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- ...ath>[ABC] = [CBH] + [ACH] </math>, so <math>AB^2 = CB^2 + AC^2 </math>. {{Halmos}} <math>AC^2 = AY \cdot AX = (AB-BC)(AB+BC) = AB^2 - BC^2 </math>. {{Halmos}}5 KB (885 words) - 22:03, 5 October 2024
- ...X&oi=print&ct=result&cd=1&cad=bottom-3results Naive Set Theory] by Paul R. Halmos.11 KB (2,019 words) - 16:20, 7 July 2024
- ...\alpha </math> is always half the measure of <math> \angle BAD </math>. {{Halmos}}3 KB (453 words) - 09:53, 24 June 2007
- ...ath>CAB</math> and <math>ACD</math> are congruent, <math>AB || CD.</math>{{Halmos}} ...</math> passes through the center of homothety, i.e. <math>P_1 </math>. {{halmos}}5 KB (833 words) - 14:23, 6 July 2024
- ...be divisible by 2. Hence <math>p </math> is divisible by an odd prime. {{Halmos}}6 KB (1,007 words) - 08:10, 29 August 2011
- or <math>f(x) = f(z) </math>, a contradiction. {{Halmos}} ...n 2, a contradiction. Thus <math>f </math> is not strictly increasing. {{Halmos}}7 KB (1,179 words) - 19:59, 28 March 2010
- ...{2}x = \frac{\pi}{2}-\gamma = \angle H_aAC </math>. The lemma follows. {{Halmos}}7 KB (982 words) - 10:45, 10 June 2007
- ...y \ge 6 </math>, so we must have <math> \displaystyle f(1) = 1 </math>. {{Halmos}} ...^2 </math>, so we know that <math> \displaystyle f(n) \neq n^2 </math>. {{Halmos}}9 KB (1,309 words) - 14:42, 13 June 2007
- ...(3,2,2,2) </math>, a contradiction, since we assumed <math>p>3 </math>. {{Halmos}} ...<math> \displaystyle p+q \ge 9/2 </math>, so <math> p+q \ge 5 </math>. {{Halmos}}3 KB (542 words) - 16:09, 19 December 2018
- ...all such subcolumns constitute a good path which lies below all others. {{Halmos}}4 KB (631 words) - 13:14, 17 June 2007
- a contradiction. {{Halmos}}2 KB (271 words) - 08:13, 29 August 2011
- ...en that the problem statement holds for nonnegative reals, we are done. {{halmos}}5 KB (733 words) - 16:57, 28 July 2007
- ...isplaystyle {} L_{2^{k-1}+1}, \dots, L_{2^k} </math> are all turned on. {{Halmos}} ...' state must be periodic and there will always be at least one lamp on. {{Halmos}}4 KB (621 words) - 18:44, 28 July 2007
- ...efore <math> \displaystyle P, Q, B, C</math> are concyclic, as desired. {{Halmos}}3 KB (409 words) - 13:45, 6 August 2007
- Hence <math> \displaystyle y </math> is rational. {{Halmos}}2 KB (365 words) - 14:03, 6 August 2007
- ...know <math>M_c \subseteq M</math>, it follows that <math>M= M_c</math>. {{halmos}} ...at <math>E</math> is an admissable set, so as before, <math>E=M</math>. {{halmos}}9 KB (1,669 words) - 18:02, 1 August 2018
- ...ath>[ABC] = [CBH] + [ACH] </math>, so <math>AB^2 = CB^2 + AC^2 </math>. {{Halmos}} <math>AC^2 = AY \cdot AX = (AB-BC)(AB+BC) = AB^2 - BC^2 </math>. {{Halmos}}6 KB (890 words) - 10:12, 23 June 2019
- ...ath>[ABC] = [CBH] + [ACH] </math>, so <math>AB^2 = CB^2 + AC^2 </math>. {{Halmos}} <math>AC^2 = AY \cdot AX = (AB-BC)(AB+BC) = AB^2 - BC^2 </math>. {{Halmos}}15 KB (2,425 words) - 08:25, 13 February 2020
- ...ne <math>CF </math> concurs with <math>AD </math> and <math>BE </math>. {{Halmos}} ...ent almost identical to that used for the first form of Ceva's theorem. {{Halmos}}5 KB (934 words) - 12:44, 6 December 2024