Plane Geometry problem: $a(PA)^2 + b(PB)^2 + c(PC)^2$ is minimum
Problem: Let $ABC$ be a fixed triangle and $P$ be a variable point in the plane of $Delta ABC$. If $a(PA)^2 + b(PB)^2 + c(PC)^2$ is minimum, then the point P with respect to $Delta ABC$ is
(A) Centroid $quad quad$ (B) Circumcentre $quad quad$ (C) Orthocentre $quad quad$ (D) Incentre
I tried this problem with pure geometry but could not solve, please help me.
triangle plane-geometry
asked Dec 2 at 0:04
prashant sharma
576
add a comment |
Problem: Let $ABC$ be a fixed triangle and $P$ be a variable point in the plane of $Delta ABC$. If $a(PA)^2 + b(PB)^2 + c(PC)^2$ is minimum, then the point P with respect to $Delta ABC$ is
(A) Centroid $quad quad$ (B) Circumcentre $quad quad$ (C) Orthocentre $quad quad$ (D) Incentre
I tried this problem with pure geometry but could not solve, please help me.
triangle plane-geometry
asked Dec 2 at 0:04
prashant sharma
576
Hint: For any $alpha, beta, gamma > 0$, the expression $alpha |PA|^2 + beta |PB|^2 + gamma |PC|^2$ is minimized at the weighted centroid $P = frac{alpha A + beta B + gamma C}{alpha + beta + gamma}$.
– achille hui
Dec 2 at 0:54
@achille hui Please elaborate.
– prashant sharma
Dec 2 at 3:07
The $P$ you seek is $frac{a A + bB + cC}{a+b+c}$, i.e. the barycentric coordinates of $P$ wrt triangle ABC is $a : b : c$. Look at its wiki entry (in particular section 2.6) for the barycentric coordinates for a list of special points. Your point is there.
– achille hui
Dec 2 at 3:16
add a comment |
Problem: Let $ABC$ be a fixed triangle and $P$ be a variable point in the plane of $Delta ABC$. If $a(PA)^2 + b(PB)^2 + c(PC)^2$ is minimum, then the point P with respect to $Delta ABC$ is
(A) Centroid $quad quad$ (B) Circumcentre $quad quad$ (C) Orthocentre $quad quad$ (D) Incentre
I tried this problem with pure geometry but could not solve, please help me.
triangle plane-geometry
asked Dec 2 at 0:04
prashant sharma
576
Problem: Let $ABC$ be a fixed triangle and $P$ be a variable point in the plane of $Delta ABC$. If $a(PA)^2 + b(PB)^2 + c(PC)^2$ is minimum, then the point P with respect to $Delta ABC$ is
(A) Centroid $quad quad$ (B) Circumcentre $quad quad$ (C) Orthocentre $quad quad$ (D) Incentre
I tried this problem with pure geometry but could not solve, please help me.
triangle plane-geometry
triangle plane-geometry
asked Dec 2 at 0:04
prashant sharma
576
asked Dec 2 at 0:04
prashant sharma
576
asked Dec 2 at 0:04
prashant sharma
576
asked Dec 2 at 0:04
prashant sharma
576
asked Dec 2 at 0:04
prashant sharma
576
576
Hint: For any $alpha, beta, gamma > 0$, the expression $alpha |PA|^2 + beta |PB|^2 + gamma |PC|^2$ is minimized at the weighted centroid $P = frac{alpha A + beta B + gamma C}{alpha + beta + gamma}$.
– achille hui
Dec 2 at 0:54
@achille hui Please elaborate.
– prashant sharma
Dec 2 at 3:07
The $P$ you seek is $frac{a A + bB + cC}{a+b+c}$, i.e. the barycentric coordinates of $P$ wrt triangle ABC is $a : b : c$. Look at its wiki entry (in particular section 2.6) for the barycentric coordinates for a list of special points. Your point is there.
– achille hui
Dec 2 at 3:16
add a comment |
Hint: For any $alpha, beta, gamma > 0$, the expression $alpha |PA|^2 + beta |PB|^2 + gamma |PC|^2$ is minimized at the weighted centroid $P = frac{alpha A + beta B + gamma C}{alpha + beta + gamma}$.
– achille hui
Dec 2 at 0:54
@achille hui Please elaborate.
– prashant sharma
Dec 2 at 3:07
The $P$ you seek is $frac{a A + bB + cC}{a+b+c}$, i.e. the barycentric coordinates of $P$ wrt triangle ABC is $a : b : c$. Look at its wiki entry (in particular section 2.6) for the barycentric coordinates for a list of special points. Your point is there.
– achille hui
Dec 2 at 3:16
Hint: For any $alpha, beta, gamma > 0$, the expression $alpha |PA|^2 + beta |PB|^2 + gamma |PC|^2$ is minimized at the weighted centroid $P = frac{alpha A + beta B + gamma C}{alpha + beta + gamma}$.
– achille hui
Dec 2 at 0:54
Hint: For any $alpha, beta, gamma > 0$, the expression $alpha |PA|^2 + beta |PB|^2 + gamma |PC|^2$ is minimized at the weighted centroid $P = frac{alpha A + beta B + gamma C}{alpha + beta + gamma}$.
– achille hui
Dec 2 at 0:54
@achille hui Please elaborate.
– prashant sharma
Dec 2 at 3:07
@achille hui Please elaborate.
– prashant sharma
Dec 2 at 3:07
The $P$ you seek is $frac{a A + bB + cC}{a+b+c}$, i.e. the barycentric coordinates of $P$ wrt triangle ABC is $a : b : c$. Look at its wiki entry (in particular section 2.6) for the barycentric coordinates for a list of special points. Your point is there.
– achille hui
Dec 2 at 3:16
The $P$ you seek is $frac{a A + bB + cC}{a+b+c}$, i.e. the barycentric coordinates of $P$ wrt triangle ABC is $a : b : c$. Look at its wiki entry (in particular section 2.6) for the barycentric coordinates for a list of special points. Your point is there.
– achille hui
Dec 2 at 3:16
add a comment |
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Hint: For any $alpha, beta, gamma > 0$, the expression $alpha |PA|^2 + beta |PB|^2 + gamma |PC|^2$ is minimized at the weighted centroid $P = frac{alpha A + beta B + gamma C}{alpha + beta + gamma}$.
– achille hui
Dec 2 at 0:54
@achille hui Please elaborate.
– prashant sharma
Dec 2 at 3:07
The $P$ you seek is $frac{a A + bB + cC}{a+b+c}$, i.e. the barycentric coordinates of $P$ wrt triangle ABC is $a : b : c$. Look at its wiki entry (in particular section 2.6) for the barycentric coordinates for a list of special points. Your point is there.
– achille hui
Dec 2 at 3:16