## Quadrilateral Inequality involving Perimeter & Diagonals

For a convex quadrilateral *ABCD* in the plane, the following inequality holds: 1/2 perimeter *ABCD* < *AC* + *BD* < perimeter *ABCD*.

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Quadrilateral Inequality involving Perimeter & Diagonals

Can you prove the result? If stuck, click the *Hint* in the sketch.

Can you provide counter-examples for the result when *ABCD* is concave or crossed?

(The problem was used in the Senior 3rd Round of the South African Mathematics Olympiad (SAMO) in 2001 as the first 'warm-up' problem. A solution is given at *SAMO Solutions: 2001*. Past SAMO papers & solutions for several years are available at *Past SAMO Papers & Solutions*.)

**Some Applications**^{1}

(1) Using the above inequality together with a familiar property of the sum of the opposite sides of a circumscribed quadrilateral *ABCD*, it follows immediately that for a convex, circumscribed *ABCD* the sum of the diagonals divided by the sum of any pair of opposite sides is always greater than 1 and less than 2. For example: 1 < (*AC* + *BD*)/(*AB* + *CD*) < 2 as shown below.

Circumscribed Quadrilateral Inequality

What do you notice about the maximum value of the ratio (*AC* + *BD*)/(*AB* + *CD*)? When does it occur and what is the precise value of its upper limit? Can you explain (prove) your observation?

(2) Actually, the lower bound of the above inequality for a convex, circumscribed quadrilateral applies to any convex quadrilateral! In other words, the sum of any pair of opposite sides of a convex quadrilateral is always less than the sum of the diagonals as shown below.

Another convex quadrilateral inequality

Can you explain (prove) why this result is true?

**Footnote**^{1} These two simple, but interesting applications were discovered by the author on 8 Dec 2010 upon reflecting on possible implications/applications of the quadrilateral diagonal-perimeter inequality at the top.

**Another Quadrilateral Inequality**

One of the most famous quadrilateral inequalities is that of Ptolemy (~ 100-168 AD), which states that for any quadrilateral *ABCD*, *AB*CD* + *BC*AD* >= *AC*BD* (and equality only holds when *ABCD* is cyclic (& convex)). Read more about the result and its applications at Ptolemy's Theorem.

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Last updated by Michael de Villiers, 19 March 2011.