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Thales semicircle theorem says that an angle inscribed in a semicircle is a right angle.

Q1. Does a cone with apex on a hemisphere and encompassing the circular base have a solid angle independent of the position of the apex?

It appears it might be true, with solid angle $(2-\sqrt{2})\pi \approx 0.59 \pi$ steradians:

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  ![ConeSolidAngle][1]

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Q2. I am seeking a proof or reference for the generalization to $d$ dimensions.

I have not found a reference, which makes me wonder if the answer to Q1 might be No...

---

**Update**. Thanks to *TMA* and @WillSawin for their analyses. Following Will, let $p=(\cos \phi, 0, \sin \phi)$ be the apex on the cone on the sphere; so $\phi=90^\circ$ places $p$ at the north pole. Below is a crude depiction of the intersection of the cone with the small green $p$-centered sphere. $\phi=(90^\circ,30^\circ,5^\circ)$ left to right:
  ![ConeSolidAngle3][2]
And here are two enlarged views of the $\phi=5^\circ$ intersection:
  ![ConeSlid5degx2][3]
I have not (yet) computed the area of the (blue) spherical polygon (which is the solid angle).

Thales semicircle theorem says that an angle inscribed in a semicircle is a right angle.

Q1. Does a cone with apex on a hemisphere and encompassing the circular base have a solid angle independent of the position of the apex?

It appears it might be true, with solid angle $(2-\sqrt{2})\pi \approx 0.59 \pi$ steradians:

---

  ![ConeSolidAngle][1]

---

Q2. I am seeking a proof or reference for the generalization to $d$ dimensions.

I have not found a reference, which makes me wonder if the answer to Q1 might be No...

---

**Update**. Thanks to *TMA* and @WillSawin for their analyses. Following Will, let $p=(\cos \phi, 0, \sin \phi)$ be the apex on the cone on the sphere; so $\phi=90^\circ$ places $p$ at the north pole. Below is a crude depiction of the intersection of the cone with the small green $p$-centered sphere. $\phi=(90^\circ,30^\circ,5^\circ)$ left to right:
  ![ConeSolidAngle3][2]
And here are two enlarged views of the $\phi=5^\circ$ intersection:
  ![ConeSlid5degx2][3]
The area of the (blue) spherical polygon is the solid angle. As Will argues, its area approaches $\frac{1}{4} (4 \pi)=\pi$ as $\phi \to 0$, so the answer to ***Q1*** is *No*.

Thales semicircle theorem says that an angle inscribed in a semicircle is a right angle.

Q1. Does a cone with apex on a hemisphere and encompassing the circular base have a solid angle independent of the position of the apex?

It appears it might be true, with solid angle $(2-\sqrt{2})\pi \approx 0.59 \pi$ steradians:

---

  ![ConeSolidAngle][1]

---

Q2. I am seeking a proof or reference for the generalization to $d$ dimensions.

I have not found a reference, which makes me wonder if the answer to Q1 might be No...

---

**Update**. Thanks to *TMA* and @WillSawin for their analyses. Following Will, let $p=(\cos \phi, 0, \sin \phi)$ be the apex on the cone on the sphere; so $\phi=90^\circ$ places $p$ at the north pole. Below is a crude depiction of the intersection of the cone with the small green $p$-centered sphere. $\phi=(90^\circ,30^\circ,5^\circ)$ left to right:
  ![ConeSolidAngle3][2]
And here two enlarged views of the $\phi=5^\circ$ intersection:
  ![ConeSlid5degx2][3]

Thales semicircle theorem says that an angle inscribed in a semicircle is a right angle.

Q1. Does a cone with apex on a hemisphere and encompassing the circular base have a solid angle independent of the position of the apex?

It appears it might be true, with solid angle $(2-\sqrt{2})\pi \approx 0.59 \pi$ steradians:

---

  ![ConeSolidAngle][1]

---

Q2. I am seeking a proof or reference for the generalization to $d$ dimensions.

I have not found a reference, which makes me wonder if the answer to Q1 might be No...

---

**Update**. Thanks to *TMA* and @WillSawin for their analyses. Following Will, let $p=(\cos \phi, 0, \sin \phi)$ be the apex on the cone on the sphere; so $\phi=90^\circ$ places $p$ at the north pole. Below is a crude depiction of the intersection of the cone with the small green $p$-centered sphere. $\phi=(90^\circ,30^\circ,5^\circ)$ left to right:
  ![ConeSolidAngle3][2]
And here are two enlarged views of the $\phi=5^\circ$ intersection:
  ![ConeSlid5degx2][3]
I have not (yet) computed the area of the (blue) spherical polygon (which is the solid angle).

Thales semicircle theorem says that an angle inscribed in a semicircle is a right angle.

Q1. Does a cone with apex on a hemisphere and encompassing the circular base have a solid angle independent of the position of the apex?

It appears it might be true, with solid angle $(2-\sqrt{2})\pi \approx 0.59 \pi$ steradians:

---

  ![ConeSolidAngle][1]

---

Q2. I am seeking a proof or reference for the generalization to $d$ dimensions.

I have not found a reference, which makes me wonder if the answer to Q1 might be No...

Thales semicircle theorem says that an angle inscribed in a semicircle is a right angle.

Q1. Does a cone with apex on a hemisphere and encompassing the circular base have a solid angle independent of the position of the apex?

It appears it might be true, with solid angle $(2-\sqrt{2})\pi \approx 0.59 \pi$ steradians:

---

  ![ConeSolidAngle][1]

---

Q2. I am seeking a proof or reference for the generalization to $d$ dimensions.

I have not found a reference, which makes me wonder if the answer to Q1 might be No...

---

**Update**. Thanks to *TMA* and @WillSawin for their analyses. Following Will, let $p=(\cos \phi, 0, \sin \phi)$ be the apex on the cone on the sphere; so $\phi=90^\circ$ places $p$ at the north pole. Below is a crude depiction of the intersection of the cone with the small green $p$-centered sphere. $\phi=(90^\circ,30^\circ,5^\circ)$ left to right:
  ![ConeSolidAngle3][2]
And here two enlarged views of the $\phi=5^\circ$ intersection:
  ![ConeSlid5degx2][3]