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inscribe
inscribe, v. (ɪnˈskraɪb) [ad. L. inscrībĕre to write in or upon, f. in- (in-2) + scrībĕre to write.] 1. trans. To write, mark, or delineate (words, a name, characters, etc.) in or on something; esp. so as to be conspicuous or durable, as on a monument, tablet, etc. (In quot. 1603, with upon in indir...
Oxford English Dictionary
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inscribe
inscribe/ɪnˈskraɪb; ɪn`skraɪb/ v[Tn, Tn.pr, Cn.n]~ A (on/in B)/~ B (with A) write (words, one's name, etc) on or in sth, esp as a formal or permanent record 在某物上写, 题(词语、 名字等)(尤指作正式的或永久性的记录) inscribe verses on a tombstone/inscribe a tombstonewith verses 在墓碑上题诗 inscribe one's name in a book/inscribe a...
牛津英汉双解词典
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NEAT - 1st Inscription Token on NEAR by inscribe.near | Near Social
NEAT is the first inscription token on NEAR blockchain and follows the NRC-20 token standard.
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in every triangle we can inscribe a circle I am trying to show that in every triangle we can inscribe a circle. I reduced it to following: in every triangle there must be a point in the interior, such that there are t...
**Hint:** the centre of the circle should be equidistant of the sides of the triangle. **Hint:** The bisector of an angle is the set of points equidistant of the sides of the angle. **Hint:** The distance between a point and a line is the length of the perpendicular segment from that point to that l...
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If you inscribe a square inside a circle what is the name of the shape in the circle but outside of the square When you inscribe a square in a circle there is part of the circle that is not in the square. There are $4...
Each of the four congruent shapes is called a minor segment of the circle. A segment is demarcated by a chord of a circle and the arc subtended by the chord. The chord divides the circle into the minor (smaller) and major (larger) segments, except when the chord is a diameter, in which case it divid...
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Proving that a line is an angle bisector > In a right triangle $\triangle ABC$ with $A=90^{\circ}$ we inscribe a square which one side of it is located on hypotenuse $BC$. Prove that the line which joins vertex $A$ to...
Let $O$ be the center of the inscribed square. Moreover let $D$ be the vertex of the square along $AB$ and let $E$ be the vertex of the square along $AC$. Then $DE$ is parallel to $BC$ and, by the law of sines applied to the triangles $\triangle ADO$ and $\triangle AEO$, we have that $$\frac{|DO|}{\...
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Is it possible to inscribe a regular tetrahedron in every convex body? Is it possible to inscribe at least one regular tetrahedron in every convex body?
The first theorem of Section 4 of this paper, mentioned by J.M. in the comments, gives an affirmative answer, citing V.V Makeev, _Inscribed simplices of a convex body_ (in Russian), Ukr. Geom. Sb. 35 (1992), 47-49 = J. Math. Sci. 72 (1994) (4), 3189-3190, MR 95d:52006: > **Theorem.** Let $K\subset\m...
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内切 [內切] - (math.) to inscribe, internally tangent - nèi qiē ...
1 to hit home on the evils of the day (idiom); fig. to hit a current political target. 2 to hit the nub of the matter.
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Inscribe A Poem On The Wall Of Nanzhuang, Capital
★ Inscribe a Poem on the Wall of Nanzhuang, Capital ☆ Poem by Cui Hu [Tang, China]☆ Translation & Dubbing by Luo Zhihai [China]☆ Music of Machengdiao in Shaoyang, Hunan, China
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Equilateral triangle inscribed in a ellipse **"Given any point on a ellipse, is it always possible to inscribe an equilateral triangle, with a vertex coincident with that point, in the ellipse?"** I thought I could u...
I have an idea, but a rigorous proof based on this idea may be a bit tough. Nevertheless, perhaps you find it useful. For each $r>0$, consider the circle centered on the given point $P$ of the ellipse with radius $r$. If the circle intersects the ellipse two points, they will be at a distance $d(r)$...
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What is the probability that a random point would lie in this inscribed circle? Take the unit square and inscribe a circle $Q$. Then take one of the corners (say the upper-left one) and inscribe a circle $Q'$ in that ...
Adding to Matti's comment. Starting out from the same symmetric top quarter, you know that we have a right triangle, with the corner angles being at $45^\circ$. Now go to the center of the smaller inscribed half-circle and draw a line from the center to the intersection with the square. You can now ...
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Icosahedron and inscribed cube We can inscribe a cube in dodecahedron (see this), where $12$ faces of dodecahedron give the $12$ edges of the cube. !enter image description here Can we inscribe cube in icosahedron?
Dodecahedron and icosahedron are dual to one another. So there would be a way where each vertex of the icosahedron corresponds to an edge of the cube. So you'd have corners of the cube in the centers of _some_ of the faces. !Cube in Icosahedron !Cube in Dodecahedron in Icosahedron Pictures created u...
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Maximizing the area of rectangle inscribed in triangle I'd like to ask if someone could help me out with this problem. Let's have a triangle with coordinates $[0,0],[4,0],[1,3]$. Inscribe a rectangle into this triang...
I assume that you want that the corners of the rectangle be along the sides of the triangles. The equations of the lines which define the triangle are y = 3 x and y = 4 - x. Now, let use define four points (x1,0), (x1,y1), (x2,0) and (x2,y2) which will define the rectangle. Since it is a rectangle, ...
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Inscribing a sphere in a parallelepiped I have a parallelepiped with sides, $a$, $b$, and $c$. $\gamma$ is the angle between $a$ and $b$, $\beta$ is the angle between $a$ and $c$, and $\alpha$ is the angle between $b$...
All faces must be equal rhombi, for a sphere to be inscribed. Just think of the three plane sections of the parallelepiped passing through its center and parallel to a couple of opposite faces. The relation between radius and edge is the same as in the case of a rhombus.
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Show that two segments are perpendicular Let $ABC$ be a triangle such that $|AC|=|BC|.$ Let $M$ be the midpoint of $AB$ and let $D$ be the midpoint of $MC$. Let $S$ be the the point obtained from projecting $M$ orthog...
It's easy to see that angle $SMB = SDC$, and $SM/MA = SD/DM$, that is $SM/MB = SD/DC$. Hence triangles $SMB, SDC$ are similar. Hence angles $BSM, CSD$ are equal. Hence angle $CSB$ equals angle $DSM$, which is a right angle by construction.
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