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Sasamine Tenjin Shrine (1862), Sasamine, Nagano City, Nagano Prefecture (01)
Sasamine Tenjin Shrine is said to be located in Sasamine, Nagano City, Nagano Prefecture, but its exact location is unknown. Problem As shown in the diagram, a perpendicular is drawn from the right angle of triangle $ABC$ to the hypotenuse, dividing it into two right triangles. Each of these triangles contains an inscribed circle. Given…
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Formula 004
[Formula] Let $a$ be the length of side $AB$ of the regular hexagon shown above. Let $O$ be its center. Then $$EO=a,$$ $$HO=\frac{\sqrt{3}}{2} a,$$ $$BF=\sqrt{3} a.$$ $$ $$ $$ $$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$…
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The Encyclopedia of Geometry (0290)
Problem Draw a circle with segment $AB$ as its diameter. From a point $O$ on $AB$, draw two rays making equal angles with $AB$, one above and one below. Let $X$ and $Y$ be the points where these rays meet the circumference of the circle. Then $$OX=OY.$$ $$ $$ $$ $$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $$ $$…
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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…