<p>Tight lower bounds for covering rectangles with right isosceles triangles are given. It is shown that a rectangle of size <InlineEquation ID="IEq1"> <EquationSource Format="TEX">$ a \times b $</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>a</mi> <mo>×</mo> <mi>b</mi> </mrow> </math></EquationSource> </InlineEquation> can be covered if the total area of the triangles is at least <InlineEquation ID="IEq2"> <EquationSource Format="TEX">$ \frac{1}{2}(a + b)^2 $</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mfrac> <mn>1</mn> <mn>2</mn> </mfrac> <msup> <mrow> <mo stretchy="false">(</mo> <mi>a</mi> <mo>+</mo> <mi>b</mi> <mo stretchy="false">)</mo> </mrow> <mn>2</mn> </msup> </mrow> </math></EquationSource> </InlineEquation> for <InlineEquation ID="IEq3"> <EquationSource Format="TEX">$ b \le a \le \sqrt{2}b $</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>b</mi> <mo>≤</mo> <mi>a</mi> <mo>≤</mo> <msqrt> <mn>2</mn> </msqrt> <mi>b</mi> </mrow> </math></EquationSource> </InlineEquation>, or <InlineEquation ID="IEq4"> <EquationSource Format="TEX">$ \frac{1}{4}(a + 2b)^2 $</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mfrac> <mn>1</mn> <mn>4</mn> </mfrac> <msup> <mrow> <mo stretchy="false">(</mo> <mi>a</mi> <mo>+</mo> <mn>2</mn> <mi>b</mi> <mo stretchy="false">)</mo> </mrow> <mn>2</mn> </msup> </mrow> </math></EquationSource> </InlineEquation> for <InlineEquation ID="IEq5"> <EquationSource Format="TEX">$ a > \sqrt{2}b $</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>a</mi> <mo>></mo> <msqrt> <mn>2</mn> </msqrt> <mi>b</mi> </mrow> </math></EquationSource> </InlineEquation>.</p>

Covering a rectangle with isosceles right triangles

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2025年11月11日

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Janusz Januszewski,
Łukasz Zielonka

摘要

Tight lower bounds for covering rectangles with right isosceles triangles are given. It is shown that a rectangle of size $ a \times b $ $a \times b$ can be covered if the total area of the triangles is at least $ \frac{1}{2}(a + b)^2 $ $\frac{1}{2} {(a + b)}^{2}$ for $ b \le a \le \sqrt{2}b $ $b \leq a \leq \sqrt{2} b$ , or $ \frac{1}{4}(a + 2b)^2 $ $\frac{1}{4} {(a + 2 b)}^{2}$ for $ a > \sqrt{2}b $ $a > \sqrt{2} b$ .