Yours is an obvious counter-example, Bleaug. Yeah, the claim is wrong. Minimal value would be a nice problem.

I wonder if k continues to shrink as the farthest point in your drawing (Point S?) continues to go further away. If that is the case, then may be k>1 holds. It'll be really surprising if the limit on k is not an integer!

I (think) I have a counter-example for fox 350 but I can't see a way of enclosing the details with my attachment. The following is a brief description of it.

Let P(3, 4)Q(0, 0)R(10, 0)S(13, 7) be the quad with SP being the longest. PR is joined and ST is drawn parallel to PR cutting QR produced at T(t, 0). Area of PQRS is the same as triangle PQT. On QR, the point H(h, 0) is located such that area of triangle PQH = A as given. Then, area of triangle PHT = kA + A. By the logic that the ratio of the areas of two triangles, having the same altitude, is equal to the ratio of their bases, we arrive at:- If the proposition (k >= 2) is true, then TH:HQ must be greater 3. (1) by equating areas, h = 70/11 (2) by equating slopes, t = 101/4 Then, TH:HQ = (101/4 - 70/11) : 70/11 = 2.9678...unfortunately.

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Sure, this is a convex quadrilateral, although not mentioned. Again let's not try to be boringly precise.

ReplyDeleteAlso, if the claim is false, there should be a counter-example.

a quick one: http://bleaug.free.fr/8foxes/8foxes350.png

ReplyDeleteI guess we could rephrase the problem: what is the minimal value of k when PS is the longest side (and PQRS is convex?)

bleaug

Yours is an obvious counter-example, Bleaug. Yeah, the claim is wrong. Minimal value would be a nice problem.

ReplyDeleteI wonder if k continues to shrink as the farthest point in your drawing (Point S?) continues to go further away. If that is the case, then may be k>1 holds. It'll be really surprising if the limit on k is not an integer!

Thank you for the correction Bleaug.

-8foxes

here is an answer under the assumption that quadrilateral must be convex:

ReplyDeletehttp://bleaug.free.fr/8foxes/8foxes350b.png

bleaug

I (think) I have a counter-example for fox 350 but I can't see a way of enclosing the details with my attachment. The following is a brief description of it.

ReplyDeleteLet P(3, 4)Q(0, 0)R(10, 0)S(13, 7) be the quad with SP being the longest.

PR is joined and ST is drawn parallel to PR cutting QR produced at T(t, 0).

Area of PQRS is the same as triangle PQT.

On QR, the point H(h, 0) is located such that area of triangle PQH = A as given. Then, area of triangle PHT = kA + A.

By the logic that the ratio of the areas of two triangles, having the same altitude, is equal to the ratio of their bases, we arrive at:-

If the proposition (k >= 2) is true, then TH:HQ must be greater 3.

(1) by equating areas, h = 70/11

(2) by equating slopes, t = 101/4

Then, TH:HQ = (101/4 - 70/11) : 70/11

= 2.9678...unfortunately.

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ReplyDelete