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- <math> \textrm{(A) } 6 \qquad \textrm{(B) } 4\sqrt {3} \qquad \textrm{(C) } 8 \qquad \textrm{(D) } 9 \qquad \textrm{(E) } ...e the result. To find the value of the base, use the line we just drew and connect it to point <math>E</math> at a right angle along line <math> \overline{DB}3 KB (447 words) - 02:49, 16 January 2021
- ...closing these <math>4</math> circles, notice that if you connect the <math>4</math> centers as a square, the diameter of the large circle will be the di2 KB (364 words) - 03:54, 16 January 2023
- ...osing which ant moves to <math>A</math>. Hence, there are <math>2 \times 2=4</math> ways the ants can move to different points. ...ath> can actually move to four different points, there is a total of <math>4 \times 20=80</math> ways the ants can move to different points.10 KB (1,840 words) - 14:01, 4 July 2024
- ..., D=(0,0), E=(2.5-0.5*sqrt(7),9), F=(6.5-0.5*sqrt(7),9), G=(4.5,9), O=(4.5,4.5); draw(A--B--C--D--A);draw(E--O--F);draw(G--O); dot(A^^B^^C^^D^^E^^F^^G^^ ...90); draw(A--B--C--D--A);draw(E--O--F);draw(G--O--J);draw(F--G,linetype("4 4")); dot(A^^B^^C^^D^^E^^F^^G^^J^^O); label("
",A,(-1,1));label(" ",B13 KB (2,080 words) - 12:14, 23 July 2024 - ...cent ordered pairs once. For example, represent (4,7),(7,3),(3,5) as <math>4,7,3,5 .</math> Label the vertices of a regular <math>n</math> -gon <math>1, { }_{2} X_{4},{ }_{4} X_{6}, \ldots,{ }_{n-2} X_{n},(n, n+1)(n+1,1)(1, n+2)(n+2,2)9 KB (1,659 words) - 17:35, 20 June 2024
- ...choose3} \cdot 42}{{45\choose4}} = \frac{10 \cdot 9 \cdot 8 \cdot 42 \cdot 4!}{45 \cdot 44 \cdot 43 \cdot 42 \cdot 3!} = \frac{16}{473}</math>. The solu ...ly the probability that the first three segments picked form a triangle by 4. We can pick any segment for the first choice, then only segments that shar3 KB (524 words) - 16:25, 17 July 2023
- ...three circles such that they form an equilateral triangle with side length 4. Obviously, the height this triangle is <math>2\sqrt{3}</math>, and the sho2 KB (287 words) - 19:21, 16 September 2024
- ...osines]] in triangle <math>MCB</math>, we get <math>MB^2 = 4\sin^2 7 + 1 - 4\sin 7(\cos 83) = 1</math>, since <math>\cos 83 = \sin 7</math>. So triangle === Solution 4 ===7 KB (1,082 words) - 01:08, 30 September 2024
- ...tersection between <math>\Delta ABC</math> and <math>\Delta A'B'C'</math>. Connect each of these points to <math>G</math>. D=(7.6667,-4);5 KB (787 words) - 16:38, 30 July 2022
- <asy>import three; currentprojection = orthographic(-1.2,-0.2,0.4); import three; currentprojection = orthographic(5,-6,4);8 KB (1,184 words) - 15:06, 13 October 2024
- ...\pi}{2}</math>. The area of the whole triangle is <math>\frac{2^2\sqrt{3}}{4}=\sqrt{3}</math>, so the area of the gap is <math>\sqrt{3}-\frac{\pi}{2}</m * [[University of South Carolina High School Math Contest/1993 Exam/Problem 4|Next Problem]]1 KB (173 words) - 16:09, 4 October 2016
- draw((0,0)--(4,0)); dot((4,0));4 KB (596 words) - 16:09, 9 May 2024
- <math>\mathrm{(A)}\, 4</math> ===Problem 4===30 KB (4,794 words) - 22:00, 8 May 2024
- pair A=(2,4), B=(1,1), C=(6,1); ...ral triangles out of the three sides from <math>\triangle ABC</math>, then connect each new vertex to each opposite vertex, as these three lines will concur a4 KB (769 words) - 15:07, 29 December 2019
- ...frac{1 + \frac{15}{17}}{1 - \frac{15}{17}}}</math>, and <math>x = \frac{r}{4}</math>. Therefore, <math>x + y + 2r = \frac{r}{4} + \frac{3r}{5} + 2r = \frac{57r}{20} = 34 \Longrightarrow r = \frac{680}{511 KB (1,853 words) - 19:10, 21 July 2024
- ...math>ABCD</math> and vertex <math>E</math> has eight edges of length <math>4</math>. A plane passes through the midpoints of <math>AE</math>, <math>BC< currentprojection = perspective(2.5,-12,4);7 KB (1,034 words) - 22:30, 18 June 2024
- <div style="text-align:center;">[[Image:2007 USAMO-4.PNG|350px]]</div> Thus, our answer is <math>4(2007) - 3 = 8025</math> cells.10 KB (1,878 words) - 13:56, 30 June 2021
- ...must have that <math>h_x^2 = \frac1{16}</math> and so <math>h_x = \frac{1}4</math> and similarly <math>h_y = \frac15</math> and <math>h_z = \frac16</ma ...\right) \left(z-\frac{1}{4}\right) \left(\frac{1}{4}\right) \left(\frac{1}{4}\right)} \6 KB (994 words) - 12:40, 3 December 2024
- ...B)}\ 3 + \frac {\sqrt {69}}{3} \qquad \text {(C)}\ 3 + \frac {\sqrt {123}}{4}\qquad \text {(D)}\ \frac {52}{9}\qquad \text {(E)}\ 3 + 2\sqrt2</math> We now need the vertical height of the centers. If we connect centers, we get a rectangular [[pyramid]] with an [[equilateral triangle]]2 KB (382 words) - 20:36, 27 November 2024
- ...on <math>ABCDEFGHIJKL</math>, as shown. Each of its sides has length <math>4</math>, and each two consecutive sides form a right angle. Suppose that <ma dotfactor=4;6 KB (867 words) - 23:17, 19 May 2023