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Let $k$ be a circle with radius $1$ and center $O(0;0)$

Points $A$ and $B$ are on the circle in the first quadrant, between $X(1;0)$ and $Y(0;1)$.

$AA_x||BB_x||YO$, $AA_y||BB_y||XO$ and points $A_x$,$B_x$ are on the $x$ axis, while points $A_y$,$B_y$ are on the $y$ axis.

What is the area of $A A_x B_x B + A A_y B_y B$ as a function of $AB$?

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I already conjectured it to be $\sqrt{x^2-\dfrac{x^4}{4}}$ but hasn't been able to prove it yet. ($x=AB$)

hjpotter92
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Ralph
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1 Answers1

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If I understand correctly, you want to add the areas of the two quadrilaterals (though your notation seems to indicate that you're adding the quadrilaterals, which would be something different).

Let $A=(\cos\phi_A,\sin\phi_A)$ and $B=(\cos\phi_B,\sin\phi_B)$, and introduce $C=(\cos\phi_A,\sin\phi_B)$ and $D=(\cos\phi_B,\sin\phi_A)$.

The triangular area $ABC$ is counted twice, and it's equal to the area of the other triangle $ABD$, so the total is equal to the area of the big rectangle $OA_yDB_x$ minus the area of the small rectangle $OB_yCA_x$. This is $\sin\phi_A\cos\phi_B-\cos\phi_A\sin\phi_B=\sin(\phi_A-\phi_B)$. Thus it depends only on the angle separating $A$ and $B$, and thus, as you had assumed, can be expressed in terms of the length of $AB$. This is a chord with angle $\phi_A-\phi_B$, and thus has length $x=2\sin((\phi_A-\phi_B)/2)$. Since $\sin \theta=2\sin(\theta/2)\cos(\theta/2)=2\sin (\theta/2)\sqrt{1-\sin^2(\theta/2)}$, your conjecture is correct.

joriki
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