317. Some interesting geometry problems for middle school.

by Virgil Nicula, Sep 15, 2011, 12:33 AM

PP1 (Sankt-Petersburg, 2003). Let $ABC$ be a triangle with the incenter $I$ and the midpoint $M$ of $[BC]$ . Prove that $\boxed{\ IB\perp IM\ \iff\ a+b=3c\ }$ .

Method 1 (Paul Stoienescu - very nice !). Denote the midpoint $N$ of the side $[AC]$ . Thus, $IB\perp IM\iff$ $m\left(\widehat{IMB}\right)=m\left(\widehat{IMN}\right)=90^{\circ}-\frac B2\iff$

$MN$ is tangent to the incircle $\iff$ $ABMN$ is circumscriptible $\iff$ $AB+MN=AN+BM\iff$ $c+\frac c2=\frac b2+\frac a2\iff a+b=3c$ .

Method 2. Denote $D\in BC\ ,\ ID\perp BC$ . Therefore, $IB\perp IM\iff IB^2=BD\cdot BM\iff$ $\frac {ac(s-b)}{s}=(s-b)\cdot\frac a2\iff s=2c\iff a+b=3c$ .

Method 3 Suppose w.l.o.g. $b>c$ . Denote $D\in BC\ ,\ ID\perp BC$ . Apply the theorem of the altitude in $\triangle BIM$ what is $I$ -right: $IB\perp IM\iff$

$ID^2=DB\cdot DM\iff$ $r^2=(s-b)\cdot \frac {b-c}{2}\iff$ $2sr^2=s(s-b)(b-c)\iff$ $2(s-a)(s-b)(s-c)=s(s-b)(b-c)\iff$

$4(s-a)(s-c)=2s(b-c)\iff$ $b^2-(a-c)^2=(a+b+c)(b-c)\iff$ $b^2-a^2+2ac-c^2=ab-ac+b^2-c^2\iff$ $3c=a+b$ .

Method 4. From $m\left(\widehat{BIC}\right)=90^{\circ}+\frac A2$ obtain that $IB\perp IM\iff$ $\widehat{MIC}\equiv\widehat{IAC}\iff$ $\triangle MIC\sim \triangle IAC\iff$

$\frac {IC}{AC}=\frac {MC}{IC}\iff$ $IC^2=AC\cdot MC\iff$ $\frac {ab(s-c)}{s}=b\cdot\frac a2\iff$ $s=2c\iff a+b=3c$ .

An interesting lemma !

Lemma. Let $\triangle ABC$ with orthocenter $H$ . Denote $D\in AH\cap BC$ , $E\in BH\cap AC$ . Then $\boxed{AD^2=\overline{BD}\cdot\overline{DC}+\overline{AE}\cdot\overline{AC}}$ .

Proof. Denote the circumcircle $w$ of $\triangle ABC$ and $\{A,L\}=AH\cap w$ . Observe that $\overline{AH}\cdot\overline {AD}=\overline {AE}\cdot\overline{AC}\iff$ $(\overline{AD}+\overline {DH})\cdot\overline {AD}=$

$\overline {AE}\cdot\overline{AC}\iff$ $AD^2=\overline{HD}\cdot\overline{AD}+\overline {AE}\cdot\overline{AC}\iff$ $AD^2=\overline{LD}\cdot\overline{DA}+\overline {AE}\cdot\overline{AC}\iff$ $AD^2=\overline{BD}\cdot\overline{DC}+\overline{AE}\cdot\overline{AC}$ .

PP2 (Mihai Miculita). In $\triangle ABC$ denote the midpoint $M$ of $[BC]$ and the projections $E$ , $F$ of $M$ on $AB$ , $AC$ respectively. Prove that $ME^2+MF^2=\overline{AE}\cdot\overline{EB}+\overline{AF}\cdot\overline{FC}$ .

Proof. Apply the upper lemma to $\triangle AMB$ and $\triangle AMC\ :$ $\left\{\begin{array}{c} ME^2=\overline{AE}\cdot\overline{EB}+\overline{MD}\cdot\overline{MB}\\\\ MF^2=\overline{AF}\cdot\overline{FC}+\overline{MD}\cdot\overline{MC}\end{array}\right|\ \bigoplus\ \implies\ ME^2+$ $MF^2=\overline{AE}\cdot\overline{EB}+\overline{AF}\cdot\overline{FC}$ since $\overline {MB}+\overline{MC}=\overline 0$ .

An easy extension. In $\triangle ABC$ denote the midpoint $M$ of $[BC]$ . For $P\in BC$ denote its projections

$E$ , $F$ on $AB$ , $AC$ respectively. Prove that $PE^2+PF^2=\overline{AE}\cdot\overline{EB}+\overline{AF}\cdot\overline{FC}+2\cdot\overline {PM}\cdot\overline{PD}$ .

Remark. If $P\in BC$ , then $\boxed{PE^2+PF^2=\overline{AE}\cdot\overline{EB}+\overline{AF}\cdot\overline{FC} \iff\ P\in\{M,D\}}$ .

PP3. Let $ABCD$ be a trapezoid with $AB\parallel CD$ and $AC\perp BD$ . Denote $O\in AC\cap BD$ . Prove that $AB\cdot CD=AO\cdot OC+BO\cdot OD$ .

Method 1 (trigonometric). Denote $\left\{\begin{array}{c} m(\angle OAB)=m(\angle OCD)=x\\\\ m(\angle OBA)=m(\angle ODC)=y\end{array}\right|$ . Thus, $x+y=90^{\circ}$ and $AB\cdot CD=AO\cdot OC+BO\cdot OD\iff$

$1=\frac {AO}{AB}\cdot \frac {OC}{CD}+\frac {BO}{AB}\cdot\frac {OD}{CD}\iff$ $1=\sin y\cos x+\sin x\cos y\iff$ $1=\sin (y+x)\iff$ $x+y=90^{\circ}$ , what is truly.

Method 2 (metric). Denote $\left\{\begin{array}{c} OA=x\ ;\ OB=y\\\\ OC=z\ ;\ OD=t\end{array}\right|$ . Thus, $AB\cdot CD=AO\cdot OC+BO\cdot OD\iff$ $AB^2\cdot CD^2=(AO\cdot OC+BO\cdot OD)^2\iff$

$\left(x^2+y^2\right)\cdot\left(z^2+t^2\right)=$ $(xz+yt)^2\iff$ $x^2t^2+y^2z^2=$ $(xz+yt)^2\iff$ $x^2t^2+y^2z^2=2xyzt\iff$ $(xt-yz)^2=0\iff$ $xt=yz$ ,

what is truly (well-known) because $[ACD]=[BCD]\iff$ $[ACD]-[COD]=[BCD]-[COD]\iff$ $[AOD]=[BOC]\iff$ $xt=yz$ .

An easy extension. Let $ABCD$ be a convex quadrilateral. Denote $O\in AC\cap BD$ and the area $[AOD]=S$ , $m(\angle AOD)=\phi$ . Prove that

$AB^2\cdot CD^2=\left[(xz+yt)+(xt+yz)\cdot\cos\phi\right]^2+(xt-yz)^2\cdot\sin^2\phi$ . Therefore, $AB\cdot CD\ge \left|(xz+yt)+(xt+yz)\cdot\cos\phi\right|$ , with equality where $\phi =90^{\circ}$ .

PP4. Let $H$ be the orthocenter of acute $\triangle ABC$ . Let the midpoint $M$ of $[BC]$ , $P\in AB$ , $Q\in AC$ so that $H\in PQ$ and $\overline {PHQ}\perp MH$ . Prove that $H$ is the midpoint of $[PQ]$ .

Proof 1. If $[AS]$ is a diameter of the circumcircle of $\triangle ABC$ , then $BHCS$ is a parallelogram and the quadrilaterals $BPHS$ , $CQHS$ are cyclically. Thus,

$\widehat {HPS}\equiv$ $\widehat {HBS}\equiv$ $\widehat {HCS}\equiv\widehat {HQS}$ , i.e. $\widehat {HPS}\equiv\widehat {HQS}$ , what means that the triangle $PSQ$ is isosceles and $\overline{SMH}\perp \overline{PHQ}\iff$ $HP=HQ$ .

Remark. In the circle with the diameter $[BC]$ can apply the butterfly's problem to the point $H$ .

Proof 2. Denote the midpoints $U$ and $V$ of $[CE]$ and $[BF]$ respectively , where $E\in BH\cap AC$ and $F\in CH\cap AB$ . Observe that $MU\perp AC$

and $MV\perp AB$ , i.e. the quadrilaterals $MHVP$ and $MHQU$ are cyclically. Since $\tan \widehat {HUM}=\tan\widehat {EHU}=\frac {EC}{2\cdot HE}=\frac 12\cdot\tan A$

(symetrically in $B$ and $C$ ) obtain that $\widehat {HPM}\equiv$ $\widehat {HVM}\equiv$ $\widehat {HUM}\equiv$ $\widehat {HQM}$ , i.e. $\widehat {HPM}\equiv\widehat {HQM}$ , what means $HP=HQ$ .

Proof 3. Denote $E\in BH\cap AC$ , $F\in CH\cap AB$ and $X\in (BE)$ , $Y\in (CF)$ so that $PX\perp PQ$ and $QY\perp PQ$ . Thus, $PEQX$ and $PFQY$ are cyclically $\iff$

$HE\cdot HX=HP\cdot HQ=HF\cdot HY$ $\implies$ $HE\cdot HX=HF\cdot HY$ $\implies$ $EFXY$ is cyclically $\implies$ $\widehat{CBE}\equiv\widehat{CFE}\equiv$ $\widehat{YFE}\equiv\widehat{YXE}$ $\implies$

$\widehat{CBE}\equiv\widehat{YXE}$ $\implies$ $XY\parallel BC$ . Since $PX\parallel MH\parallel QY$ obtain that $PQYX$ is a right trapezoid and $Z\in XY\cap MH$ is the midpoint of $[XY]$ , i.e. $HP=HQ$ .

An easy extension. Let $H$ be the orthocenter of an acute $\triangle ABC$ . For $M\in [BC]$ denote $P\in AB$ , $Q\in AC$ so that $H\in PQ$ and $\overline {PHQ}\perp MH$ . Prove that $\frac {HP}{HQ}=\frac {MB}{MC}$ .

Proof. Denote $E\in BH\cap AC$ , $F\in CH\cap AB$ and $X\in (BE)$ , $Y\in (CF)$ so that $PX\perp PQ$ and $QY\perp PQ$ . Thus, $PEQX$ and $PFQY$ are cyclically $\implies$

$HE\cdot HX=HP\cdot HQ=$ $HF\cdot HY\implies$ $HE\cdot HX=HF\cdot HY\implies$ $EFXY$ is cyclically $\implies$ $\widehat{CBE}\equiv\widehat{CFE}\equiv$ $\widehat{YFE}\equiv\widehat{YXE}\implies$ $\widehat{CBE}\equiv\widehat{YXE}$ $\implies$

$XY\parallel BC$ . Since $PX\parallel MH\parallel QY$ obtain that $PQYX$ is a right trapezoid and if $Z\in XY\cap MH$ , then $\frac {HP}{HQ}=\frac {ZX}{ZY}=\frac {MB}{MC}$ , i.e. $\frac {HP}{HQ}=\frac {MB}{MC}$ .

PP5. Let $\triangle ABC$ with $B=2C$ . Denote the bisector $[AD$ of $\widehat{BAC}$ where $D\in BC$ and $E\in (AC)$ for which $CE=BD$ . Prove that $m\left(\widehat{CBE}\right)=\frac C2$ .

Proof. It is well-known or prove ( $***$ ) easily that $B=2C\iff b^2=c(c+a)\ (1)$ . Thus, $CE=BD=\frac {ac}{b+c}\implies$

$AE=AC-CE=b-\frac {ac}{b+c}=$ $\frac {b^2+bc-ac}{b+c}\stackrel{(1)}{=}\frac {c^2+ac+bc-ac}{b+c}=c\implies$ $AE=AB\implies$ $m\left(\widehat{CBE}\right)=2C-\frac {3C}{2}=\frac C2$ .

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( $***$ ) Denote $F\in (AC)$ so that $[BF$ is the bisector of $\widehat{ABC}$ . Since $\widehat{ABF}\equiv\widehat {ACB}$ obtain that $AB^2=AC\cdot AF\iff$ $c^2=b\cdot \frac {bc}{a+c}\iff$ $b^2=c(a+c)$ .

PP6. Let $ABC$ be a triangle. Its $A$ -bisector meet again its circumcircle $w$ in the point $L$ . Suppose that exists $E\in (AB)$ so that $AE=\frac {b+c}{2}$ . Prove that $LE\perp AB$ .

Proof. Get $BE=\frac {c-b}{2}$ and $\boxed{EA^2-EB^2=bc}$ . If $D\in AL\cap BC$ , then $\left\{\begin{array}{cccc} \triangle LBD\sim LAB & \iff & LB^2=LA\cdot LD & (1)\\\\ \triangle LBA\sim\triangle CDA & \iff & AD\cdot AL=bc & (2)\end{array}\right|\implies$

$LA^2-LB^2\stackrel{(1)}{=}LA^2-LA\cdot LD=$ $LA\cdot (LA-LD)=$ $AL\cdot AD\stackrel{(2)}{\implies}$ $\boxed{LA^2-LB^2=bc}\implies$ $LA^2-LB^2=EA^2-EB^2\implies$ $LE\perp AB$ .

Remark 1. $LA^2-LB^2=$ $4R^2\left[\sin^2\left(B+\frac A2\right)-\sin^2\frac A2\right]=$ $4R^2\sin (A+B)\sin B=$ $4R^2\sin B\sin C\implies$ $LA^2-LB^2=bc$ .

Remark 2. If $M$ is the midpoint of the side $[BC]$ , then can prove easily that $EM\perp \overline {ADL}$ .

PP7 (Laurentiu Panaitopol). Let $ABCD$ be a convex quadrilateral. Consider the points $M\in(BC)\ ,\ N\in(CD)$ so that $\frac{MB}{MC}=m\ ,\ \frac{ND}{NC}=n$ .

Denote $P\in AM\cap BN$ , $\frac{PA}{PM}=x$ and $\frac{PB}{PN}=y$ . Prove that $ABCD$ is a parallelogram $\iff \frac {x+1}{y+1}=\frac {n+1}{y}=\frac {m}{m-y}$ .

PP8. Let $\triangle ABC$ with $B=2C$ and $\angle A>90^\circ$ . Let $D\in AB$ such that $CD\perp AC$ and let $M$ be the midpoint of $BC$ . Prove that $\angle AMB=\angle DMC$ .

Proof. Let $P\in AB$ so that $PB=PC$ . Then $\widehat{PCB}\equiv\widehat{ABC}$ $\Longrightarrow$ $CA$ is the bisector of $\widehat{PCB}$ $\Longrightarrow$ $CD \perp CA$ and it is the external bisector of $\widehat{PCB}$ $\Longrightarrow$

cross ratio $(P,B,A,D)$ is harmonic. Hence, from $MP \perp MB$ we deduce that $MP$ , $BC$ bisect $\widehat{AMD}$ internally and externally $\Longrightarrow$ $\widehat{AMB}\equiv\widehat{DMC}$ .

PP9. Let $ABCD$ be a square. For $M\in (BC)$ the bisector of $\widehat {BAM}$ meet $BC$ at $E$ and the bisector of $\widehat{DAM}$ meet $CD$ at $F$ . Prove that $AM \perp EF$ .

Proof 1. Note that $m(\angle EAF)=45^{\circ}$ . Extend $CB$ to $G$ such that $\triangle ABG\equiv \triangle ADF$ . Then $AF=AG$

and $m(\angle FAE)=m(\angle GAE)=45^{\circ}$ . Thus, $\triangle AGE\equiv \triangle AFE$ and $AB\perp GE\implies AM\perp EF$ .

Proof 2. Denote $H\in AM$ so that $AH=AB$ . Note that $\left\{\begin{array}{ccc} \triangle ABE\equiv\triangle AHE & \implies & EH\perp AH\\\\ \triangle ADF\equiv\triangle AHF & \implies & FH\perp AH\end{array}\right|\implies H\in EF$ and $AM\perp EF$ .

Proof 3. Suppose w.l.o.g. $AB=1$ and denote $\left\{\begin{array}{c} m\left(\widehat{EAB}\right)=m\left(\widehat{EAM}\right)=x\\\\ m\left(\widehat{FAD}\right)=m\left(\widehat{FAM}\right)=y\end{array}\right|$ . Note $x+y=\frac {\pi}{2}$ and $BE=\tan x$ . Thus, $\left\{\begin{array}{c} BM=\tan 2x\ ;\ MC=1-\tan 2x\\\\ DF=\tan y\ ;\ FC=1-\tan y\end{array}\right|$ .

Hence $AM\perp EF\iff$ $AE^2+FM^2=AF^2+EM^2\iff$ $AB^2+BE^2+FC^2+MC^2=$ $AB^2+DF^2+EM^2\iff$ $BE^2+\left(MC^2-ME^2\right)=$

$DF^2-FC^2\iff$ $\tan ^2x+(1-\tan x)(1+\tan x-2\tan 2x)=$ $2\tan y-1\iff$ $1=\tan y+\tan 2x(1-\tan x)\iff$ $1=\tan y+\frac {2\tan x}{1+\tan x}\iff$

$\tan y=\frac {1-\tan x}{1+\tan x}\iff$ $\tan y=\tan\left(45^{\circ}-x\right)\iff$ $x+y=45^{\circ}$ .

PP10. Let $\triangle ABC$ with the incircle $w=C(I)$ and $\left\{\begin{array}{ccc} E\in AC\cap w & ; & F\in AB\cap w\\\\ X\in EF\cap BI & ; & Y\in EF\cap CI\end{array}\right|$ . Prove that $X$ and $Y$ belong to the circle with the diameter $[BC]$ .

Proof. Denote $D\in BC\cap w$ . Show easily that $IEXC$ is inscribed in the circle $w_1$ with the diameter $[IC]$ and $IFYB$ is inscribed in the circle $w_2$

with the diameter $[BI]$ . Since $D\in w_1\cap w_2$ obtain that the quadrilaterals $DIXC$ and $DIFB$ are cyclically, i.e. $XB\perp XC$ and $YC\perp YB$ .

Application. Let a tangential $ABCD$ with incircle $w=(I)$ . Let $\left\{\begin{array}{ccc} E\in BC\cap w & ; & F\in AD\cap w\\\\ K\in AI\cap EF & ; & N\in DI\cap EF\end{array}\right|$ and $M\in BK\cap CN$ . Prove that $IKMN$ is cyclically.

PP11. Let a cyclical convex $ABCD$ with $I\in AC\cap BD$ . Denote the projections $(K,L,M,N)$ of $I$ on $AB$ ,

$BC$ , $CD$ and $DA$ respectively. Prove that $KLMN$ is tangential $\iff IK+IM=IL+IN$ .

Proof. Let $ABCD$ be the cyclic quadrilateral, $I\in AC\cap BD$ , the projections $K,L,M,N$ of $I$ on $AB$ , $BC$ , $CD$ , $DA$ respectively. The quadrilaterals $IKBL$ and $IKAN$ are

cyclically $\iff$ $\widehat{IKL}\equiv\widehat{IBL}$ and $\widehat{IAN}\equiv\widehat{IKN}$ . The quadrilateral $ABCD$ is cyclic $\iff$ $\widehat{DBC}=\widehat{DAC}$ . Therefore, $\widehat{IKL}=\widehat{IKN}$ . So $I$ belongs to the angle bisector of $\widehat{NKL}$

and analogously show that $I$ belongs to the other angle bisectors of $\widehat{KNM}$ , $\widehat{LMN}$ , $\widehat{MNK}$ . In conclusion, $KL+MN=ML+KN$ , using the Pithot's reciprocal theorem.

PP12. Let $ABC$ be a triangle with the incircle $w=C(I,r)$ . Denote $D\in BC\cap w$ , $E\in CA\cap w$ and $D_1\in (BC)$ , $E_1\in (CA)$ for

which $D_1C=DB$ , $E_1C=EA$ . Define the points $P\in AD_1\cap BE_1$ and $\{P,Q\}=AD_1\cap w$ . Prove that $AQ=PD_1$ .

Proof. It is well-known that $DB=D_1C=s-b$ and $EA=E_1C=s-a$ . Apply the Menelaus' theorem to the transversal $\overline {BPE_1}$ for $\triangle ACD_1\ :$

$\frac {BD_1}{BC}\cdot\frac {E_1C}{E_1A}\cdot\frac {PA}{PD_1}=1\iff$ $\frac {PD_1}{PA}=\frac {s-c}{a}\cdot\frac {s-a}{s-c}\iff$ $\boxed{\frac {PA}{PD_1}=\frac {a}{s-a}}$ . It is well-known that $[DQ]$ is a diameter of $w$ , i.e. $I\in DQ$ .

Thus, $\frac {D_1Q}{D_1A}=\frac {2r}{h_a}=$ $\frac {2ar}{ah_a}=\frac {2ar}{2sr}=\frac as$ , i.e $\frac {D_1Q}{a}=\frac {D_1A}{s}=\frac {QA}{s-a}\iff$ $\boxed{\frac {QA}{QD_1}=\frac {s-a}{a}}$ . Therefore, $\frac {QA}{QD_1}=\frac {PD_1}{PA}$ , i.e. $AQ=PD_1$ .

PP13. Let $\frac{x^2+y^2}{x^2-y^2} + \frac{x^2-y^2}{x^2+y^2} = k$ . Compute the following expression in terms of $k\ :\ \ E= \frac{x^8 + y^8}{x^8-y^8} - \frac{ x^8-y^8}{x^8+y^8}\ .$

Proof 1. $k=\frac {2\left(x^4+y^4\right)}{x^4-y^4}\iff$ $\frac {k+2}{k-2}=\left(\frac xy\right)^4\implies$ $E=\frac {4x^8y^8}{x^{16}-y^{16}}=$ $\frac {4}{\left(\frac xy\right)^8-\left(\frac yx\right)^8}=$ $\frac {4}{\left(\frac {k+2}{k-2}\right)^2-\left(\frac {k-2}{k+2}\right)^2}$ $\implies$ $E=\frac {\left(k^2-4\right)^2}{4k\left(k^2+4\right)}$ .

PP14. Let $ABC$ be a triangle with the incenter $I$ and $b<c$ . Denote the midpoint $M$ of the side $[BC]$ and the diameter $[NS]$ of

the circumcircle of $\triangle ABC$ so that $NS\perp BC$ and the line $BC$ separates $I$ and $S$ . Prove that $m(\angle ANI)=m( \angle CMI)$ .

Proof 1. Denote $D\in BC$ so that $ID\perp BC$ . Observe that $m(\angle ANI)=m( \angle (IMC)\iff$ $\tan\widehat {ANI}=\tan\widehat{IMC}\iff$ $\frac {AI}{AN}=\frac {DI}{DM}\iff$

$AI\cdot DM=AN\cdot DI\iff$ $\frac {r}{\sin\frac A2}\cdot \frac {c-b}{2}=2R\sin\frac {C-B}{2}\cdot r\iff$ $2R(\sin C-\sin B)=2R\cdot 2\cdot \sin\frac {C-B}{2}\cos\frac {C+B}{2}$ , what is truly.

Proof 2. I"ll use same notations from the previous proof. It is well known that $S$ is the circumcenter of $\triangle BIC$ . From this, since $CS\perp CN$ we get that $NB$ and $NC$

are the tangents from the point $N$ to the circumcircle of $\triangle BIC$ . Now denote $E\in NI\cap BC$ . Therefore, $IE$ is the $I$ -symmedian in $\triangle BIC$ so that $\widehat{MIS}\equiv\widehat {EIC}$ .

Let $F$ be the second intersection of $NI$ with the circumcircle of $\triangle BIC$ . Since $m(\angle IAN) = m(\angle IFS) = 90^{\circ}$ we have that $\angle INA = \angle ISF$ . On the other side

$\angle IMC = \angle MBI + \angle MIB = \angle CBI + \angle FIC = \angle ISF = \angle INA$ .

PP15. Let $ABCD$ and $BEGF$ be two parallelograms so that $E\in (BC)$ and $BC$ separates $A\ ,\ F$ . Prove that $AE\cap DG\cap CF\ne\emptyset$ .

Proof 1. Let $AB=a\ ,\ AD=b\ ,\ BF=c\ \,\ BE=d$ , $H\in CD\cap FG\ ,\ K\in CF\cap AE\ ,\ L\in CF\cap DG$ . Apply Menelaus' theorem to transversals/triangles :

$\left\{\begin{array}{ccccccc} \overline{AEK}/\triangle BCF\ : & \frac {AB}{AF}\cdot\frac {KF}{KC}\cdot\frac {EC}{EB}=1 & \implies & \frac {a}{a+c}\cdot \frac {KF}{KC}\cdot\frac {b-d}{d}=1 & \implies & \frac {KC}{KF}=\frac {a(b-d)}{d(a+c)}\\\\ \overline{DGK}/\triangle HCF\ : & \frac {DC}{DH}\cdot\frac {GH}{GF}\cdot\frac {LF}{LC}=1 & \implies & \frac {a}{a+c}\cdot \frac {b-d}{d}\cdot\frac {LF}{LC}=1 & \implies & \frac {LC}{LF}=\frac {a(b-d)}{d(a+c)}\end{array}\right\|$ $\implies\frac {KC}{KF}=\frac {LC}{LF}\implies$ $K\equiv L\implies$ $AE\cap DG\cap CF\ne\emptyset$ .

Proof 2. Let $\left\{\begin{array}{c} AB=a\ ,\ AD=b\ ,\ BF=c\ \,\ BE=d\\\\ K\in CF\cap AE\ ,\ L\in CF\cap DG\\\\ X\in AE\cap FG\ ,\ Y\in DG\cap BC\end{array}\right\|$ . Thus, $\frac {XF}{EB}=\frac {AF}{AB}$ $\implies$ $\boxed{XF=\frac {d(a+c)}{a}}\ (1)$ and $\frac {YC}{YE}=\frac {DC}{EG}$ $\implies$

$\frac {YC}{a}=$ $\frac {YE}{c}=$ $\frac {b-d}{a+c}$ $\implies$ $\boxed{YC=\frac {a(b-d)}{a+c}}\ (2)\implies$ $\frac {KC}{KF}=\frac {CE}{XF}\stackrel{(1)}{\implies}$ $\boxed{\frac {KC}{KF}=\frac {a(b-d)}{d(a+c)}}\ (3)$ and $\frac {LC}{LF}=\frac {YC}{GF}$ $\stackrel{(2)}{\implies}$ $\boxed{\frac {LC}{LF}=\frac {a(b-d)}{d(a+c)}}\ (4)$ .

From $(3)$ and $(4)$ obtain that $\frac {KC}{KF}=\frac {LC}{LF}$ , i.e. $K\equiv L\implies$ $AE\cap DG\cap CF\ne\emptyset$ .

PP16. Let $\triangle ABC$ with $A>90^{\circ}$ and circumcircle $w$ . Denote the reflection $E$ of $B$ w.r.t. the midpoint $M$ of $[AC]$ and the diameter $[BD]$ of $w$ . Prove that $AC\perp DE$ .

Proof. If $O$ is the center of $w$ , then $OM\perp AC$ and $MB=ME\implies OM\parallel DE\implies AC\perp DE$ . Remark that the point $C$ is the orthocenter of $\triangle ADE$ .

PP17. Let $ABC$ be a triangle with orthocentre $H$ and circumcircle $w=C(O)$ . Denote $\left\{\begin{array}{c} X\in AB\ ,\ HX\parallel AC\\\\ Y\in AC\ ,\ HY\parallel AB\end{array}\right\|$ . Prove that $OX = OY$ .

Proof. $AXHY$ is a parallelogram, i.e. $\left\{\begin{array}{c} HY=XA\\\ HX=YA\end{array}\right\|$ and $m\left(\widehat {HXB}\right)=m\left(\widehat{HYC}\right)=A$ $\implies$ $H$ -right $\triangle HXB\ ,\ \triangle HYC$ are

similarly $\boxed{\frac {HX}{XB}=\frac{HY}{YC}}\ (*)$ $\implies$ $\frac {YA}{XB}=\frac {HX}{XB}\stackrel{(*)}{=}$ $\frac{HY}{YC}=\frac {XA}{YC}\implies$ $XA\cdot XB=YA\cdot YC$ $\implies$ $p_w(X)=p_w(Y)\implies OX=OY$ .

Remark. If $E\in AH\cap XY$ , then $OE\perp XY$ . It is well-known that $[EM]$ is the diameter of the Euler's circle for $\triangle ABC$ .

PP18. For $C(A,r)$ and $C(B,r)$ exists $w=C(I,x)$ tangent to both $\stackrel{\frown}{AC}$ , $\stackrel{\frown}{BC}$ and to $AB$ . Prove that if the length of $\stackrel{\frown}{BC}$ is $l$ , then the perimeter of $w$ is $\frac {9l}{4}$ .

Proof. $\pi r=3l\implies \boxed{r=\frac {3l}{\pi}}\ (*)$ . So $(r-x)^2=x^2+\left(\frac r2\right)^2$ $\implies$ $x=\frac {3r}{8}$ . Perimeter $L$ of $w$ is $2\pi\cdot\frac {3r}{8}=\frac {3\pi }{4}\cdot r\stackrel{(*)}{=}\frac {3\pi}{4}\cdot\frac {3l}{\pi}=\frac {9l}{4}$ . For $l:=12$ get $L=27$ .

$\blacktriangleright$ Lemma. Let $ABCD$ be a convex quadrilateral for which denote $O\in AB\cap CD$ . Find the geometrical locus of the mobile point $L$ which

belongs to the interior of the angle $\widehat{AOD}$ for which $\alpha\cdot [AOB]+\beta\cdot [OCD]=k$ (constant), where $\alpha$ and $\beta$ are two given positive numbers.

Proof. Denote $\left\{\begin{array}{c} E\in (OA\ ,\ OE=\alpha\cdot AB\\\\ F\in (OD\ ,\ OF=\beta\cdot CD\end{array}\right\|$ . Observe that $[LOE]=\alpha\cdot [LAB]$ and $[LOF]=\beta\cdot [LDC]$ and $k=\alpha\cdot [AOB]+\beta\cdot [OCD]=$

$[LOE]+[LOF]=$ $[LEOF]=[EOF]+[ELF]$ $\implies$ $[ELF]=k-[EOF]$ is constant because the triangle $EOF$ is fixed. Since the segment $[EF]$ is fixed

(with constant length) obtain that $\delta_{EF}(L)$ is constant, i.e. the geometrical locus of $L$ is a parallel segment with $EF$ and which prop up the rays $(OA$ and $(OD$ .

PP19. Let $\triangle ABC$ , $D\in BC$ , $E\in CA$ , $F\in AB$ so that $AD\perp BC$ and $AD\cap BE\cap CF\ne\emptyset$ . Prove that $\widehat{ADF}\equiv\widehat{ADE}$ .

Proof 1. Let $\left\{\begin{array}{c} G\in DE\\\ H\in DF\end{array}\right\|$ so that $A\in GH\parallel BC$ . Ceva's $\implies\ 1=\frac {DB}{DC}\cdot\frac {EC}{EA}\cdot\frac {FA}{FB}=$ $\frac {DB}{DC}\cdot\frac {DC}{AG}\cdot \frac {AH}{DB}=$ $\frac {AG}{AH}\implies \left\{\begin{array}{c} AG=AH\\\\ AD\perp HG\end{array}\right\|\implies$ $\widehat{ADF}\equiv\widehat{ADE}$

Proof 2. Denote $I\in AD\cap EF$ and $J\in BC\cap EF$ . Since the division $(E,F;I,J)$ is harmonically and $DI\perp DJ$ , from an well-known property obtain that $\widehat{ADF}\equiv\widehat{ADE}$ .

PP20. Let $ABC$ be a triangle with the $A$ -bisector $[AD$ , where $D\in (BC)$ . Suppose $m\left(\widehat {ADC}\right)=60^{\circ}$ and $DB^2=DA\cdot DC$ . Prove that $AD=BC$ .

Proof 1 (synthetic). Let $E\in (AD)$ be a point so that $DE=DC$ . Observe that $\triangle CDE$ is equilateral and $DB^2=DA\cdot DC\iff$

$DB^2=DA\cdot DE\iff$ $\triangle DBE\sim\triangle DAB\implies$ $\widehat{CBE}\equiv\widehat {DAB}\equiv\widehat {DAC}\implies$ $\triangle CBE\equiv\triangle DAC\implies$ $AD=BC$ .

Proof 2 (metric).

PP21. Let $\triangle ABC$ and $P\in (AC)\ ,\ Q\in (AB)$ for which denote the circumcircles $p\ ,\ q$ of $\triangle ABP\ ,\triangle ACQ$ respectively,

$M\in BC\cap q\ ,\ N\in BC\cap p$ so that $M\in BN$ and $\left\{\begin{array}{c} I\in BP\cap CQ\\\\ m\left(\widehat{BIC}\right)=\phi\end{array}\right\|$ . Prove that $m\left(\widehat{MAN}\right)=A+\phi -\pi$ .

Proof. Observe that $m\left(\widehat{MAN}\right)=A-\left[m\left(\widehat{MAB}\right)+m\left(\widehat{NAC}\right)\right]=$ $A-\left[m\left(\widehat{BCI}\right)+m\left(\widehat{CBI}\right)\right]=$ $A-\left[\pi -m\left(\widehat{BIC}\right)\right]\implies$ $m\left(\widehat{MAN}\right)=A+\phi -\pi$ .

PP22. Let $ABC$ be a $A$ -right-angled triangle with $C=15^{\circ}$ and $c=k$ . Ascertain without trigonometry $a$ and $b$ .

Proof 1. Suppose w.l.o.g. $k=1$ , i.e. $\boxed{c=1}$ and denote $D\in (AC)$ so that $m\left(\widehat{CBD}\right)=15^{\circ}\implies$ $m\left(\widehat{ABD}\right)=60^{\circ}$ and $DC=DB=2\cdot AB=2$ , i.e. $DC=2$ . Thus,

$AD^2=BD^2-BA^2\implies$ $AD=\sqrt 3\implies$ $\boxed{b=2+\sqrt 3}$ and $BC^2=AB^2+AC^2=$ $1+\left(2+\sqrt 3\right)^2\implies$ $a^2=4\left(2+\sqrt 3\right)\implies$ $a=2\sqrt{2+\sqrt 3}\implies$ $\boxed{a=\sqrt 2+\sqrt 6}$ .

Proof 2. Suppose w.l.o.g. $k=1$ , i.e. $\boxed{c=1}$ . Denote $E\in (AB)$ so that the line $BC$ separates the points $A$ , $E$ and $m\left(\widehat{BCE}\right)=15^{\circ}$ .

Denote $AE=x$ , i.e. $BE=AE-c=x-1\ ,\ CE=2x$ and $\boxed{b=x\sqrt 3}$ . Apply the theorem of the $C$ -angled bisector in $\triangle ACE\ :$ $\frac {CA}{CE}=\frac {BA}{BE}\implies$

$\frac {b}{2x}=\frac {1}{x-1}\implies$ $\boxed{b=\frac {2x}{x-1}}$ . Thus, $\frac {2x}{x-1}=x\sqrt 3\implies$ $x=\frac {3+2\sqrt 3}3$ $\implies$ $\left\{\begin{array}{c} a=\sqrt 2+\sqrt 6\\\\ b=2+\sqrt 3\\\\ c=1\end{array}\right\|$ .

PP23. Let $\triangle ABC$ such that $\frac {a+b}{c}< \lambda$ , where $\lambda>1$ . Prove that $\tan\frac{A}{2}\tan\frac{B}{2}< \frac{\lambda -1}{\lambda +1}$ .

Proof. Prove easily that $\boxed{\tan\frac A2\tan\frac B2=\frac{s-c}{s}}\ (*)$ and the function $f(\lambda )=\frac{\lambda -1}{\lambda +1}=1-\frac {2}{\lambda +1}\ ,\ \lambda >1$ is increasing.

Therefore, $\frac {a+b}{c}< \lambda\iff$ $f\left(\frac {a+b}{c}\right)<f\left(\lambda\right)\iff$ $\frac {s-c}{s}<\frac{\lambda -1}{\lambda +1}\stackrel{(*)}{\implies}$ $\tan\frac{A}{2}\tan\frac{B}{2}< \frac{\lambda -1}{\lambda +1}$ .

PP24. $\triangle ABC$ is inscribed in $w=C(O,r)$ . The vertices of the triangle divide $w$ in 3 arcs of lengths $\mathrm{arc}(BC)=3$ , $\mathrm{arc}(CA)=4$ and $\mathrm{arc}(AB)=5$ . What is the area of the triangle ?

Proof. Perimeter of the circle is $2\pi r=3+4+5=12\implies \boxed{r=\frac {6}{\pi}}$ . Thus, $\frac A3=\frac B4=$ $\frac C5=\frac {\pi}{12}$ $\implies$ $S=[ABC]=$ $2r^2\sin A\sin B\sin C\implies$

$S=\frac {72}{\pi^2}\sin\frac {\pi}{4}$ $\sin\frac {\pi}{3}\sin\frac {5\pi}{12}=$ $\frac {36\sqrt 3}{\pi^2}\left(2\sin\frac {3\pi}{12}\sin\frac {5\pi}{12}\right)=$ $\frac {36\sqrt 3}{\pi^2}\left(\cos \frac {\pi}6-\cos\frac {2\pi}3\right)\implies$ $\boxed{S=\frac {9\sqrt 3\left(\sqrt 3+1\right)}{\pi^2}}$ .

PP25. Let $\triangle ABC$ with circumcircle $w$ , $M\in (AB)$ , $N\in (AC)$ so that $BM=CN$ . Let midpoint $P$ of arc $\stackrel{\frown}{BC}$ which contains $A$ . Prove that $PM=PN$ , $\widehat {MPN}\equiv\widehat {BAC}$ .

Proof. Prove easily that $\triangle BPM\stackrel{(L.U.L)}{\equiv}\triangle CPN$ . Hence $PM=PN$ and $\widehat{MPB}\equiv\widehat{NPC}$ $\iff$ $\widehat {MPN}=\widehat{MPB}+\widehat{BPN}=$ $\widehat{NPC}+\widehat{NPB}=$ $\widehat{BPC}$ $\implies$ $\widehat{MPN}\equiv\widehat{BPC}$ .

PP26. Let a rhombus $ABCD$ and $M\in (AC)$ , $N\in (BC)$ so that $MN=MD$ . Denote $P\in AC\cap DN$ and $R\in AB\cap DM$ . Prove that $PR=PD$ .

Proof. $MB=MD=MN$ $\implies$ $MB=MN$ $\implies$ $\widehat{MNB}\equiv\widehat{MBN}\equiv$ $\widehat{MDC}\implies$ $\widehat{MNB}\equiv\widehat{MDC}\implies$

$CDMN$ is cyclically $\implies$ $\widehat{RAP}\equiv\widehat{NCM}\equiv$ $\widehat{RDP}\implies$ $ARPD$ is cyclically and $\widehat{PAR}\equiv\widehat{PAD}$ $\implies$ $PR=PD$ .

PP27 (Balkan 2005). Let $ABC$ be an acute-angled triangle whose inscribed circle touches $AB$ and $AC$ at $D$ and $E$ respectively. Let $X$ and $Y$ be the points of intersection

of the bisectors of $\angle ACB$ and $\angle ABC$ with the line $DE$ respectively and let $Z$ be the midpoint of $BC$ . Prove that the triangle $XYZ$ is equilateral if and only if $\angle A = 60^\circ$ .

Proof. Let $I$ be the incenter. An angle chase shows that $B,I,X,D$ are concyclic, so, since $ID\perp AB$ , it means that $BX\perp CI$ . This means that $CXB$ is a right triangle,

so $ZX=ZB=ZC$ . In the exactly same way we show that $ZY=ZB$ , so $XYZ$ is always isosceles. Since $ZX=ZC$ , we have $\widehat{ZXC}\equiv\widehat{ZCX}\equiv\widehat{ACX}$ ,

so $ZX\|AC$ . We also have $ZY\|AB$ . From the above we conclude that $XYZ$ is equilateral iff $\angle XZY=60^{\circ}$ which is equivalent to $\angle A=60^{\circ}$ .

PP28. Let a regular $ABCDEF$ with $AB=1$ and $M\in (AC)$ and $N\in (CE)$ so that $B\in MN$ and $AM=CN=r$ . Find $r$ .

Proof 1 (trigonometric). Let $m\left(\widehat{CBM}\right)=\phi$ . Thus, $AC=\sqrt 3$ , $\tan \phi =\frac {CN}{CB}=r$ and $\frac {CM}{\sin\widehat{MBC}}=\frac {CB}{\sin\widehat {BMC}}\iff$

$\frac {\sqrt 3-r}{\sin\phi}=\frac 1{\sin\left(\phi +30^{\circ}\right)}\iff$ $\frac {\sqrt 3-r}{\tan\phi}=\frac 2{\tan\phi+\sqrt 3}\iff$ $\frac {\sqrt 3-r}{r}=\frac 2{r+\sqrt 3}\iff$ $3-r^2=2r^2\iff$ $r=1$ .

Proof 2 (trigonometric). Denote $\left\{\begin{array}{c} m(\angle CBN)=v\\\\ m(\angle EBN)=w\\\\ \left(v+w=60^\circ\right)\end{array}\right|$ . Apply theorem of Ratio : $\left\{\begin{array}{cc} BM/\triangle ABC\ : & \frac {MC}{MA}=\frac {\sin v}{\sin (w+60^\circ)}\\\\ BN/\triangle CBE\ : & \frac {NE}{NC}=\frac {2\sin w}{\sin v}\end{array}\right|$ . Since $\frac {MC}{MA}=\frac {NE}{NC}$ obtain that

$\sin^2v=2\sin w\sin (w+60^\circ )\iff$ $\sin^2v=\cos 60^\circ -\cos (2w+60^\circ )\iff$ $1-\cos 2v=$ $1-2\cos (2w+60^\circ )\iff$ $\cos 2v=2\cos (2w+60^\circ )$ . From

$2v+\left(2w+60^{\circ}\right)=180^{\circ}$ obtain that $\cos 2v=-2\cos 2v\iff$ $\cos 2v=0\iff$ $v=45^{\circ}\iff$ $\left\{\begin{array}{c} v=45^\circ\\\ w=15^\circ\end{array}\right|$ . Prove easily that $AC=\sqrt 3$ . In conclusion,

$\frac {MC}{MA}=\frac {\sin 45}{\sin 75}=$ $\frac {2}{1+\sqrt 3}=$ $\sqrt 3-1$ , i.e. $\frac {MC}{\sqrt 3-1}=$ $\frac {MA}{1}=\frac {AC}{\sqrt 3}=1\implies$ $MA=r=1$ .

Remark. Theorem of Ratio : Let $\triangle ABC$ and a point $M$ what belongs to the sideline $BC$ . Then $\boxed{\frac {MB}{MC}=\frac {AB}{AC}\cdot\frac {\sin\widehat{MAB}}{\sin\widehat{MAC}}}$ .

Proof 3. Pythagoras' theorem to $:\ \left\{\begin{array}{cccc} \triangle CMN\ : & MN^2=MN^2=(\sqrt 3-r)^2=r^2-r(\sqrt 3-r) & \implies & \boxed{MN^2=3\left(r^2-r\sqrt 3+1\right)}\\\\ \triangle CMB\ : & MB^2=(\sqrt 3-r)^2+1-\sqrt 3(\sqrt 3-r) & \implies & \boxed{MB^2=r^2-r\sqrt 3+1}\\\\ \triangle BCN\ : & NB^2=CN^2+CB^2 & \implies & \boxed{NB^2=r^2+1}\end{array}\right\|$ . Thus, $B\in MN\iff$

$MN+MB=NB\iff$ $\left(1+\sqrt 3\right)\sqrt {r^2-r\sqrt 3+1}=\sqrt {r^2+1}\iff$ $\left(4+2\sqrt 3\right)\left(r^2-r\sqrt 3+1\right)=r^2+1\iff$ $\left(3+2\sqrt 3\right)(r-1)^2=0\iff r=1$ .

PP29. Let a cyclical $ABCD$ and let incenters $I$ , $J$ of $\triangle ABC$ , $\triangle ACD$ respectively. Prove that $BIJD$ is cyclically $\iff$ $IJ\perp AC\iff ABCD$ is tangential.

Proof. Let the tangent points $X$ , $Y$ of $C(I)$ , $C(J)$ respectively with $AC$ . Suppose w.l.o.g. that $\boxed{X\in (AY)}\ (1)$ . Thus,

$\left\{\begin{array}{ccc} AX+BC=CX+BA & \implies & AX-CX=BA-BC\\\\ AY+DC=CY+DA & \implies & AY-CY=DA-DC\end{array}\right\|\ (*)$ . Let $\left\{\begin{array}{cc} m\left(\widehat{BAC}\right)=2x\ ; &m\left(\widehat{BCA}\right)=2y\\\\ m\left(\widehat{DAC}\right)=2u\ ; & m\left(\widehat{DCA}\right)=2v\end{array}\right\|\ ,$

$x+y+u+v=90^{\circ}$ $\implies$ $\left\{\begin{array}{ccc} m\left(\widehat{BIX}\right)=m\left(\widehat{BIC}\right)+m\left(\widehat{CIX}\right)=\left(90^{\circ}+x\right)+\left(90^{\circ}-y\right) & \implies & m\left(\widehat{BIX}\right)=180^{\circ}+x-y\\\\ m\left(\widehat{BDJ}\right)=m\left(\widehat{ADB}\right)-m\left(\widehat{ADJ}\right)=2y-\left(90^{\circ}-u-v\right) & \implies & m\left(\widehat{BDJ}\right)=y-x\end{array}\right\|$ .

$\blacktriangleright\ ABCD$ is tangential $\iff AB+CD=AD+BC\iff$ $BA-BC=DA-DC\stackrel{(*)}{\iff}$

$AX-CX=AY-CY\iff$ $AX-AY=CX-CY\stackrel{(1)}{\iff}$ $-XY=XY\iff$ $IJ\perp AC$ .

$\blacktriangleright\ IJ\perp AC\iff$ $X\equiv Y\iff$ $m\left(\widehat{BIX}\right)+m\left(\widehat{BDJ}\right)=\left(180^{\circ}+x-y\right)+\left(y-x\right)=180^{\circ}\iff$ the quadrilateral $BIJD$ is cyclically.

PP30. Prove that in any $A$ -rightangled $\triangle ABC$ the line $IG\not\parallel BC$ , but is possibly that $IG\parallel AB\ \vee\ IG\parallel AC$ (standard notations).

Proof. Let $\triangle ABC$ for which I"ll characterize the particularity $IG\parallel BC$ . Thus, $IG\parallel BC\iff 3r=h_a\iff 3ar=2pr\iff b+c=2a\ .$ Therefore,

$\boxed{IG\ \parallel\ BC\iff b+c=2a}\ .$ Suppose $AB\perp AC$ , i.e. $\triangle ABC$ is $A$ - rightangled. In this case $(b+c)^2\le 2\left(b^2+c^2\right)$ , i.e. $\boxed{b+c\le a\sqrt 2}\ (*)$ . Appears three cases:

$\odot\ IG\parallel BC\iff$ $b+c=2$ $a\ \stackrel{(*)}{\iff}\ 2a\le a\sqrt 2$ , what isn't truly. Prove easily that $\boxed{IG\perp BC\iff b+c=3a}$ . In conclusion,

$\odot\ a+b=3c\iff IG\perp AB\iff IG\parallel AC\iff$ $a+c=2b\iff a=2b-c\iff b^2+c^2=4b^2-4bc+c^2\iff$ $\frac a5=\frac b4=\frac c3\ .$

$\odot\ a+c=3b\iff IG\perp AC\iff IG\parallel AB\iff$ $a+b=2c\iff a=2c-b\iff b^2+c^2=4c^2-4bc+b^2\iff$ $\frac a5=\frac b3=\frac c4\ .$

This post has been edited 195 times. Last edited by Virgil Nicula, Nov 20, 2015, 7:47 AM

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You seem to have solved every existing problem.
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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!
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Thank you a lot dear Virgil for an interesting and instructive blog.
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Thanks to all !
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El_Ectric:
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You're welcome.
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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!
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Dear Virgil !

Congratulations for this blog !
Babis stergiou - Greece
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My (math)blog

317. Some interesting geometry problems for middle school.

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