SAT1000 - P889

Geometry Level pending

Let $P$ be an arbitrary point in $\triangle ABC$ , $\lambda_1 = \dfrac{S_{\triangle PBC}}{S_{\triangle ABC}}$ , $\lambda_2 = \dfrac{S_{\triangle PCA}}{S_{\triangle ABC}}$ , $\lambda_3 = \dfrac{S_{\triangle PAB}}{S_{\triangle ABC}}$ .

Define $f: \mathbb R^2 \mapsto \mathbb R^3$ , $f(P)=(\lambda_1,\lambda_2,\lambda_3)$ .

If point $G$ is the centroid of $\triangle ABC$ , $f(Q)=(\dfrac{1}{2},\dfrac{1}{3},\dfrac{1}{6})$ , then which choice is true ?

$A.\ \textup{Q is always in}\ \triangle GAB.$

$B.\ \textup{Q is always in}\ \triangle GBC.$

$C.\ \textup{Q is always in}\ \triangle GCA.$

$D.\ \textup{Q is concurrent with G}.$

Note: $S_{\triangle ABC}$ denotes the area of $\triangle ABC$ .

Have a look at my problem set: SAT 1000 problems

B

A

C

D

2 solutions

David Vreken
Jun 27, 2020

$\triangle ABC$ can be stretched and skewed and still maintain its ratio of areas, so stretch and skew it so that $A$ is at $A(6, 0)$ , $B$ is at $B(0, 6)$ , and $C$ is at $C(0, 0)$ .

Let the coordinates of $Q$ be $Q(x, y)$ . Then $S_{\triangle QBC} = 3x$ and $S_{\triangle QCA} = 3y$ , and since $f(Q) = (\frac{1}{2}, \frac{1}{3}, \frac{1}{6})$ , $\frac{3x}{18} = \frac{1}{2}$ and $\frac{3y}{18} = \frac{1}{3}$ , which solves to $x = 3$ and $y = 2$ , so the coordinates of $Q$ are $Q(3, 2)$ .

Let $M$ be the midpoint of $BC$ and $N$ be the midpoint of $AB$ .

Then $M$ is at $M(3, 0)$ and $N$ is at $(3, 3)$ , and the equation of medians $AM$ and $CN$ are $y = -2x + 6$ and $y = x$ , which intersect at the centroid $G(2, 2)$ .

Since $Q$ is to the right of $G$ , $Q$ is always in $\triangle GAB$ , so option A is correct.

X X
Jun 26, 2020

Apply barycentric coordinates and observe the cevians.

SAT1000 - P889

2 solutions

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