# Search results

• pair M=midpoint(A--Ep); pair C=intersectionpoints(Circle(M,2.5),Circle(A,3))[1];
2 KB (296 words) - 19:15, 25 December 2020
• ..._{}[/itex] is not divisible by the square of any prime number. Find $m+n+d.$ ...], so $\angle OEB = \angle OED = 60^{\circ}$. Thus $\angle BEC = 60^{\circ}$, and by the [[Law of Cosines]],
3 KB (443 words) - 18:31, 4 July 2013
• ...th>\angle A = 37^\circ[/itex], $\angle D = 53^\circ$, and $M$ and $N$ be the [[midpoint]]s of $\overline{BC}</math pair[] M = intersectionpoints(N--E,B--C); 8 KB (1,206 words) - 15:43, 17 June 2021 • ...is there at least 1 positive integer [itex]n$ such that $mn \le m + n$? ...er of $\triangle AED$ is twice the perimeter of $\triangle BEC$. Find $AB$.
11 KB (1,733 words) - 10:04, 12 October 2021
• ...er of $\triangle AED$ is twice the perimeter of $\triangle BEC$. Find $AB$. ...er of $\triangle AED$ is twice the perimeter of $\triangle BEC$.
2 KB (317 words) - 14:24, 13 March 2021
• ...nd $m$ minutes, with $0 < m < 60$, what is $h + m$? ...llowing is equal to $12^{mn}$ for every pair of integers $(m,n)$?
13 KB (2,105 words) - 12:13, 12 August 2020
• ...llowing is equal to $12^{mn}$ for every pair of integers $(m,n)$? \mathrm{(B)}\ P^nQ^m
14 KB (2,130 words) - 10:32, 7 November 2021
• ...e from point $E$ of $\triangle AEB$, $\triangle BEC$, $\triangle CED$, $\triangle DEA$. ...EWX+m\angle EYX+m\angle EYZ+m\angle EWZ=[/itex]$360^\circ-m\angle CED-m\angle AEB=180^\circ$.
2 KB (439 words) - 06:54, 19 July 2016
• ...rtitions the trapezoid into rectangle $ADEB$ and triangle $BEC$. We will use the triangle to solve for $x \times y$ using
5 KB (746 words) - 23:56, 31 May 2021
• $\angle ABE$ is external to $\triangle BEC$ at $\angle B$. Therefore it is equal to the sum: $\a The two angles sum to [itex]102^\circ$, thus $m\angle ECB < 51^\circ$
721 bytes (117 words) - 18:46, 17 July 2011
• ...ath> and $n$ are relatively prime positive integers. Find $m+n$. ...and $\omega$ are swapped. Thus points $F$ and $M$ map to each other, and are isogonal. It follows that $AF$
10 KB (1,676 words) - 21:39, 4 August 2021
• ...ngles $\triangle ABD$ and $\triangle BCE$. Let $M$ be the midpoint of $\overline{AE}$, and $N$ be ...0, 0), B = (16, 0), C = (20, 0), D = (8, 8*sqrt(3)), EE = (18, 2*sqrt(3)), M = (9, sqrt(3)), NN = (14, 4*sqrt(3));
4 KB (686 words) - 22:51, 31 August 2021
• <cmath>\frac{1}{2}(AE)(AD)(\sin(m\angle EAD)).</cmath> But, note that $\sin(m\angle EAD)=\sin(m\angle CAD)=\frac{O}{H}=\frac{3}{5}$. Thus, we see that
4 KB (630 words) - 16:50, 29 October 2021
• Drawing it out, we see $\angle BDC$ and $\angle BEC$ are right angles, as they are inscribed in a semicircle. Using the f <cmath>\angle BEC+\angle EBC+\angle ECB=180^{\circ}\implies90^{\circ}+70^{\circ}+\angle ECB=1
4 KB (641 words) - 01:57, 20 September 2021
• ...math>EP=EQ=2 \sqrt{5}[/itex]. Now by the Law of Cosines in $\triangle BEC$, we have $\cos{\left(\angle BEC\right)}=\frac{EB^2+EC^2-BC^2}{2 \cdot EB \cdot EC}=\frac{5}{\sqrt{30}}</mat 5 KB (871 words) - 00:09, 7 November 2021 • pair A,B,C,DD,EE,FF, M; B = (0,0); C = (3,0); M = (1.45,0); 24 KB (3,745 words) - 18:04, 6 October 2021 • We could use the famous m-n rule in trigonometry in [itex]\triangle ABC$ with Point $E</ma <cmath>[AEB] + [BEC] = \frac{1}{2}(x)(5)\sin(\angle DEC) + \frac{1}{2}(x)(15)\sin(180-\angle DE 11 KB (1,658 words) - 21:53, 6 November 2021 • ...ath> and [itex]n$ are relatively prime positive integers. Find $m+n.$ ...ta.[/itex] It follows that $\angle CED=\theta$ and $\angle BEC=\angle DEA=180^\circ-\theta.$
12 KB (1,929 words) - 07:01, 4 October 2021
• ...> where $m$ and $n$ are relatively prime, find $m+n.$ .../math> We claim that $EF$ is the angle bisector of $\angle BEC$.
7 KB (744 words) - 20:18, 11 July 2021