Proof of Interesting Geometric Vector TheoremDate: 09/17/2005 at 09:50:25 From: James Subject: Proof of a geometric theorem O is the centre of the circumscribing circle of triangle ABC and H is its orthocentre. Prove that vector OH is equal to the sum of the vectors OA, OB and OC. Combining vectors OB and OC is OL where OL is a diagonal of the rhombus OCLB on OB and OC as adjacent sides. Combining vectors OL and OA to give vector OH would complete the proof. I am unable to do it. I need to show that the quadrilateral AOLH is a parallelogram. Now AH is parallel to OL (both perpendicular to BC) but I cannot show that AH = OL which would prove AOLH is a parallelogram. Nor can I show, alternatively, that AO is parallel to HL. Hope you can help. Date: 09/23/2005 at 11:25:07 From: Doctor George Subject: Re: Proof of a geometric theorem Hi James, I have come up with two proofs of your conjecture. One uses analytic geometry, and the other uses vector algebra. Analytic Geometry Proof ----------------------- Position the circumscribed circle with O at (0,0). Now define A as (a,b), B as (-a,b) and C as (c,d). This convention is perfectly general because any triangle can be positioned and oriented to meet these conditions. First, add the three vectors. OA + OB + OC = (c, d+2b) Now construct the orthocenter as the intersection of two altitudes of the triangle. The altitude from C is on the line x = c. The altitude from B is on a line with slope m = (a-c)/(d-b). To find the orthocenter we need the intersection of x = c and y - b = [(a-c)/(d-b)](x+a) If we substitute x = c we find that the y coordinate of the orthocenter is y = [(a-c)/(d-b)](c+a) + b = (a^2-c^2)/(d-b) + b Now substitute a^2 = r^2 - b^2 and c^2 = r^2 - d^2, where 'r' is the radius of the circumscribed circle. y = (r^2- b^2 - r^2 + d^2)/(d-b) + b = (d^2 - b^2)/(d-b) + b = [(d+b)(d-b)]/(d-b) + b = d + 2b Therefore, the orthocenter is (c, d+2b) as was to be shown, so the proof is complete. Vector Algebra Proof -------------------- Let's start with the following observation. For any vectors U and V, if |U| = |V| then U+V is perpendicular to U-V. We show this using the dot product. (U+V).(U-V) = |U|^2 - U.V + V.U - |V|^2 = 0 Now for simplicity we will use different notation from the first proof. Let the vectors from the center of the circumscribed circle to the vertices of the triangle be A, B and C. Note that |A| = |B| = |C| and vectors A-B, B-C and C-A represent the three sides of the triangle. We want to show that the orthocenter is the point A+B+C. Now construct the orthocenter as the intersection of two altitudes of the triangle. Since B+C is perpendicular to B-C the altitude from A is on the line with parametric equation L1 = A + a(B+C) where 'a' is the parameter. Also, since A+B is perpendicular to A-B the altitude from C is on the line with parametric equation L2 = C + b(A+B) where 'b' is the parameter. When L1 and L2 intersect A + a(B+C) = C + b(A+B) a(B+C) = C - A + b(A+B) Now take the dot product of both sides with A-B. Note that this causes 'b' to drop out, which allows us to solve for 'a'. a(B+C).(A-B) = (C-A).(A-B) a(B.A - |B|^2 + C.A - C.B) = C.A - C.B - |A|^2 + A.B Since |B| = |A| the quantity in the parentheses is equal to the right hand side. It can be shown that this quantity is non-zero, so we can divide by it to arrive at a = 1 The orthocenter is the value of L1 when a = 1. Therefore, the orthocenter is the point A+B+C as was to be shown, so the proof is complete. Thanks for writing. This has been a fun problem. Write again if you need more help. - Doctor George, The Math Forum http://mathforum.org/dr.math/ Date: 09/25/2005 at 10:01:59 From: James Subject: Thank you (Proof of a geometric theorem) Dr. George - Thank you very much indeed for your elegant vector algebra solution and also your solution by analytical geometry. I can sleep much more easily now that I know the conjecture/theorem can be proved true. I came across the problem on page 32 of the Penguin Dictionary of Curious and Interesting Geometry by David Wells, first published 1991. It's well worth a read but it has given me a few headaches doing so! Thanks again for your help and expertise. - James Date: 09/25/2005 at 21:34:25 From: Doctor George Subject: Re: Thank you (Proof of a geometric theorem) Hi James, I'm glad that you were able to follow the proofs. The vector algebra proof can actually be shortened. Instead of solving for 'a' as I did, the system L1 = A + a(B+C) L2 = C + b(A+B) can be solved by inspection to give a = 1 and b = 1. We now have a very concise proof for a theorem that is not intuitive. Thanks for the book reference. Since not many people read such books it seems that math is one of your hobbies. Write again anytime. - Doctor George, The Math Forum http://mathforum.org/dr.math/ |
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