=(sqrt(3), 1), point=B; fill(Arc(O, 2, 0, 90)--O--cycle, mediumgray); clip(B--Arc(O, 2, 30, 60)--cycle); draw(Circle(origin, 2)); draw((-2,0)--(2,0)^^(0,-2)--(0,2)); draw(A--D^^C--E); label("$A$", A, dir(point--A)); label("$C$", C, dir(point--C)); label("$O$", O, dir(point--O)); label("$D$", D, dir(point--D)); label("$E$", E, dir(point--E)); label("$B$", B, SW);[/asy]$

Proof. $\left\{\begin{array}{c} CE=\sqrt 3\\\\ BE=\sqrt 3-1\\\\ m\left(\widehat{COE}\right)=60^{\circ}\end{array}\right|$ . Shaded $BDE = \mathrm{area\ sector}\ DOE\ -\ 2\cdot [OBE]=$ $\frac 12\cdot r^2\cdot m\left(\widehat{DOE}\right)-$ $2\cdot \frac {OC\cdot BE}2=$

$\frac 12\cdot 4\cdot \frac {\pi}{6}-\left(\sqrt 3-1\right)=$ $\frac {\pi}{3}-\left(\sqrt 3-1\right)=$ $1+\frac {\pi}{3}-\sqrt 3$ . Let $w=C(O,\rho)$ be a circle and $\{X,Y\}\subset w$ so that $m\left(\widehat{XOY}\right)=\phi$

(radians). Remark that the length of the arc $XY$ is $\boxed{l(\phi )=\rho\phi}$ and the area of the sector $XOY$ is $\boxed{s(\phi )=\frac {l(\phi )\rho}{2}=\frac {\rho^2\phi}{2}}$ . See here

PP19. Let a convex $ABCD$ with $\left\{\begin{array}{ccc} O\in AC\cap BD\\\\ \{M,P\}\subset (AB)\\\\ \{N,Q\}\subset (A,D)\end{array}\right\|$ so that $\frac {MB}{MC}=\frac {PC}{PB}=\frac {ND}{NA}=\frac {QA}{QD}$ . Prove that $[MON]=[POQ]=\frac {MB\cdot MC}{BC^2}\cdot [ABCD]$ .

Proof. Let $S=[ABCD]$ and $\frac {MB}{m}=\frac {MC}{1}=\frac {BC}{m+1}$ and $\left\{\begin{array}{ccc} \nearrow\ [AOB]=a & ; & [BOC]=b\\\\ \searrow\ [COD]=c & ; & [DOA]=d\end{array}\right\|$ . Thus, $a+b+c+d=S$ . Let $R\in (AB)$ and $S\in (CD)$ so that

$\left\{\begin{array}{c} \nearrow\ MR\parallel AC\parallel NS\\\\ \searrow\ NR\parallel BD\parallel MS\end{array}\right\|$ . Thus, $\left\{\begin{array}{ccccc} \nearrow\ \frac {[MOS]}{[BCD]}=\frac {MS\cdot\delta_{BD}(M)}{BD\cdot\delta_{BD}(C)} & \implies & \frac {[MOS]}{[BCD]}=\frac {MS}{BD}\cdot\frac {BM}{BC}=\frac {CM}{CB}\cdot \frac {BM}{BC} & \implies & \frac {[MOS]}{[BCD]}=\frac {m}{(m+1)^2}\\\\ \searrow\ \frac {[NOS]}{[ACD]}=\frac {NS\cdot\delta_{AC}(S)}{AC\cdot\delta_{AC}(D)} & \implies & \frac {[NOS]}{[ACD]}=\frac {NS}{AC}\cdot\frac {CS}{CD}=\frac {BM}{BC}\cdot \frac {CM}{CB} & \implies & \frac {[NOS]}{[ACD]}=\frac {m}{(m+1)^2}\end{array}\right\|$ , Prove easily that

$\left\{\begin{array}{ccc} \nearrow\ \frac {[MCS]}{[BCD]}=\frac {[NAR]}{[DAB]}=\frac 1{(m+1)^2} & \implies & [MCS]+[NAR]=\frac S{(m+1)^2}\\\\ \searrow\ \frac {[SDN]}{[CDA]}=\frac {[MBR]}{[CBA]}=\frac {m^2}{(m+1)^2} & \implies & [SDN]+[MBR]=\frac {m^2S}{(m+1)^2}\end{array}\right\|$ $\implies$ $[MSN]=\frac 12\cdot [MSNR]=\frac 12\cdot \left\{S-\left([MCS]+[NAR]+[SDN]+[MBR]\right)\right\}=$

$\frac 12\cdot \left[S-\frac S{(m+1)^2}-\frac {m^2S}{(m+1)^2}\right]=$ $\frac S2\cdot\left[1-\frac 1{(m+1)^2}-\frac {m^2}{(m+1)^2}\right]\implies$ $[MSN]=\frac {mS}{(m+1)^2}$ . In conclusion, $[MON]=[MOS]+[NOS]-[MSN]=$

$\frac {m}{(m+1)^2}\cdot\left([BCD]+[ACD]-S\right)\implies$ $[MON]=\frac {m}{(m+1)^2}\cdot\left|[OCD]-[OAB]\right|$ . If denote $f(m)=\frac m{(m+1)^2}$ , then $f\left(\frac 1m\right)=f(m)\implies$

$\boxed{[MON]=[POQ]= \frac {m}{(m+1)^2}\cdot\left|[OCD]-[OAB]\right|}$ .

PP20. Let $ABC$ be an $A$ -right triangle with the incenter $I$ and $D\in AI\cap BC$ . Prove that the area $[DIC]=r^2\iff B=30^{\circ}$ .

Proof. $[DIC]=r^2\iff$ $r\cdot DC=2r^2\iff$ $DC=2r\iff$ $\frac {ab}{b+c}=b+c-a\iff$ $ab=(b+c)^2-a(b+c)\iff$

$ab=a^2+2bc-a(b+c)\iff$ $a^2-(2b+c)a+2bc=0\iff$ $(a-c)(a-2b)=0\left\{\begin{array}{cccc} \nearrow & a=c & \implies & \emptyset\\\\ \searrow & a=2b & \implies & B=30^{\circ}\end{array}\right\|$ .

P21 (Nguyen Viet Hung, Vietnam) <= click.

Proof (elementary). Suppose w.l.o.g. $c<b\ ,$ denote the midpoint $M$ of $[BC]\ ,$ $D\in AI\cap BC$ and $S\in AK\cap BC\ .$ Thus, $\left\{\begin{array}{cccc} \frac {SB}{c^2}=\frac {SC}{b^2}=\frac a{b^2+c^2} & \implies & \boxed{\ SB=\frac {ac^2}{b^2+c^2}\ } & (1)\\\\ \frac {DB}c=\frac {DC}b=\frac a{b+c} & \implies & \boxed{\ DB=\frac {ac}{b+c}\ } & (2)\end{array}\right\|$ $\implies$

$\left\{\begin{array}{cccc} SD=BD-BS\ \stackrel{1\wedge 2}{=}\ \frac {ac}{b+c}-\frac {ac^2}{b^2+c^2} & \implies & \boxed{\ SD=\frac {abc(b-c)}{(b+c)\left(b^2+c^2\right)}\ } & (3)\\\\ DM=BM-BD\ \stackrel{2}{=}\ \frac a2-\frac {ac}{b+c} & \implies & \boxed{\ DM=\frac {a(b-c)}{2(b+c)}\ } & (4)\end{array}\right\|$ $\implies$ $SM=SD+DM\ \stackrel{3\wedge 4}{=}\ \frac {abc(b-c)}{(b+c)\left(b^2+c^2\right)}+\frac {a(b-c)}{2(b+c)}=$ $\frac {a(b-c)}{b+c}\cdot\left(\frac {bc}{b^2+c^2}+\frac 12\right)=$

$\frac {a(b-c)(b+c)^2}{2(b+c)\left(b^2+c^2\right)}=$ $\frac {a\left(b^2-c^2\right)}{2\left(b^2+c^2\right)}\implies$ $\boxed{\ SM=\frac {a\left(b^2-c^2\right)}{2\left(b^2+c^2\right)}\ }\ (5)\ .$ With the van Aubel's theorem can prove easily $\left\{\begin{array}{cccc} \frac {KA}{b^2+c^2}=\frac {KS}{a^2}=\frac {AS}{a^2+b^2+c^2} & \implies & \boxed{\ \frac {AK}{AS}=\frac {b^2+c^2}{a^2+b^2+c^2}\ } & (6)\\\\ \frac {IA}{b+c}=\frac {ID}a=\frac {AD}{a+b+c} & \implies & \boxed{\ \frac {AI}{AD}=\frac {b+c}{a+b+c}\ } & (7)\end{array}\right\|\ .$

Thus, $\boxed{\ [GIK]=[AKI]+[AIG]-[AKG]\ }\ (*)\ ,$ where $\ :\ \left\{\begin{array}{cccc} \frac {[AKI]}{[ABC]}=\frac {[AKI]}{[ASD]}\cdot\frac {[ASD]}{[ABC]}=\frac {AK}{AS}\cdot\frac {AI}{AD}\cdot\frac {SD}{BC}\ \stackrel{3\wedge 6\wedge 7}{=}\ \frac {b^2+c^2}{a^2+b^2+c^2}\cdot\frac {b+c}{a+b+c}\cdot\frac {bc(b-c)}{(b+c)\left(b^2+c^2\right)} & \implies & \frac {[AKI]}{[ABC]}=\frac {bc(b-c)}{\sum a\cdot\sum a^2} & (8)\\\\ \frac {[AIG]}{[ABC]}=\frac {[AIG]}{[ADM]}\cdot\frac {[ADM]}{[ABC]}=\frac {AI}{AD}\cdot\frac {AG}{AM}\frac {DM}{BC}\ \stackrel{4\wedge 7}{=}\ \frac {b+c}{a+b+c}\cdot\frac 23\cdot\frac {b-c}{2(b+c)}=\frac {b-c}{3(a+b+c)} & \implies & \frac {[AIG]}{[ABC]}=\frac {b-c}{3\sum a} & (9)\\\\ \frac {[AKG]}{[ABC]}=\frac {[AKG]}{[ASM]}\cdot \frac {[ASM]}{[ABC]}=\frac {AK}{AS}\cdot\frac {AG}{AM}\cdot\frac {SM}{BC}\ \stackrel{5\wedge 6}{=}\ \frac {b^2+c^2}{a^2+b^2+c^2}\cdot \frac 23\cdot\frac {b^2-c^2}{2\left(b^2+c^2\right)}=\frac {b^2-c^2}{3\left(a^2+b^2+c^2\right)} & \implies & \frac {[AIG]}{[ABC]}=\frac {b^2-c^2}{3\sum a^2} & (10)\end{array}\right\|\ .$

The relation $(*)$ implies $\frac{[GIK]}{[ABC]}=\left|\ \frac {[AKI]}{[ABC]}+\frac {[AIG]}{[ABC]}-\frac{[AKG]}{[ABC]}\ \right|\ \stackrel{8\wedge 9\wedge 10}{=}\ \left|\ \frac {bc(b-c)}{\sum a\cdot\sum a^2}+\frac {b-c}{3\sum a}-\frac {b^2-c^2}{3\sum a^2}\ \right|=$ $\frac {|b-c|\cdot\left|3bc+a^2+b^2+c^2-(b+c)(a+b+c)\right|}{3(a+b+c)\left(a^2+b^2+c^2\right)}=$

$\frac {|b-c|\cdot\left|3bc+a^2+b^2+c^2-(b+c)^2-a(b+c)\right|}{3(a+b+c)\left(a^2+b^2+c^2\right)}=$ $\frac {|b-c|\cdot\left|a^2-a(b+c)+bc\right|}{3(a+b+c)\left(a^2+b^2+c^2\right)}=$ $\frac {|b-c|\cdot\left|(a-b)(a-c)\right|}{3(a+b+c)\left(a^2+b^2+c^2\right)}$ $\implies$ $\boxed{\frac{[GIK]}{[ABC]}=\frac{|(a-b)(b-c)(c-a)|}{3(a+b+c)\left(a^2+b^2+c^2\right)}}\ .$

This post has been edited 238 times. Last edited by Virgil Nicula, Jan 21, 2018, 8:59 AM

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263. Newton's sums. Applications.

262. IMO Shortlist 2005, Geometry 6th.

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258. Some difficult and nice problems with complex numbers.

257. Seeking roots of 2th equation with complex coef.

256. An extremum problems with complex numbers.

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255. A line which is parallelly with OH.

254. An interesting C. Mateescu's geometrical inequality.

253. An old algebraic inequality.

252. OI=f(A,R) in certain conditions.

251. Problems of Analytical Geometry.

250. An application of the Euler's relation.

249. An usual and nice metrical relations in quadrilateral.

248. Pagina instructiva.

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246. R1-R2=OI (nice).

245. An inequality in an acute triangle.

244. Some properties of cyclic orthodiagonal quadrilaterals.

243. Nice ! (a+b) is constant.

242. A general geometrical problems.

241. Areas in a triangle.

240. Some interesting geometrical inequalities.

239. Algebraic marathon.

238. Some metrical problems.

237. Some problems with ... problems.

236. A locus with the metrical relation.

235. Proofs of some geometry problems with complex numbers.

234. Some simple and nice problems from contests.

233. An interesting identity and an easy & nice inequality.

February 2011

232. Some geometrical inequalities (own).

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231. A conditioned minimum value.

230. Some usual synthetical properties in a triangle.

229. Concurs online MATEFBC, ed. 5 -subiect 2.

228. OJM - Romania, 2010 problem 3.

227. IMO ShortList 2003 (problem 5).

226. Some inequalities from Calculus (Real Analysis).

225. An interesting problems from Kunny.

224. Some nice and easy inequalities in a triangle.

223. An inequality with complex numbers-RMO shortlist 2010.

222. An extension of probl. 2 from IMO 2009.

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219. A very nice geometrical inequality.

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216. Nice ! Find the length of hypotenuse (Belarus - 2010)

215. Limit of the unique root for polynominal equations.

214. A range of some functions.

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212. A simple applications of the Casey's theorem.

211. Pretty hard problem !

210. Tangential circle of the triangle ABC.

209. Nice problem, nice proof !

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207. Some interesting problems of Toshio Seimiya, Japan.

206. Four problems with extremum in a quadrilateral.

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204. Saraghin 2010 (three geometry problems).

203. An application of the Pascal's permutation theorem.

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201. Applications of Desarques' and Pappus' theorems.

200. Some nice applications of the inversion.

199. Some properties of symmedian. Two extensions.

198. Generalized Carnot's lemma.

197. Triangles outside of a triangle and concurrence.

195. Semicircle. Nice application of harmonical division.

194. The Lalescu's sequence. Properties.

193. An inequality with n! .

192. An easy and nice problem of Valentin Vornicu.

191. Six nice problems with the incircle of a triangle.

190. Very nice ! Coculescu's Contest seniors 2010.

189. Pascal's theorem and some nice and easy applications.

188. Some problems from vectorial geometry.

187. Distancies.

186. Without the l'Hospital's rules.

185. The Durrande's relation.

184. An interesting geometry problem from RMO 2010.

183. A remarkable characterization of equilateral triangle.

182. Another two nice problems with complex numbers (own).

181. Two special inequalities from CRUX (own).

180. Own proposed problem from CRUX (with complex numbers).

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178. Some nice and simple geometry problems.

177. A nice proof of one identity in a triangle.

176. Another nice problems for middle school.

175. A Geometric Proof (without trig.) of Heron's Formula.

174. Some interesting properties of modulus (middle school).

173. An interesting trigonometric equations.

172. The Zelinsky's problem.

171. Another "slicing" proposed problems (middle school).

October 2010

170. A nice "slicing" proposed problem for middle school.

169. Some metrical problems in a circle (III).

168. Some properties of the Lemoine's point.

167. Nice problems - OIM, 1995 and ... Toshio Seimyia.

166. Some constant geometrical expressions.

165. Some metrical problems (5) in a circle (II).

164. Some metrical problems (5) in a circle.

163. Applications of Ptolemy's inequality. Ex. - China,1998.

162. Generalization of proposed problem from Mongolia, 2000.

161. A difficult inequality (barycentrical coordinates)

160. A very nice "slicing" problem.

159. Multiple derivatives in a point.

158. Very nice and difficult limits.

157. Steinbart's theorem. Applications.

156. Some problems with perpendicularities.

155. A constant ratio in a triangle ABC.

154. Difficult inequality (without derivatives eventually).

153. Geometrical minimum/maximum.

152. Concurrence, cyclically, metrics, symmedian a.s.o.

151. Miscellaneous problems for the admission at math.

150. Some (5) problems with fixed points.

149. Some (8) problems with collinearity or concurrence.

148. A special class of systems.

147. A problem with two polynominals.

146. Synthetic, metric, trigonometric and analitycal proofs.

145. Spain MO P5 - Point on median.

144. A conditioned concurrency in a triangle.

143. Assess of a ratio (nice and difficult problem).

142. Inscribed/circumscribed quadrilateral w.r.t. ABC.

141. N is interior <==> 1/3<a/b<3 (CRUX Mathematicorum).

140. A synthetical property <==> linear/angled relation.

139. A caracterization of a right angle in a triangle.

138. Generalized "slicing" problems (own).

137. Stewart's relation.

136. A geometrical interpretation of a system (own).

135. Three strong geometrical inequalities.

September 2010

134. Some problems with projections/perpendiculars (2).

133. Some problems with projections/perpendiculars (1).

132. Asupra unei inegalitati intr-un triunghi.

131. A stronger Form of the Steiner-Lehmus Theorem (own).

130. A triangle and a fixed point.

129. BX=CY and three properties.

128. Value of IG and two inequalities in neighbourhood of 0.

127. Pagina educativa.

126. Routh's theorem.

125. I like very much this Darij Grinberg's message.

124. A nice and easy problem with a right-angled triangle.

123. My extension and its proof.

122. An identity for an equilateral triangle. Extension.

121. Viete's relations with cos A , cos B , cos C.

120. Two equal neighbouring angles/sides in a quadrilateral.

119. Inspired by a Miculita's problem.

118. An usual remarkable geometrical locus.

117. Some applications of harmonic division/quadrilateral.

116. The length of AE (nine methods !).

115. An difficult equation with one radical.

114. A property of tangential triangle for ABC.

113. Range for a function.

112. The equivalence relation ".s.s."

111. Simple, but interesting ...

110. German TST 2004 - Exam V , Problem 1

109. Two equal areas ==> find A.

108. Old, short and nice proof of Euler's relation.

107. XY parallel BC and X,Y are isogonal/izotomic points.

106. OLM - Bucuresti (geometrie).

105. A problem with three circles.

104. Concursul Revistei de Matematica din Galati (RMG).

103. Metrical usual relations in triangle. Applications.

102. Characterization of a metrical relation in a triangle.

101. O problema cu numere complexe.

100. Bobiller's theorem.

99. Some simple and nice geometry problems.

98. Symmedian & median (metrical relations).

97. Problem of butterfly.

August 2010

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95. Three circles in an angle.

94. Equations of angle bisectors (interior and exterior).

93. The Appolonius' construct problems.

92. CA , CB tangent to parabola - OIMU 2008 Problem 4.

91. JBMO 2010, Problem 3.

90. Nice and difficult inequality in ABC (own).

89. Similarity (parallel/antiparallel).

88. A nice geometric locus conditioned metrically.

87. An interesting vectorial identity.

86. A very nice maxmin.

85. Problem "slicing" : 60-24-12.

84. IRAN national math olympiad - 2010

83. A metrical characterization of concurrence.

82. AA' , BN , CM concur in Gergonne point.

81. Sawayama and Thebault’s theorem.

80. Three concurrent perpendiculars (the Carnot's lemma).

79. A quadrilateral with many perpendicularities.

78. Jakobi's theorem.

77. Two equivalent perpendicularities.

76. A parallelogram and a circle (nice !).

75. Extended Steinbart Theorem.

74. Inegalitatea Erdos-Mordell.

73. A nice problems with a square and a perpendicularity.

July 2010

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70. Colinearity with pole/polar - H\in PQ .

71. An angle questions (synthetic/trigonometric proofs).

69*1. Slicing problem 3.

69. Position of x1, x2 w.r.t. a given real number k.

68. Some nice applications of "pole/polar".

67. Remarkable algebraic/geometrical inequalities (1).

66. Remarkable perpendicularities in a triangle.

65. Parallel to BC through I .

64. Problems of the type "instant" (at most one minute).

63. A nice problem with radical center (livetolove212).

62. Nice and easy problem with a right triangle.

61. IMO Shortlist 2009 (very nice problem and its proof).

60. Mediteranean M.C. 2010 Problem 3.

59. IMO 2010, problem 4.

58. A cevian and two incircles.

57. Opinii si comentarii relativ la inv. rom. II

56. A extremum problem with areas.

55. Two important circles - mixtilinear incircle/excircle.

54. Cinci grade de libertate.

53. An remarkable concurrency.

52. Harmonical division.

51*1. JBTST-3/2010 Aplicatii la proprietatea fluturelui.

51. A nice problem with perpendicularity from Jaime.

50. IMAC 2010 seniors, the first day.

49. The midpoint of a cevian in an acute triangle.

June 2010

48. An inequality in a triangle : h_a+max{b,c} ...

47. Interesting and difficult identities in a triangle.

46. Outside of a quadrilateral.

45. \cos B+\cos C=1 <==> nice property.

44. Relatia IA=IO sau IA=IM intr-un triunghi ABC etc.

43. A "slicing" problem with a parameter.

42. IMAC 2010 Problema 2.

41. ONM 2010 Calarasi Problema 3.

May 2010

40. Lema lui Haruki.

39. A nice and remarkable problem with Brocard's point.

38. Minimum area of triangle touching with semicircle.

37. Own extension of the Steiner-Lehmus' problem.

36. ONM (juniori) - 2010 Iasi, problema 4.

35. Nice problem from Arqady (Eduard_Tur).

34. IMO Shortlist 2002 geometry problem 7.

April 2010

32. Linkuri diverse.

31.Some nice metrical problems.

29. Inegalitate integrala de la Kunny.

28. A fost odată ...

27.Problem of Darij Grinberg (I - centroid of triangle DEF).

26. Outside of a triangle.

25. Geometric equations.

24. A fine and cool inequalities.

23. A very nice exponential inequality (own - from CRUX).

22. Some interesting limits (sequence/function).

21 bis. Some problems with complex numbers.

21. Probleme de geometrie (Yetty & I).

20. O ineg.cu suma de AG/AI .

19. Own proposed problems in CRUX Mathematicorum - Canada.

18. O problema a lui Toshio Seimiya.

17. O inegalitate "tare" intr-un triunghi.

16. Length of the (interior) angle bisector (10 proofs).

15. Inequalities stronger than Panaitopol's.

14. Opinii si comentarii relativ la inv. rom.

13. Inegalitatea de la Sorin Borodi.

12. Inegalitatea I. Maftei & S. Radulescu (2).

11. Inegalitatea I. Maftei & S. Radulescu (1).

10. Walker's inequality and its extension.

9. Inegalitatea HO+HA+HB+HC >= 3R .

8. The first problem Shortlist ONM 2010 Romania.

7. Some interesting "slicing" problems.

6. Interesting problems from JBMO.

5. Theoretical preliminary for inequalities.

4. Remarkable geometrical inequalities (2).

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godly blog. opopop
by OlympusHero, Dec 30, 2020, 6:08 PM
long blog
by MrMustache, Nov 11, 2020, 4:52 PM
372554 views!
by mrmath0720, Sep 28, 2020, 1:11 AM
wow... i am lost.

369302 views!

-piphi
by piphi, Jun 10, 2020, 11:44 PM
That was a lot! But, really good solutions and format! Nice blog!!!!
by CSPAL, May 27, 2020, 4:17 PM
impressive
awesome. 358,000 visits?????
by OlympusHero, May 14, 2020, 8:43 PM
Nice blog.
The element of Mathematics is dense here.
I love the style of your posts.
Thanks for writing the blog! I can learn Geometry from here too.
by epsilonist, Jun 27, 2011, 6:00 AM
Thank you very much for this blog.
by Affine, May 7, 2011, 6:20 PM
You seem to have solved every existing problem.
by Goutham, May 2, 2011, 10:59 AM
I love your posts, man (those which I understand)! It's interesting how you punish integrals with recurrence relations and sometimes make seemingly unrelated integrals evaluate each other!
by Caspar, Apr 11, 2011, 2:36 PM
Thank you a lot dear Virgil for an interesting and instructive blog.
by vittasko, Mar 12, 2011, 9:42 AM
Interesting Blog
by ShahinBJK, Mar 12, 2011, 5:07 AM
Interesant, d-le Nicula.
by silent_control, Mar 3, 2011, 6:00 PM
Yours is the largest blog full of math equations as far as I have known! Congrats!
by Goutham, Jan 30, 2011, 2:06 PM
15001th view.
by fries4guys, Jan 20, 2011, 4:57 PM
Thanks to all !
by Virgil Nicula, Dec 11, 2010, 8:32 PM
El_Ectric:
You can either copy some parts from the blog post, paste it at "Search Blogs" (which is lower on the page, after "Blog Stats") and find the whole post.

You're welcome.
by Thomas_, Dec 8, 2010, 6:42 PM
Thanks Thomas_.
by El_Ectric, Dec 8, 2010, 4:21 PM
El_Ectric: Try to open this blog with Internet Explorer (browser). If it still doesn't work, just copy the post you're interested in, and than paste it into Word Document, or almost any kind of reading documents and you'll have all the details.

Nice blog!
by Thomas_, Nov 24, 2010, 4:59 PM
Dear Virgil !

Congratulations for this blog !
Babis stergiou - Greece
by stergiu, Nov 12, 2010, 6:15 AM

72 shouts

My (math)blog

351. Problems with areas.

by Virgil Nicula, Jul 23, 2012, 1:31 AM

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