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2006 AMC 12B Problems
\text {(A) } 4 \qquad \text {(B) } 12 \qquad \text {(C) } 16 \qquad \text {(D) } 24 \qquad \text {(E) } 48 Joe and JoAnn each bought 12 ounces of coffee in a 16-ounce cup. Joe drank 2 ounces of his coffee and t

13 KB (2,058 words) - 12:36, 4 July 2023
2004 AMC 12A Problems
On the AMC 12, each correct answer is worth <math>6</math> points, each incorrect answer == Problem 12 ==

13 KB (1,953 words) - 00:31, 26 January 2023
2003 AMC 12A Problems
<math> \mathrm{(A) \ } 4.5\qquad \mathrm{(B) \ } 9\qquad \mathrm{(C) \ } 12\qquad \mathrm{(D) \ } 18\qquad \mathrm{(E) \ } 24 </math> <math> \mathrm{(A) \ } 10\qquad \mathrm{(B) \ } 11\qquad \mathrm{(C) \ } 12\qquad \mathrm{(D) \ } 13\qquad \mathrm{(E) \ } 14 </math>

13 KB (1,955 words) - 21:06, 19 August 2023
2002 AMC 12A Problems
...} 4\qquad \mathrm{(B) \ } 6\qquad \mathrm{(C) \ } 9\qquad \mathrm{(D) \ } 12\qquad \mathrm{(E) \ }</math> infinitely many ...files onto disks, each with 1.44 MB space. 3 of the files take up 0.8 MB, 12 of the files take up 0.7 MB, and the rest take up 0.4 MB. It is not possibl

12 KB (1,792 words) - 13:06, 19 February 2020
2002 AMC 12B Problems
\qquad\mathrm{(E)}\ 12</math> Let <math>n</math> be a positive integer such that <math>\frac 12 + \frac 13 + \frac 17 + \frac 1n</math> is an integer. Which of the followi

10 KB (1,547 words) - 04:20, 9 October 2022
2006 AMC 12B Problems/Problem 13
<math> \overline{AB}</math> and <math> \overline{AD}</math> are equal in length. To find the area of the smaller rhombus, we on ...icely into 2/3 of equilateral triangle ABD. Thus the area of BFDE is (2/3)*12=8.

3 KB (447 words) - 03:49, 16 January 2021
2006 AMC 12B Problems/Problem 15
...are on the circle centered at <math> P</math>, such that <math> \overline{AD}</math> and <math> \overline{BC}</math> are common external tangents to the O[2] = (12,0);

3 KB (458 words) - 16:40, 6 October 2019
Power of a Point Theorem/Introductory Problem 4
...at <math>E</math> . Given that <math> BE = 16, DE = 4, </math> and <math> AD = 5 </math>, find <math> CE </math>. pair C = (0,-12);

1 KB (177 words) - 02:14, 26 November 2020
2006 AMC 10A Problems
draw((6,4)--(6,0)--(12,0)--(12,-4)); == Problem 12 ==

13 KB (2,028 words) - 16:32, 22 March 2022
2006 AMC 10A Problems/Problem 17
...t <math>\overline{HE}</math>. In addition, <math>AH=AC=2</math>, and <math>AD=3</math>. What is the area of quadrilateral <math>WXYZ</math> shown in the ...math>[WXYZ] = \left(\frac{1}{\sqrt{2}}\right)^2 =\boxed{\textbf{(A) }\frac 12}</math>.

6 KB (1,066 words) - 00:21, 2 February 2023
2005 AIME I Problems
In quadrilateral <math> ABCD, BC=8, CD=12, AD=10, </math> and <math> m\angle A= m\angle B = 60^\circ. </math> Given that ...f 70. The coordinates of <math> B </math> and <math> C </math> are <math> (12,19) </math> and <math> (23,20), </math> respectively, and the coordinates o

6 KB (983 words) - 05:06, 20 February 2019
2005 AIME I Problems/Problem 7
In [[quadrilateral]] <math> ABCD,\ BC=8,\ CD=12,\ AD=10, </math> and <math> m\angle A= m\angle B = 60^\circ. </math> Given that label("$12$",(10.935,7.794),S);

4 KB (567 words) - 20:20, 3 March 2020
2005 AIME I Problems/Problem 11
...the perpendicular is <math>\frac{r}{\sqrt{2}}</math>. Note also that <math>AD</math> is equal to the length of that perpendicular plus the radius to the ...ath> in half, and so by symmetry the distance from <math>O</math> to <math>AD</math> is <math>\frac{a}{2}</math> and so the distance from <math>O</math>

4 KB (707 words) - 11:11, 16 September 2021
2005 AIME II Problems/Problem 14
...H = \frac{56}{5}</math> and <math>HC = \frac{42}{5} \Rightarrow HD = \frac{12}{5}</math>. .../math> we have that <math>AD^2 = \left(\frac{56}{5}\right)^2 + \left(\frac{12}{5}\right)^2 = \frac{656}{5}</math>. By Law of Sines on <math>\triangle ADC

14 KB (2,340 words) - 16:38, 21 August 2024
2004 AIME II Problems/Problem 13
...th> be a [[convex]] [[pentagon]] with <math> AB \parallel CE, BC \parallel AD, AC \parallel DE, \angle ABC=120^\circ, AB=3, BC=5, </math> and <math>DE = ...ne{CE}</math> be <math>F</math>. Since <math>AB \parallel CE, BC \parallel AD, </math> it follows that <math>ABCF</math> is a [[parallelogram]], and so <

3 KB (486 words) - 22:15, 7 April 2023
2004 AIME II Problems
...er <math> B </math> is matched with point <math> B' </math> on edge <math> AD. </math> The crease is <math> EF, </math> where <math> E </math> is on <mat pair point=( 12.5, 35/3 );

9 KB (1,410 words) - 05:05, 20 February 2019
1983 AIME Problems
...<math>\log_xw=24</math>, <math>\log_y w = 40</math> and <math>\log_{xyz}w=12</math>. Find <math>\log_zw</math>. == Problem 12 ==

7 KB (1,104 words) - 03:13, 27 May 2024
1989 AIME Problems
4x_1 + 9x_2 + 16x_3 + 25x_4 + 36x_5 + 49x_6 + 64x_7 &= 12, \\ == Problem 12 ==

7 KB (1,045 words) - 20:47, 14 December 2023
1991 AIME Problems
...ne {P_kQ_k}</math>. Repeat this construction on the sides <math>\overline {AD}</math> and <math>\overline {CD}</math>, and then draw the diagonal <math>\ for (int i=0; i<12; ++i)

7 KB (1,106 words) - 22:05, 7 June 2021
1992 AIME Problems
...<math>AB^{}_{}</math> is drawn tangent to <math>BC^{}_{}</math> and <math>AD^{}_{}</math>. Given that <math>AP^{}_{}=\frac mn</math>, where <math>m^{}_{ == Problem 12 ==

8 KB (1,117 words) - 05:32, 11 November 2023

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