# Proving $2500 \pi-100<\sqrt{1 \cdot 199}+\sqrt{2 \cdot 198}+\cdots+\sqrt{99 \cdot 101}<2500\pi$ by geometric methods

How to prove this inequality by geometric methods?

$$2500 \pi-100<\sqrt{1 \cdot 199}+\sqrt{2 \cdot 198}+\cdots+\sqrt{99 \cdot 101}<2500\pi$$

• See math.stackexchange.com/questions/312263/… for a very similar problem. – Martin R Jan 5 at 9:43
• Thanks. However, it is necessary to solve by geometric methods – Vertum Jan 5 at 9:56
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• To start with I would draw a circle with diameter 200 and recall the power of a point theorem. Then I would try to play with Riemann sums... – user Jan 5 at 10:24
• – Martin Sleziak Jan 5 at 11:41

Note that $$\sqrt{k(200-k)}=\sqrt{100^2-(k-100)^2}\qquad(1\leq k\leq99)\ .$$ This means that you should look at the circular disc $$(x-100)^2+y^2\leq100^2$$. Your sum is a rectangle approximation to the area of the upper left quarter of this disc.
• Thank. I used the similar idea $\sqrt{k(200-k)}$ is the height lowered from a right angle in a right triangle. The hypotenuse relies on diameter $d=200$ – Vertum Jan 5 at 10:44