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Difference between revisions of "1993 AHSME Problems"

(→‎Problem 1)
(→‎Problem 2)
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== Problem 2 ==
 
== Problem 2 ==
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In <math>\triangle ABC</math>, <math>\angle A=55\deg</math>, <math>\angle C=75\deg</math>, <math>D</math> is on side <math>\overbar{AB}</math> and <math>E</math> is on side <math>\overbar{BC}</math> If <math>DB=BE</math>, then <math>\angle BED=</math>
 +
 +
<math>\text{(A)}\ 50\deg \\
 +
\text{(B)}\ 55\deg \\
 +
\text{(C)}\ 60\deg \\
 +
\text{(D)}\ 65\deg \\
 +
\text{(E)}\ 70\deg </math>
  
  

Revision as of 21:58, 28 February 2011

Problem 1

For integers $a, b$ and $c$, define $\boxed{a,b,c}$ to mean $a^b-b^c+c^a$. Then $\boxed{1,-1,2}$ equals

$\text{(A)} \ -4 \qquad \text{(B)} \ -2 \qquad \text{(C)} \ 0 \qquad \text{(D)} \ 2 \qquad \text{(E)} \ 4$

Solution

Problem 2

In $\triangle ABC$, $\angle A=55\deg$, $\angle C=75\deg$, $D$ is on side $\overbar{AB}$ (Error compiling LaTeX. Unknown error_msg) and $E$ is on side $\overbar{BC}$ (Error compiling LaTeX. Unknown error_msg) If $DB=BE$, then $\angle BED=$

$\text{(A)}\ 50\deg \\ \text{(B)}\ 55\deg \\ \text{(C)}\ 60\deg \\ \text{(D)}\ 65\deg \\ \text{(E)}\ 70\deg$


Solution

Problem 3

Solution

Problem 4

Solution

Problem 5

Solution

Problem 6

Solution

Problem 7

Solution

Problem 8

Solution

Problem 9

Solution

Problem 10

Solution

Problem 11

Solution

Problem 12

Solution

Problem 13

Solution

Problem 14

Solution

Problem 15

Solution

Problem 16

Solution

Problem 17

Solution

Problem 18

Solution

Problem 19

Solution

Problem 20

Solution

Problem 21

Solution

Problem 22

Solution


Problem 23

Solution

Problem 24

Solution

Problem 25

Solution

Problem 26

Solution

Problem 27

Solution

Problem 28

Solution

Problem 29

Solution

Problem 30

Solution

See also

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