Let $ABCD$ be a convex quadrilateral such that \[\begin{array}{rl} E,F \in [AB],& AE = EF = FB \\ G,H \in [BC],& BG = GH = HC \\ K,L \in [CD],& CK = KL = LD \\ M,N \in [DA],& DM = MN = NA \end{array}\] Let \[[NG] \cap [LE] = \{P\}, [NG]\cap [KF] = \{Q\},\] \[{[}MH] \cap [KF] = \{R\}, [MH]\cap [LE]=\{S\}\] Prove that $Area(ABCD) = 9 \cdot Area(PQRS)$ $NP=PQ=QG$
Problem
Source: Turkey TST 1990 - P4
Tags: geometry, analytic geometry, linear algebra, matrix, rotation, geometry proposed
nikoma
15.09.2013 18:04
Consider barycentric coordinates $A = (1,0,0)$, $B = (0,1,0)$, $C = (0,0,1)$, $D = (p,q,1 - p - q)$. Then \[E = \left(\frac{2}{3},\frac{1}{3},0\right) \wedge F = \left(\frac{1}{3},\frac{2}{3},0\right) \wedge G = \left(0,\frac{2}{3},\frac{1}{3}\right) \wedge H = \left(0,\frac{1}{3},\frac{2}{3}\right)\] and \[K = \left(\frac{p}{3},\frac{q}{3}, \frac{3 - p - q}{3}\right) \wedge L = \left(\frac{2p}{3},\frac{2q}{3},\frac{3 - 2p - 2q}{3}\right)\] \[M = \left(\frac{1 + 2p}{3},\frac{2q}{3},\frac{2 - 2p - 2q}{3}\right) \wedge N = \left(\frac{2 + p}{3},\frac{q}{3},\frac{1 - p - q}{3}\right)\] Hence equation of line $NG$ is \[(2-2p-3q)x+(p+2)y-(2p+4)z = 0\] The equation of line $MH$ is \[(2p + 6q - 2)x - (4p + 2)y + (2p + 1)z = 0\] The equation of line $EL$ is \[(2p + 2q - 3)x - (4p + 4q - 6)y + (2p - 4q)z = 0\] And the equation of line $FK$ is \[(2p + 2q - 6)x + (3 - p - q)y + (2p - q)z = 0\] Intersecting respective lines we get \[P = \frac{1}{9}(2p + 4,2q + 2,-2p -2q + 3)\] \[Q = \frac{1}{9}(p + 2,q + 4,-p - q + 3)\] \[R = \frac{1}{9}(2p + 1,2q + 2,-2p - 2q + 6)\] \[S = \frac{1}{9}(4p + 2,4q + 1,-4p-4q+6)\] Then \[\overrightarrow{NP} = \frac{1}{9}(p + 2,q - 2,-p-q)\] \[\overrightarrow{PQ} = \frac{1}{9}(p + 2,q - 2,-p - q)\] \[\overrightarrow{QG} = \frac{1}{9}(p + 2,q - 2,-p - q)\] i.e. $\boxed{NP = PQ = QK}$. Now note that \begin{align*}[ABCD] = [ABC] + [ACD] = [ABC] + [ABC] \cdot \begin{vmatrix}
1 & 0 & 0\\
0 & 0 & 1\\
p & q & 1 - p - q
\end{vmatrix} = (1 - q) \cdot [ABC] \end{align*}
and \[[PQRS] = [PQR] + [PRS]\] Note that \[[PQR] = [ABC] \cdot \frac{1}{729} \cdot \begin{vmatrix}
2p + 4 & 2q + 2 & -2p - 2q + 3\\
p + 2 & q + 4 & -p - q + 3\\
2p + 1 & 2q + 2 & -2p - 2q + 6
\end{vmatrix} = [ABC] \cdot \frac{2 - q}{27}\]
\[[PRS] = [ABC] \cdot \frac{1}{729} \cdot \begin{vmatrix}
2p + 4 & 2q + 2 & -2p - 2q + 3\\
2p + 1 & 2q + 2 & -2p - 2q + 6\\
4p + 2 & 4q + 1 & -4p - 4q + 6
\end{vmatrix} = [ABC] \cdot \frac{1 - 2q}{27}\]
Hence \[9 \cdot [PQRS] = 9 \cdot [ABC] \cdot \frac{3 - 3q}{27} = [ABC] \cdot (1 - q) = [ABCD]\]
As wanted.
Vo Duc Dien
18.09.2013 19:19
a) Let I be the midpoint of AB. From I draw the segments ID’ || AD, ID’ = AD, IL’ || EL, IL’ = EL, IC’ || BC, IC’ = BC, IK’ || FK, IK’ = FK. We have LL’ = EI = ½AE || DD’ and LL’ = IF = KK’ = ½DD’ = ½CC’. Now let J be the intersection of CD and C’D’. It is also seen that the two triangles CC’J and DD’J are congruent and thus J is the midpoint of CD. Since LL’ || DD’ || CC’ || KK’, LL’ = ½DD’ = ½CC’ = KK’ and LJ = ½DL = ½CK = JK, D’, L’, J, K’ and C’ are collinear, and we have L’J = ½D’L’. Now pick points N’, M’ on ID’ such that IN’ = N’M’ = M’D’, and points G’, H’ on IC’ such that IG’ = G’H’ = H’C’. With the same analogy above, we also conclude that N’G’ intersects NG at its midpoint; let this midpoint be X. The same can be said about M’H’ with midpoint of MH being denoted Y. Observe that since C’D’ || M’H’ || N’G’, the midpoints X, Y and J must be on a straight line. Proceed by letting S’ = IL’ ∩ M’H’. Because MM’ || LL’, there is only one single line starting from S’ that parallels both MM’ and LL’. Let this line intersect EL at a temporary point Z with S’Z = LL’ = ½MM’. Because S’Y/S’M’ = L’J/D'L' = 1/2, S’Z = ½MM’ and, therefore, Z must also be on MY, or Z is the common point of EL and MY and Z coincides with S. Furthermore, since LJ = JK = ½LK = ½DL, SY = ½MS. Similarly, on the right side RY = ½HR, or MS = SR = RH. Applying this procedure for segment NG and we would also have NP = PQ = QG. Also observe that if we rotate the configuration so that A → B, B → C, C → D, D → A, etc… we would also easily see that FQ = QR = RK and EP = PS = SL. b) Now it is easily seen that (PQRS) = 1/3 of (MNGH) - reader to check that out - and (MNGH) = 1/3 of (ABCD). Therefore, (PQRS) = 1/9 of (ABCD) where () denotes area.