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The Encyclopedia of Geometry (0289)
Problem Let $O$ be a point inside a circle, and let $A, \ B$, and $C$ be points on the circumference. If $$OA=OB=OC,$$ then $O$ is the center of the circle. $$ $$ $$ $$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ Solution…
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Formula 003 (6)
[Proof] 6. $$BG=\frac{\sqrt{5}-1}{4} a.$$ Since $$\frac{BG}{BC}=\sin 18°,$$ we obtain $$BG=BC×\sin 18°=\frac{\sqrt{5}-1}{4} a.$$ 7. $$\frac{(1+\sqrt{5}) \sqrt{2+\sqrt{0.8}}}{4}=\sqrt{1+\sqrt{0.8}}.$$ We compute: $$\frac{(1+\sqrt{5})\sqrt{\,2+\sqrt{0.8}\,}}{4}=\frac{\sqrt{\,6+2\sqrt{5}\,}\,\sqrt{\,2+\sqrt{0.8}\,}}{4} = \frac{\sqrt{\,16+4\sqrt{5}+6\sqrt{0.8}\,}}{4}.$$ Since $$4 \sqrt{5}=4 \sqrt{6.25×0.8}=10 \sqrt{0.8},$$ we have $$\frac{\sqrt{16+10 \sqrt{0.8}+6 \sqrt{0.8}}}{4}=\frac{4 \sqrt{1+\sqrt{0.8}}}{4}=\sqrt{1+\sqrt{0.8}}.$$ Thus the identity holds: $$\frac{(1+\sqrt{5}) \sqrt{2+\sqrt{0.8}}}{4}=\sqrt{1+\sqrt{0.8}}.$$ [note] This is a purely algebraic identity. It does not rely on any geometric properties of the…
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Formula 003 (5)
[Proof] 5. $$CG=\frac{1+\sqrt{5}}{2 \sqrt{2+\sqrt{0.8}}} a.$$ Since $$\frac{CG}{BC}=\cos 18°,$$ we obtain $$CG=BC×\cos 18°=\frac{\sqrt{10+2 \sqrt{5}}}{4} a.$$ Thus $$CG=\frac{10+2\sqrt{5}}{4\sqrt{10+2\sqrt{5}}}\,a =\frac{10+2\sqrt{5}}{4\sqrt{5}\,\sqrt{2+0.4\sqrt{5}}}\,a =\frac{10+10\sqrt{5}}{20\sqrt{2+\sqrt{0.8}}}\,a =\frac{1+\sqrt{5}}{2\sqrt{2+\sqrt{0.8}}}\,a.$$ [Note: Computation of $\cos 18°$] The value of $\cos 18°$ is calculated as follows: From the calculation in $(1)$, $$\sin 18°=\frac{\sqrt{5}-1}{4}.$$ Using $$\cos^2 θ+\sin^2 θ=1,$$ we have $$\cos^2 18°=1-\left( \frac{\sqrt{5}-1}{4} \right)^2=1-\frac{3-\sqrt{5}}{8}=\frac{5+\sqrt{5}}{8}.$$ Therefore $$\cos 18°=\frac{\sqrt{5+\sqrt{5}}}{\sqrt{8}}=\frac{\sqrt{5+\sqrt{5}}}{2 \sqrt{2}}=\frac{\sqrt{10+2 \sqrt{5}}}{4}.$$…
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Formula 003 (2)
[Proof] 2. $$DO=\frac{\sqrt{2+\sqrt{0.8}}}{2} a.$$ Since $$\frac{HD}{DO}=\sin 36°=\cos 54°,$$ we obtain $$DO=\frac{a}{2}\times\frac{1}{\cos 54^\circ} =\frac{a}{2}\times \frac{4}{\sqrt{10-2\sqrt{5}}} =\frac{2}{\sqrt{10-2\sqrt{5}}}\,a.$$ Rationalizing the denominator, $$\frac{2}{\sqrt{10-2 \sqrt{5}}}=\frac{2 \sqrt{10-2 \sqrt{5}}}{10-2 \sqrt{5}}=\frac{\sqrt{10-2 \sqrt{5}}}{5-\sqrt{5}}=\frac{(5+\sqrt{5})\sqrt{10-2 \sqrt{5}}}{20}.$$ Thus $$DO=\frac{\sqrt{(5+\sqrt{5})^{2}(10-2\sqrt{5})}}{20}\,a =\frac{\sqrt{200+40\sqrt{5}}}{20}\,a =\frac{10\sqrt{\,2+0.4\sqrt{5}\,}}{20}\,a =\frac{\sqrt{\,2+\sqrt{0.8}\,}}{2}\,a.$$ [Note: Computation of $\cos 54°$] From $(1)$, we already know $$\cos 36°=\frac{1+\sqrt{5}}{4}.$$ Using $\cos^2 θ+\sin^2 θ=1$, we compute $$\sin^2 36°=1-\left( \frac{1+\sqrt{5}}{4} \right)^2=1-\frac{3+\sqrt{5}}{8}=\frac{5-\sqrt{5}}{8}.$$ Hence $$\sin 36^\circ…
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Formula 003
[Formula] Let $a$ be the side length $AB$ of the regular pentagon shown above. Then the following relations hold: $$BE=\frac{1+\sqrt{5}}{2} a,$$ $$DO=\frac{2+\sqrt{0.8}}{2} a,$$ $$HO=\frac{(2+\sqrt{0.8})\sqrt{2+\sqrt{0.8}}}{8 \sqrt{0.8}} a,$$ $$AF=\frac{1}{\sqrt{2+\sqrt{0.8}}} a,$$ $$CG=\frac{1+\sqrt{5}}{2 \sqrt{2+\sqrt{0.8}}} a,$$ $$BG=\frac{\sqrt{5}-1}{4} a,$$ $$\frac{(1+\sqrt{5}) \sqrt{2+\sqrt{0.8}}}{4}=\sqrt{1+\sqrt{0.8}}.$$ $$ $$ $$ $$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$…
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The Encyclopedia of Geometry (0288)
Problem Let $AB$ be a chord of circle $O$, and let $C$ be the midpoint of arc $\hat {AB}$. Then $$∠CAB=∠CBA.$$ Moreover, $OC$ is perpendicular to $AB$ and bisects it. $$ $$ $$ $$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ Solution…
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Eiryu-ji Temple (1864), Shiokawa Town, Suzaka City, Nagano Prefecture (01)
Eiryu-ji Temple is said to be located in Shiokawa-machi, Suzaka City, Nagano Prefecture, but its exact location is unknown. Problem As shown in the figure, consider a right‑angled triangle $ABC$ with $BC=a$ (constant) and $AC=x$ (variable). Draw an arc of radius $x$ centered at vertex $A$. Let $y$ be the side length of the square…
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Formula 002
[Formula] Let $a$ denote the length of side $AB$ in the square shown above. Then $$AC=\sqrt{2} a.$$ $$ $$ $$ $$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ [Proof] $$AC=\sqrt{AB^2+BC^2}=\sqrt{a^2+a^2}=\sqrt{2a^2}=\sqrt{2}a.$$ Alternatively, using the ratio of the side and…
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The Encyclopedia of Geometry (0287)
Problem The perpendicular from the center $O$ of the circle to the chord $AB$ bisects both the major and minor arcs determined by the chord. $$ $$ $$ $$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ Solution Let $H$ and $K$ be the…
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The Encyclopedia of Geometry (0286)
Problem A straight line drawn from the center of a circle to the midpoint of a chord is perpendicular to that chord. $$ $$ $$ $$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ Solution Let $O$ be the center of the circle,…