Let $ Q$ be the centre of the inscribed circle of a triangle $ ABC.$ Prove that for any point $ P,$ \[ a(PA)^2 + b(PB)^2 + c(PC)^2 = a(QA)^2 + b(QB)^2 + c(QC)^2 + (a + b + c)(QP)^2, \] where $ a = BC, b = CA$ and $ c = AB.$
Problem
Source: IMO ShortList 1988, Problem 23, Singapore 2, Problem 69 of ILL
Tags: geometry, circumcircle, trigonometry, incenter, analytic geometry, IMO Shortlist
Virgil Nicula
09.11.2005 23:09
.Let $w=C(I,r)$ be the incircle of the triangle $ABC$. Then there are the following wellknown relations: $\blacksquare \ 1^{\circ}.\ a\cdot IA^2+b\cdot IB^2+c\cdot IC^2=abc.$ $\blacksquare \ 2^{\circ}.\ \left(\forall\right) P,\ a\cdot PA^2+b\cdot PB^2+c\cdot PC^2=(a+b+c)\cdot PI^2+abc.$ Examples. $Ex\ 1.\ \ P: =I\Longrightarrow 1^{\circ}.$ $Ex\ 2.\ \ P: =O\Longrightarrow 2sR^2=2s\cdot OI^2+4Rsr\Longrightarrow OI^2=R^2-2Rr.$ Let $ABC$ be a fixed triangle, a fixed point $F(u,v,w)$ and the mobile point $M: =M(x,y,z)$ (with barycentrical coordinates). I note the power $p(X)$ of the point $X$ w.r.t. the circumcircle of the triangle $ABC$. Then for $\left(\forall\right) M$ there are the following relations: $1^{\circ}.\ p(M)=-(yza^2+zxb^2+xyc^2).$ $2^{\circ}.\ x\cdot MA^2+y\cdot MB^2+z\cdot MC^2=-p(M).$ $3^{\circ}.\ u\cdot MA^2+v\cdot MB^2+w\cdot MC^2=MF^2-p(F).$ $4^{\circ}.\ FM^2=-[(y-v)(z-w)a^2+(z-w)(x-u)b^2+(x-u)(y-v)c^2]\ge0.$ The remarcable points for a triangle $ABC: \ A(1,0,0),\ B(0,1,0),\ C(0,0,1);$ $G(\frac 13,\frac 13,\frac 13),\ I\left(\frac{a}{2s},\frac{b}{2s},\frac{c}{2s}\right),\ N\left(\frac{s-a}{s},\frac{s-b}{s},\frac{s-c}{s}\right), I_a\left(\frac{-a}{2(s-a)},\frac{b}{2(s-a)},\frac{c}{2(s-a)}\right);$ $H(\cot B\cot C,\coth\cot A,\cot A\ cot B),\ O\left(\frac{R^2}{2S} \sin 2A,\frac{R^2}{2S} \sin 2B,\frac{R^2}{2S}\sin 2C\right);$ $\Gamma\left(\frac{(s-b)(s-c)}{r(4R+r)},\frac{(s-c)(s-a)}{r(4R+r},\frac{(s-a)(s-b)}{r(4R+r)}\right)$ (the Gergogne's point); $L\left(\frac{a^2}{a^2+b^2+c^2},\frac{b^2}{a^2+b^2+c^2},\frac{c^2}{a^2+b^2+c^2}\right)$ (the Lemoine's point or the simedian center). Remark. You observe that: $x+y+z=1,\ \frac{[MBC]}{|x|}=\frac{[AMC]}{|y|}=\frac{[ABM]}{|z|}$, where $[XYZ]$ is the area of the triangle $ABC.$
cosinator
07.07.2009 16:56
Here is a solution with vectors. The claim is clearly true for $ P=Q$. No we shall show that given a point $ P$ satisfying the property, another point $ R$ will satisfy the property as well. That is, we must prove $ a(PA)^{2}+b(PB)^{2}+c(PC)^{2}= a(QA)^{2}+b(QB)^{2}+c(QC)^{2}+(a+b+c)(QP)^{2}$ $ \Rightarrow a(RA)^{2}+b(RB)^{2}+c(RC)^{2}= a(QA)^{2}+b(QB)^{2}+c(QC)^{2}+(a+b+c)(QR)^{2}$ So we must prove the respective differences are the same: $ \sum a(PA^2-RA^2)=(a+b+c)(QP^2-QR^2)$ We use vectors: $ \sum a(PA^2-RA^2)=(a+b+c)(QP^2-QR^2)$ $ \Leftrightarrow \sum a(PA-RA)(PA+RA)=(a+b+c)(QP-QR)(QP+QR)$ $ \Leftrightarrow \sum a(P-R)(P+R+2A)=(a+b+c)(P-R)(P+R+2Q)$ $ \Leftrightarrow \sum a(P+R+2A)=(a+b+c)(P+R+2Q)$ $ \Leftrightarrow \sum a(2A)=(a+b+c)(2Q)$ Now we simply have the vector definition of Q, the incenter of the triangle.
jt314
19.03.2019 05:09
Place masses $a, b, c$ on vertices $A, B, C$. It is clear that the centre of gravity is $Q$. By parallel axis theorem, the solution follows.
proshi
19.03.2019 06:48
My solution uses $I$ for incenter instead of $Q$: Let $I$ be the origin to all the vectors I will work with. First off, note that $\vec{I}=0$ if we take the origin as $I.$ We know that $\vec{PA}=\vec{IP}+\vec{IA}.$ The same holds true for the rest as well. So, the left hand side of our equation becomes $$a(\vec{IP}+\vec{IA})^2+b(\vec{IP}+\vec{IB})^2+c(\vec{IC}+\vec{IP})^2.$$We expand that to get $$(a \cdot PA^2 + b \cdot PB^2 + c \cdot PC^2 = a \cdot IA^2 + b \cdot IB^2 + c \cdot IC^2 + (a + b + c) \cdot IP^2)+2\vec{IP}(a\vec{IA}+b\vec{IB}+c\vec{IC}).$$All that remains is showing that $2\vec{PI}(a\vec{IA}+b\vec{IB}+c\vec{IC})=0.$ First off, since we took $I$ as the origin, we know that $a\vec{IA}+b\vec{IB}+c\vec{IC}$ is the same as $a\vec{A}+b\vec{B}+c\vec{C}.$ Now, this reminds us of the barycentric coordinates of incenter of $\triangle ABC.$ Namely, we know that $$\dfrac{a}{a+b+c}\vec{A}+\dfrac{b}{a+b+c}\vec{B}+\dfrac{c}{a+b+c}\vec{C}=\vec{I}=0.$$Multiplying through by $a+b+c$ gives $a\vec{A}+b\vec{B}+c\vec{C}=0$ as desired. So, we have shown that $a \cdot PA^2 + b \cdot PB^2 + c \cdot PC^2 = a \cdot IA^2 + b \cdot IB^2 + c \cdot IC^2 + (a + b + c) \cdot IP^2.$
DottedCaculator
15.11.2021 17:05
Let $P$ be the origin, and let $I=Q$. Then, we need to show $aA\cdot A+bB\cdot B+cC\cdot C=aIA^2+bIB^2+cIC^2+(a+b+c)I\cdot I$. We have
\begin{align*}
IA^2&=\frac1{(a+b+c)^2}(bB+cC-(b+c)A)\cdot(bB+cC-(b+c)A)\\
&=\frac1{(a+b+c)^2}((b+c)cb^2+(b+c)bc^2-bca^2)\\
&=\frac1{(a+b+c)^2}{bc(b+c)^2-bca^2}\\
&=\frac{bc(b+c+a)(b+c-a)}{(a+b+c)^2}\\
&=\frac{bc(b+c-a)}{a+b+c}.\end{align*}Therefore, $aIA^2+bIB^2+cIC^2=abc$. Now, we have
\begin{align*}
(a+b+c)I\cdot I&=\frac1{a+b+c}(aA+bB+cC)\cdot(aA+bB+cC)\\
&=\frac1{a+b+c}(a^2A\cdot A+b^2B\cdot B+c^2C\cdot C+2abA\cdot B+2bcB\cdot C+2caC\cdot A)\\
&=\frac1{a+b+c}(A\cdot A(a^2+ab+ac)+B\cdot B(ba+b^2+bc)+C\cdot C(ca+cb+c^2)-abc(a+b+c))\\
&=aA\cdot A+bB\cdot B+cC\cdot C-abc.\end{align*}Therefore, we get $aA\cdot A+bB\cdot B+cC\cdot C=aIA^2+bIB^2+cIC^2+(a+b+c)I\cdot I$.
pinkpig
01.02.2022 21:48
Let $I=Q.$
Note that $$a\cdot IA^2+b\cdot IB^2+c\cdot IC^2=a\cdot(\overrightarrow{A}-\overrightarrow{I})^2+b\cdot(\overrightarrow{B}-\overrightarrow{I})^2+b\cdot(\overrightarrow{C}-\overrightarrow{I})^2=$$$$a\cdot(\overrightarrow{A}\cdot\overrightarrow{A}-2\overrightarrow{A}\cdot\overrightarrow{I}+\overrightarrow{I}\cdot\overrightarrow{I})+b\cdot(\overrightarrow{B}\cdot\overrightarrow{B}-2\overrightarrow{B}\cdot\overrightarrow{I}+\overrightarrow{I}\cdot\overrightarrow{I})+c\cdot(\overrightarrow{C}\cdot\overrightarrow{C}-2\overrightarrow{C}\cdot\overrightarrow{I}+\overrightarrow{I}\cdot\overrightarrow{I})=$$$$a\cdot|\overrightarrow{A}|^2+b\cdot|\overrightarrow{B}|^2+c\cdot|\overrightarrow{C}|^2-2(a\cdot\overrightarrow{A}\cdot\overrightarrow{I}+b\cdot\overrightarrow{B}\cdot\overrightarrow{I}+c\cdot\overrightarrow{C}\cdot\overrightarrow{I})+(a+b+c)\cdot\overrightarrow{I}\cdot\overrightarrow{I}=$$$$a\cdot|\overrightarrow{A}|^2+b\cdot|\overrightarrow{B}|^2+c\cdot|\overrightarrow{C}|^2-2\cdot\overrightarrow{I}\cdot(a\cdot\overrightarrow{A}+b\cdot\overrightarrow{B}+c\cdot\overrightarrow{C})+(a+b+c)\cdot\overrightarrow{I}\cdot\left(\frac{a\cdot\overrightarrow{A}+b\cdot\overrightarrow{B}+c\cdot\overrightarrow{C}}{a+b+c}\right)=$$$$a\cdot|\overrightarrow{A}|^2+b\cdot|\overrightarrow{B}|^2+c\cdot|\overrightarrow{C}|^2-\overrightarrow{I}\cdot(a\cdot\overrightarrow{A}+b\cdot\overrightarrow{B}+c\cdot\overrightarrow{C})$$Also, $$(a+b+c)IP^2=(a+b+c)|\overrightarrow{IP}|^2=(a+b+c)|\overrightarrow{P}-\overrightarrow{I}|^2=$$$$(a+b+c)(\overrightarrow{P}\cdot\overrightarrow{P}-2\cdot\overrightarrow{P}\cdot\overrightarrow{I}+\overrightarrow{I}\cdot\overrightarrow{I})=$$$$(a+b+c)(\overrightarrow{P}\cdot\overrightarrow{P})-2\cdot\overrightarrow{P}(a\cdot\overrightarrow{A}+b\cdot\overrightarrow{B}+c\cdot\overrightarrow{C})+(a+b+c)\cdot\overrightarrow{I}\cdot\overrightarrow{I}=$$$$(a+b+c)\cdot\overrightarrow{P}\cdot\overrightarrow{P}-2\cdot\overrightarrow{P}(a\cdot\overrightarrow{A}+b\cdot\overrightarrow{B}+c\cdot\overrightarrow{C})+(a+b+c)\cdot\overrightarrow{I}\cdot\left(\frac{a\cdot\overrightarrow{A}+b\cdot\overrightarrow{B}+c\cdot\overrightarrow{C}}{a+b+c}\right)=$$$$(a+b+c)\cdot\overrightarrow{P}\cdot\overrightarrow{P}-2\cdot\overrightarrow{P}(a\cdot\overrightarrow{A}+b\cdot\overrightarrow{B}+c\cdot\overrightarrow{C})+\overrightarrow{I}\cdot(a\cdot\overrightarrow{A}+b\cdot\overrightarrow{B}+c\cdot\overrightarrow{C}).$$In addition, $$a\cdot PA^2+b\cdot PB^2+c\cdot PC^2=a\cdot|\overrightarrow{PA}|^2+b\cdot|\overrightarrow{PB}|^2+c\cdot|\overrightarrow{PC}|^2=$$$$a\cdot|\overrightarrow{P}-\overrightarrow{A}|^2+b\cdot|\overrightarrow{P}-\overrightarrow{B}|^2+c\cdot|\overrightarrow{P}-\overrightarrow{C}|^2=$$$$a\cdot(\overrightarrow{P}\cdot\overrightarrow{P}-2\cdot\overrightarrow{P}\cdot\overrightarrow{A}+\overrightarrow{A}\cdot\overrightarrow{A})+b\cdot(\overrightarrow{P}\cdot\overrightarrow{P}-2\cdot\overrightarrow{P}\cdot\overrightarrow{B}+\overrightarrow{B}\cdot\overrightarrow{B})+c\cdot(\overrightarrow{P}\cdot\overrightarrow{P}-2\cdot\overrightarrow{P}\cdot\overrightarrow{C}+\overrightarrow{C}\cdot\overrightarrow{C})=$$$$(a+b+c)\cdot\overrightarrow{P}\cdot\overrightarrow{P}-2\cdot\overrightarrow{P}(a\cdot\overrightarrow{A}+b\cdot\overrightarrow{B}+c\cdot\overrightarrow{C})+(a\cdot|\overrightarrow{A}|^2+b\cdot|\overrightarrow{B}|^2+c\cdot|\overrightarrow{C}|^2).$$We conclude that $$a\cdot IA^2+b\cdot IB^2+c\cdot IC^2=a\cdot(\overrightarrow{A}-\overrightarrow{I})^2+b\cdot(\overrightarrow{B}-\overrightarrow{I})^2+b\cdot(\overrightarrow{C}-\overrightarrow{I})^2+(a+b+c)IP^2=$$$$a\cdot|\overrightarrow{A}|^2+b\cdot|\overrightarrow{B}|^2+c\cdot|\overrightarrow{C}|^2-\overrightarrow{I}\cdot(a\cdot\overrightarrow{A}+b\cdot\overrightarrow{B}+c\cdot\overrightarrow{C})+(a+b+c)\cdot\overrightarrow{P}\cdot\overrightarrow{P}-2\cdot\overrightarrow{P}(a\cdot\overrightarrow{A}+b\cdot\overrightarrow{B}+c\cdot\overrightarrow{C})+\overrightarrow{I}\cdot(a\cdot\overrightarrow{A}+b\cdot\overrightarrow{B}+c\cdot\overrightarrow{C})=$$$$(a+b+c)\cdot\overrightarrow{P}\cdot\overrightarrow{P}-2\cdot\overrightarrow{P}(a\cdot\overrightarrow{A}+b\cdot\overrightarrow{B}+c\cdot\overrightarrow{C})+(a\cdot|\overrightarrow{A}|^2+b\cdot|\overrightarrow{B}|^2+c\cdot|\overrightarrow{C}|^2)=$$$$a\cdot PA^2+b\cdot PB^2+c\cdot PC^2=a\cdot|\overrightarrow{PA}|^2+b\cdot|\overrightarrow{PB}|^2+c\cdot|\overrightarrow{PC}|^2,$$as desired.