A geometry problem by Relue Tamref

Geometry Level 4

The left diagram below shows that, given the green parallelogram, 4 right triangles have been constructed--two of them each having leg lengths $a$ and $b,$ and the other two $x$ and $y,$ where $a,b,x,y$ are all distinct real numbers.

The right diagram shows exactly the same except that the given green figure is a rectangle, not a parallelogram.

Find the relationship between the area of the parallelogram $(P)$ and the area of the rectangle $(R).$

Note: The diagrams may not be drawn to scale.

P<R

P\le R

P=R

P\ge R

P>R

1 solution

Relue Tamref
Sep 2, 2017

This is Cauchy-Schwarz Inequality

As you can see $(y+a)(x+b) - yx - ab = ax + by$ it's equal to the parallelogram area, but it's also $b \times \sqrt{x^2 + y^2}$ , and the rectangle area it's $\sqrt{b^2 + a^2} \times \sqrt{x^2 + y^2}$ so $ax + by = b \times \sqrt{x^2 + y^2} \leq \sqrt{b^2 + a^2} \times \sqrt{x^2 + y^2} \iff b \leq \sqrt{b^2 + a^2} \Rightarrow b^2 \leq b^2 + a^2$ wich is true.

A geometry problem by Relue Tamref

1 solution

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