It is on the circle PK, where PK is the diameter.
Since angle PAK=90, angle PAK is inscribed in the semicircular arc PK"
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1 KB (207 words) - 10:14, 20 August 2025
==2025 II Problem 10 1==
[[File:Tur 2 2025 10 1.png|300px|right]]
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122 KB (21,381 words) - 13:47, 3 August 2025
...is congruent to angle <math>CKP</math>. Segments <math>MK</math> and <math>PK</math> are congruent as stated in the problem. Segments <math>AK</math> and
Also, it is given that <math>AK=CK</math> and <math>PK=MK</math>.
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9 KB (1,367 words) - 20:11, 12 September 2025
...</math> ordered from left to right. Set the values <math>BP=x</math>,<math>PK=x</math>,<math>BQ=4y</math> and <math>QK=y</math>. Setting both sides segme
...-\frac 9 2 : \frac 9 2 - \frac {18} 5 : \frac {18} 5= \frac 3 2 : \frac 9 {10} : \frac {18} 5 = 5:3:12.</math> Hence <math>r,s,t=5,3,12</math> and <math>
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9 KB (1,396 words) - 16:39, 28 September 2025
...ath>\tfrac 35</math>, so <math>RS=\tfrac 35s</math> and <math>RQ=\tfrac{1}{10}s</math>.
...>\tfrac 35 s^2</math>, the area of <math>CGPF</math> is <math>s^2(\tfrac 3{10}-\tfrac 1{400}\sin 2\theta)</math>.
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9 KB (1,404 words) - 20:07, 13 October 2023
C=(10,0);
B=(10,10);
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5 KB (812 words) - 20:43, 19 September 2024
pair A=(7,6), B=(0,0), C=(10,0), P=(4,0), Q=(6,8), R;
draw((0,0)--(10,0)--(7,6)--(0,0),blue+0.75);
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5 KB (824 words) - 07:04, 13 October 2025
...}{13}\equiv12!\cdot14\cdot15\cdot16\cdot...\cdot23\equiv(-1)(1)(2)(3)(...)(10)\equiv\boxed{7}\mod13</cmath>
...Let <math>{p} = 101</math>. The sum <cmath>\sum_{k=1}^{10} \frac{1}{\binom pk}</cmath> can be written in the form <math>\frac{a}{p!}</math>, where <math>
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4 KB (688 words) - 11:17, 8 August 2025
a = pk
With 0 < m, n, t, j < 10
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817 bytes (173 words) - 21:34, 11 April 2019
