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<imgclass="center-fit" src="polygonApproximation.png" alt="A slice of a circle with angle=2x and the corresponding isosceles triangle split in half">
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</figure>
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<br>
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<p>Now look at the isosceles triangle formed by joining the endpoints of the arc to the center.
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<br><br>
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Focus on half of this configuration: a right triangle with</p>
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<ulstyle="margin:6px">
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<li>one leg = r</li>
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<li>the other leg = half the polygon side</li>
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<li>hypotenuse = half the polygon diagonal</li>
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</ul>
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<br>
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<p>This right triangle always contains the corresponding half‑sector of the circle, so its area is larger.
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<br>
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But the key detail is the ratio of these areas.</p>
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<br>
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<mathxmlns="http://www.w3.org/1998/Math/MathML">
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<mrow>
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<mi>Area of the half‑sector</mi>
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<mo>=</mo>
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<mfrac>
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<mrow>
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<mi>x</mi>
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<mo>×</mo>
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<msup>
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<mi>r</mi>
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<mn>2</mn>
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</msup>
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</mrow>
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<mn>2</mn>
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</mfrac>
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</mrow>
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</math>
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<br><br>
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<mathxmlns="http://www.w3.org/1998/Math/MathML">
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<mrow>
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<mi>Area of the right triangle</mi>
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<mo>=</mo>
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<mfrac>
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<mrow>
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<mi>sin</mi>
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<mo>⁡</mo>
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<mo>(</mo><mi>x</mi><mo>)</mo>
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</mrow>
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<mrow>
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<mn>2</mn>
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<mo>×</mo>
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<mi>cos</mi>
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<mo>⁡</mo>
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<mo>(</mo><mi>x</mi><mo>)</mo>
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</mrow>
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</mfrac>
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<mo>×</mo>
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<msup>
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<mi>r</mi>
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<mn>2</mn>
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</msup>
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</mrow>
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</math>
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<br><br>
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<p>Since</p>
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<br>
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<mathxmlns="http://www.w3.org/1998/Math/MathML">
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<mrow>
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<mi></mi>
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<mi>cos</mi>
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<mo>⁡</mo>
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<mo>(</mo><mi>x</mi><mo>)</mo>
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<mo><</mo>
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<mn>1</mn>
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</mrow>
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</math>
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<br><br>
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<p>the inequality</p>
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<p>Imagine the half diagonals as two rigid plates and the side as a straight lid wedged between them.
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<br>
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<mathxmlns="http://www.w3.org/1998/Math/MathML">
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<mrow>
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<mfrac>
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<mrow>
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<mi>sin</mi>
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<mo>⁡</mo>
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<mo>(</mo><mi>x</mi><mo>)</mo>
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</mrow>
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<mrow>
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<mn>2</mn>
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<mo>×</mo>
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<mi>cos</mi>
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<mo>⁡</mo>
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<mo>(</mo><mi>x</mi><mo>)</mo>
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</mrow>
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</mfrac>
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<mo>></mo>
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<mi>x</mi>
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</mrow>
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</math>
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<br><br>
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<p>can hold even when</p>
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If we bend the lid into a curve, it can slip lower between the plates because the distance between its endpoints decreases, while its length doesn't change.
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<br>
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<mathxmlns="http://www.w3.org/1998/Math/MathML">
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<mrow>
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<mi>x</mi>
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<mo>></mo>
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<mi>sin</mi>
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<mo>⁡</mo>
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<mo>(</mo><mi>x</mi><mo>)</mo>
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</mrow>
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</math>
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The polygon still encloses more area, but the arc sits lower — even with an equal perimeter.
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<br><br>
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<p>This means that a polygon can enclose more area than the circle even when its perimeter is smaller than the circle’s circumference.
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A “circumscribed polygon” is not guaranteed to be an upper bound in every sense.
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<br><br>
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As the number of sides increases, the polygon’s vertices flatten toward 180°.
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<br>
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The polygon begins to enclose area in a fundamentally different way than the circle.
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<br>
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The usual assumption — that a tangent circumscribed polygon must have a larger perimeter — is not guaranteed.
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<br><br>
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A circle has constant curvature. A straight line has zero curvature. Treating the two as interchangeable is a category error disguised as approximation.
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<br><br>
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Imagine two rigid plates with a straight lid wedged between them.
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<br>
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If we bend the lid into a curve, it can slip lower between the plates because the distance between its endpoints decreases — even if the curved lid is longer.
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<br>
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The polygon still encloses more area, but the arc sits lower.
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<br><br>
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These are the overlooked geometric facts.</p>
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This is the overlooked geometric fact.</p>
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<br><br>
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<p>Why this matters
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<br>
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<br>
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<br><br>
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The classical argument assumes:</p>
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<ulstyle="margin:6px">
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<li>the inscribed polygon is always an under‑estimate</li>
@@ -2772,8 +2658,7 @@ <h4>Archimedes and the Illusion of Limits</h4>
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<br>
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<p>But the geometry shows:</p>
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<ulstyle="margin:6px">
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<li>a tangent polygon can have smaller perimeter than the circle, yet still enclose larger area</li>
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<li>so “circumscribed” and “upper bound” are not equivalent concepts</li>
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<li>“circumscribed” and “upper bound” are not equivalent concepts</li>
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<li>The geometric ordering required by the polygon‑approximation method is not structurally guaranteed.</li>
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