https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Bryanc921&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T08:13:46ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=1992_AHSME_Problems/Problem_12&diff=660021992 AHSME Problems/Problem 122014-11-22T18:57:56Z<p>Bryanc921: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<br />
Let <math>y=mx+b</math> be the image when the line <math>x-3y+11=0</math> is reflected across the <math>x</math>-axis. The value of <math>m+b</math> is<br />
<br />
<math>\text{(A) -6} \quad<br />
\text{(B) } -5\quad<br />
\text{(C) } -4\quad<br />
\text{(D) } -3\quad<br />
\text{(E) } -2</math><br />
<br />
== Solution ==<br />
<math>\fbox{C}</math><br />
First we want to put this is slope-intercept form, so we get <math>y=\dfrac{1}{3}x+\dfrac{11}{3}</math>. When we reflect a line across the x-axis, the line's x-intercept stays the same, while the y-intercept and the slope are additive inverses of the original line. Since <math>m+b</math> is the sum of the slope and the y-intercept, we get <math>-\dfrac{1}{3}-\dfrac{11}{3}=\boxed{-4}</math>.<br />
<br />
== See also ==<br />
{{AHSME box|year=1992|num-b=11|num-a=13}} <br />
<br />
[[Category: Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Bryanc921https://artofproblemsolving.com/wiki/index.php?title=1991_AHSME_Problems/Problem_5&diff=659981991 AHSME Problems/Problem 52014-11-21T22:52:31Z<p>Bryanc921: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<asy><br />
draw((0,0)--(2,2)--(2,1)--(5,1)--(5,-1)--(2,-1)--(2,-2)--cycle,dot);<br />
MP("A",(0,0),W);MP("B",(2,2),N);MP("C",(2,1),S);MP("D",(5,1),NE);MP("E",(5,-1),SE);MP("F",(2,-1),NW);MP("G",(2,-2),S);<br />
MP("5",(2,1.5),E);MP("5",(2,-1.5),E);MP("20",(3.5,1),N);MP("20",(3.5,-1),S);MP("10",(5,0),E);<br />
</asy><br />
<br />
In the arrow-shaped polygon [see figure], the angles at vertices <math>A,C,D,E</math> and <math>F</math> are right angles, <math>BC=FG=5, CD=FE=20, DE=10</math>, and <math>AB=AG</math>. The area of the polygon is closest to<br />
<math>\text{(A) } 288\quad<br />
\text{(B) } 291\quad<br />
\text{(C) } 294\quad<br />
\text{(D) } 297\quad<br />
\text{(E) } 300</math><br />
<br />
== Solution ==<br />
<math>\fbox{E}</math><br />
Since they tell us that AB=AG and that angle A is a right angle, triangle ABG is a 45-45-90 triangle. Thus we can find the legs of the triangle by dividing <math>20</math> (the hypotenuse) by <math>\sqrt2</math>. Both legs have length <math>10\sqrt2</math>, so the area of the right triangle is <math>100</math>. The rectangle is simple, just <math>20\times10</math>, so the area is 200. Adding these areas, we get <math>300</math> as the total area.<br />
<br />
== See also ==<br />
{{AHSME box|year=1991|num-b=4|num-a=6}} <br />
<br />
[[Category: Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Bryanc921https://artofproblemsolving.com/wiki/index.php?title=1991_AHSME_Problems/Problem_5&diff=659971991 AHSME Problems/Problem 52014-11-21T22:51:53Z<p>Bryanc921: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<asy><br />
draw((0,0)--(2,2)--(2,1)--(5,1)--(5,-1)--(2,-1)--(2,-2)--cycle,dot);<br />
MP("A",(0,0),W);MP("B",(2,2),N);MP("C",(2,1),S);MP("D",(5,1),NE);MP("E",(5,-1),SE);MP("F",(2,-1),NW);MP("G",(2,-2),S);<br />
MP("5",(2,1.5),E);MP("5",(2,-1.5),E);MP("20",(3.5,1),N);MP("20",(3.5,-1),S);MP("10",(5,0),E);<br />
</asy><br />
<br />
In the arrow-shaped polygon [see figure], the angles at vertices <math>A,C,D,E</math> and <math>F</math> are right angles, <math>BC=FG=5, CD=FE=20, DE=10</math>, and <math>AB=AG</math>. The area of the polygon is closest to<br />
<math>\text{(A) } 288\quad<br />
\text{(B) } 291\quad<br />
\text{(C) } 294\quad<br />
\text{(D) } 297\quad<br />
\text{(E) } 300</math><br />
<br />
== Solution ==<br />
<math>\fbox{E}</math><br />
Since they tell us that AB=AG and that angle A is a right angle, triangle ABG is a 45-45-90 triangle. Thus we can find the legs of the triangle by dividing <math>20</math> by <math>\sqrt2</math>. Both legs have length <math>10\sqrt2</math>, so the area of the right triangle is <math>100</math>. The rectangle is simple, just <math>20\times10</math>, so the area is 200. Adding these areas, we get <math>300</math> as the total area.<br />
<br />
== See also ==<br />
{{AHSME box|year=1991|num-b=4|num-a=6}} <br />
<br />
[[Category: Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Bryanc921https://artofproblemsolving.com/wiki/index.php?title=1991_AHSME_Problems/Problem_4&diff=659961991 AHSME Problems/Problem 42014-11-21T22:45:28Z<p>Bryanc921: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<br />
Which of the following triangles cannot exist?<br />
<br />
(A) An acute isosceles triangle (B) An isosceles right triangle (C) An obtuse right triangle (D) A scalene right triangle (E) A scalene obtuse tr<br />
iangle<br />
<br />
== Solution ==<br />
<math>\fbox{C}</math><br />
A right triangle has an angle of 90 degrees, so any obtuse angle would make the sum of those two angles over 180. This contradicts the angle sum theorem (all triangles have angles that sum to 180) since negative angles don't exist.<br />
<br />
== See also ==<br />
{{AHSME box|year=1991|num-b=3|num-a=5}} <br />
<br />
[[Category: Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Bryanc921https://artofproblemsolving.com/wiki/index.php?title=1990_AHSME_Problems/Problem_14&diff=659901990 AHSME Problems/Problem 142014-11-21T01:24:09Z<p>Bryanc921: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<asy><br />
draw(circle((0,0),1),black);<br />
draw((0,1)--(cos(pi/14),-sin(pi/14))--(-cos(pi/14),-sin(pi/14))--cycle,dot);<br />
draw((-cos(pi/14),-sin(pi/14))--(0,-1/cos(3pi/7))--(cos(pi/14),-sin(pi/14)),dot);<br />
draw(arc((0,1),.25,230,310));<br />
MP("A",(0,1),N);MP("B",(cos(pi/14),-sin(pi/14)),E);MP("C",(-cos(pi/14),-sin(pi/14)),W);MP("D",(0,-1/cos(3pi/7)),S);<br />
MP("x",(0,.8),S);<br />
</asy><br />
<br />
An acute isosceles triangle, <math>ABC</math>, is inscribed in a circle. Through <math>B</math> and <math>C</math>, tangents to the circle are drawn, meeting at point <math>D</math>. If <math>\angle{ABC=\angle{ACB}=2\angle{D}</math> and <math>x</math> is the radian measure of <math>\angle{A}</math>, then <math>x=</math><br />
<br />
<math>\text{(A) } \frac{2\pi}{7}\quad<br />
\text{(B) } \frac{4\pi}{9}\quad<br />
\text{(C) } \frac{5\pi}{11}\quad<br />
\text{(D) } \frac{6\pi}{13}\quad<br />
\text{(E) } \frac{7\pi}{15}</math><br />
<br />
== Solution ==<br />
<math>\fbox{A}</math><br />
We can make two equations (assume angle D is y): <math>y+2x=180</math> and <math>4y+x=180</math>. We find that <math>x=\dfrac{360}{7}</math>. Now we have to convert this to radians. 360 degrees is <math>2\pi</math> radians, so since we have <math>\dfrac{360}{7}</math> degrees, the answer is <math>\dfrac{2\pi}{7}</math>.<br />
<br />
== See also ==<br />
{{AHSME box|year=1990|num-b=13|num-a=15}} <br />
<br />
[[Category: Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Bryanc921https://artofproblemsolving.com/wiki/index.php?title=1990_AHSME_Problems/Problem_14&diff=659891990 AHSME Problems/Problem 142014-11-21T01:20:01Z<p>Bryanc921: /* Problem */</p>
<hr />
<div>== Problem ==<br />
<asy><br />
draw(circle((0,0),1),black);<br />
draw((0,1)--(cos(pi/14),-sin(pi/14))--(-cos(pi/14),-sin(pi/14))--cycle,dot);<br />
draw((-cos(pi/14),-sin(pi/14))--(0,-1/cos(3pi/7))--(cos(pi/14),-sin(pi/14)),dot);<br />
draw(arc((0,1),.25,230,310));<br />
MP("A",(0,1),N);MP("B",(cos(pi/14),-sin(pi/14)),E);MP("C",(-cos(pi/14),-sin(pi/14)),W);MP("D",(0,-1/cos(3pi/7)),S);<br />
MP("x",(0,.8),S);<br />
</asy><br />
<br />
An acute isosceles triangle, <math>ABC</math>, is inscribed in a circle. Through <math>B</math> and <math>C</math>, tangents to the circle are drawn, meeting at point <math>D</math>. If <math>\angle{ABC=\angle{ACB}=2\angle{D}</math> and <math>x</math> is the radian measure of <math>\angle{A}</math>, then <math>x=</math><br />
<br />
<math>\text{(A) } \frac{2\pi}{7}\quad<br />
\text{(B) } \frac{4\pi}{9}\quad<br />
\text{(C) } \frac{5\pi}{11}\quad<br />
\text{(D) } \frac{6\pi}{13}\quad<br />
\text{(E) } \frac{7\pi}{15}</math><br />
<br />
== Solution ==<br />
<math>\fbox{A}</math><br />
<br />
== See also ==<br />
{{AHSME box|year=1990|num-b=13|num-a=15}} <br />
<br />
[[Category: Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Bryanc921