Geometry Problem

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 824.

## Tuesday, November 27, 2012

### Problem 824: Desargues' Theorem, Triangles in Perspective, Concurrent Lines, Collinear Points, Center, Axis, Perspector, Perspectrix

Labels:
axis,
center,
collinear,
concurrent,
Desargues' theorem,
perspective,
perspector,
perspectrix,
triangle

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Apply Menelaus’s theorem to triangle OAB and transversal DB’A’ we have

ReplyDelete(DA/DB).(B’B/B’O).(A’O/A’A)=-1

Apply Menelaus’s theorem to triangle OBC and transversal FC’B’ we have

(FB/FC).(C’C/C’O).(B’O/B’B)=-1

Apply Menelaus’s theorem to triangle OAC and transversal EC’A’ we have

(EC/EA).(A’A/A’O).(C’O/C’C)=-1

Multiplying the three equalities we obtain after simplification.

(DA/DB).(FB/FC).(EC/EA)=-1

So D,E,F are collinear per converse of Menelaus’s theorem

Using homogeneous coordinates in projective geometry.

ReplyDeleteWLOG, we may let A=[1,0,0], B=[0,1,0], C=[0,0,1] and O=[1,1,1].

Then we have

line OA: y=z

line OB: x=z

line OC: x=y

Thus we may let A'=[a,1,1], B'=[1,b,1], C'=[1,1,c].

Now we have

line AB: z=0

line A'B': (1−b)x−(a-1)y+(ab−1)z=0

∴ D = AB∩A'B' = [a−1,1−b,0]

line AC: y=0

line A'C': (c−1)x−(ac-1)y+(a−1)z=0

∴ E = AC∩A'C' = [a−1,0,1−c]

line BC: x=0

line B'C': (bc−1)x−(c-1)y+(1−b)z=0

∴ F = BC∩B'C' = [0,b−1,1−c]

Now consider the determinant

|a−1 1−b 0 |

|a−1 0 1−c| = −(a−1)(b−1)(1−c)−(a−1)(1−b)(1−c) = 0

| 0 b−1 1−c|

Hence, D,E,F are collinear.

Another proof using projective geometry but without coordinates.

ReplyDeleteSince O lies on AA', we can write O=mA+nA'.

Similarly, we have O=pB+qB' and O=rC+sC'.

So we have

mA+nA' = pB+qB' ⇒ mA−pB = −nA'+qB' = D

mA+nA' = rC+sC' ⇒ mA−rC = −nA'+sC' = E

pB+qB' = rC+sC' ⇒ pB−rC = −qB'+sC' = F

D+E+F = (mA−pB)+(mA−rC)+(pB−rC) = 0

So D,E,F are collinear.