https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Giraffefun&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T10:29:27ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=1969_AHSME_Problems/Problem_32&diff=987791969 AHSME Problems/Problem 322018-11-21T18:24:34Z<p>Giraffefun: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<br />
Let a sequence <math>\{u_n\}</math> be defined by <math>u_1=5</math> and the relationship <math>u_{n+1}-u_n=3+4(n-1), n=1,2,3\cdots.</math>If <math>u_n</math> is expressed as a polynomial in <math>n</math>, the algebraic sum of its coefficients is:<br />
<br />
<math>\text{(A) 3} \quad<br />
\text{(B) 4} \quad<br />
\text{(C) 5} \quad<br />
\text{(D) 6} \quad<br />
\text{(E) 11} </math><br />
<br />
== Solution ==<br />
<br />
Note that the first differences create a linear function, so the sequence <math>{u_n}</math> is quadratic.<br />
<br />
The first three terms of the sequence are <math>5</math>, <math>12</math>, and <math>23</math>. From there, a [[system of equations]] can be written.<br />
<cmath>a+b+c=5</cmath><br />
<cmath>4a+2b+c=12</cmath><br />
<cmath>9a+3b+c=23</cmath><br />
Solve the system to get <math>a=2</math>, <math>b=1</math>, and <math>c=2</math>. The sum of the coefficients is <math>\boxed{\textbf{(C) } 5}</math>.<br />
<br />
Note: solving the system is extra work, as the answer is described by the first equation.<br />
<br />
== See also ==<br />
{{AHSME 35p box|year=1969|num-b=31|num-a=33}} <br />
<br />
[[Category: Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Giraffefunhttps://artofproblemsolving.com/wiki/index.php?title=1989_AHSME_Problems/Problem_7&diff=982191989 AHSME Problems/Problem 72018-10-21T22:04:32Z<p>Giraffefun: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<br />
In <math> \triangle ABC</math>, <math>\angle A = 100^\circ</math>, <math>\angle B = 50^\circ</math>, <math>\angle C = 30^\circ</math>, <math>\overline{AH}</math> is an altitude, and <math>\overline{BM}</math> is a median. Then <math>\angle MHC=</math><br />
<br />
<asy><br />
draw((0,0)--(16,0)--(6,6)--cycle);<br />
draw((6,6)--(6,0)--(11,3)--(0,0));<br />
dot((6,6));<br />
dot((0,0));<br />
dot((11,3));<br />
dot((6,0));<br />
dot((16,0));<br />
label("A", (6,6), N);<br />
label("B", (0,0), W);<br />
label("C", (16,0), E);<br />
label("H", (6,0), S);<br />
label("M", (11,3), NE);</asy><br />
<br />
<math> \textrm{(A)}\ 15^\circ\qquad\textrm{(B)}\ 22.5^\circ\qquad\textrm{(C)}\ 30^\circ\qquad\textrm{(D)}\ 40^\circ\qquad\textrm{(E)}\ 45^\circ </math><br />
<br />
==Solution==<br />
We are told that <math>\overline{BM}</math> is a median, so <math>\overline{AM}=\overline{MC}</math>. Drop an altitude from <math>M</math> to <math>\overline{HC}</math>, adding point <math>N</math>, and you can see that <math>\triangle ACH</math> and <math>\triangle MCN</math> are similar, implying <math>\overline{HN}=\overline{NC}</math>, implying that <math>\triangle MNH</math> and <math>\triangle MNC</math> are congruent, so <math>\angle MHC=\angle C=30^\circ</math>.<br />
<br />
<asy><br />
draw((0,0)--(16,0)--(6,6)--cycle);<br />
draw((6,6)--(6,0)--(11,3)--(0,0));<br />
draw((11,3)--(11,0));<br />
dot((6,6));<br />
dot((0,0));<br />
dot((11,3));<br />
dot((6,0));<br />
dot((16,0));<br />
dot((11,0));<br />
label("A", (6,6), N);<br />
label("B", (0,0), W);<br />
label("C", (16,0), E);<br />
label("H", (6,0), S);<br />
label("M", (11,3), NE);<br />
label("N", (11,0), S);</asy><br />
<br />
== See also ==<br />
{{AHSME box|year=1989|num-b=6|num-a=8}} <br />
<br />
[[Category: Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Giraffefunhttps://artofproblemsolving.com/wiki/index.php?title=LaTeX:Symbols&diff=93416LaTeX:Symbols2018-03-23T03:41:06Z<p>Giraffefun: </p>
<hr />
<div>{{Latex}}<br />
<br />
This article will provide a short list of commonly used LaTeX symbols. <br />
<br />
== Common Symbols ==<br />
<br />
=== Operators ===<br />
*<math>\div</math><br />
*<math>\frac{2}{1}</math><br />
*<math>+</math><br />
*<math>-</math><br />
*<math>\dfrac{1}{2}</math><br />
<br />
== Finding Other Symbols ==<br />
<br />
Here are some external resources for finding less commonly used symbols:<br />
<ul><br />
<li><br />
[http://detexify.kirelabs.org/classify.html Detexify] is an app which allows you to draw the symbol you'd like and shows you the <math>\LaTeX</math> code for it!<br />
<br/><br/></li><br />
<br />
<li><br />
MathJax (what allows us to use <math>\LaTeX</math> on the web) maintains a [http://docs.mathjax.org/en/latest/tex.html#supported-latex-commands list of supported commands].<br />
<br/><br/></li><br />
<br />
<li><br />
[http://mirrors.ctan.org/info/symbols/comprehensive/symbols-a4.pdf The Comprehensive LaTeX Symbol List].<br />
<br/><br/></li><br />
</ul><br />
<br />
-------------------------------------------------------------------------------------------------------------<br />
<br />
==Operators==<br />
{| class="latextable"<br />
!Symbol!!Command!!Symbol!!Command!!Symbol!!Command<br />
|-<br />
|<math>\pm</math>||\pm||<math>\mp</math>||\mp||<math>\times</math>||\times<br />
|-<br />
|<math>\div</math>||\div||<math>\cdot</math>||\cdot||<math>\ast</math>||\ast<br />
|-<br />
|<math>\star</math>||\star||<math>\dagger</math>||\dagger||<math>\ddagger</math>||\ddagger<br />
|-<br />
|<math>\amalg</math>||\amalg||<math>\cap</math>||\cap||<math>\cup</math>||\cup<br />
|-<br />
|<math>\uplus</math>||\uplus||<math>\sqcap</math>||\sqcap||<math>\sqcup</math>||\sqcup<br />
|-<br />
|<math>\vee</math>||\vee||<math>\wedge</math>||\wedge||<math>\oplus</math>||\oplus<br />
|-<br />
|<math>\ominus</math>||\ominus||<math>\otimes</math>||\otimes||<math>\circ</math>||\circ<br />
|-<br />
|<math>\bullet</math>||\bullet||<math>\diamond</math>||\diamond||<math>\lhd</math>||\lhd<br />
|-<br />
|<math>\rhd</math>||\rhd||<math>\unlhd</math>||\unlhd||[[Image:Unrhd.gif]]||\unrhd<br />
|-<br />
|<math>\oslash</math>||\oslash||<math>\odot</math>||\odot||<math>\bigcirc</math>||\bigcirc<br />
|-<br />
|<math>\triangleleft</math>||\triangleleft||<math>\Diamond</math>||\Diamond||<math>\bigtriangleup</math>||\bigtriangleup<br />
|-<br />
|<math>\bigtriangledown</math>||\bigtriangledown||<math>\Box</math>||\Box||<math>\triangleright</math>||\triangleright<br />
|-<br />
|<math>\setminus</math>||\setminus||<math>\wr</math>||\wr||<math>\sqrt{x}</math>||\sqrt{x}<br />
|-<br />
|<math>x^{\circ}</math>||x^{\circ}||<math>\triangledown</math>||\triangledown||<math>\sqrt[n]{x}</math>||\sqrt[n]{x}<br />
|-<br />
|<math>a^x</math>||a^x||<math>a^{xyz}</math>||a^{xyz}<br />
|}<br />
<br />
==Relations==<br />
{| class="latextable"<br />
!Symbol !! Command !!Symbol !! Command!!Symbol !! Command<br />
|-<br />
| <math>\le</math>||\le||<math>\ge</math>||\ge||<math>\neq</math>||\neq<br />
|-<br />
| <math>\sim</math>||\sim||<math>\ll</math>||\ll||<math>\gg</math>||\gg<br />
|-<br />
| <math>\doteq</math>||\doteq||<math>\simeq</math>||\simeq||<math>\subset</math>||\subset<br />
|-<br />
| <math>\supset</math>||\supset||<math>\approx</math>||\approx||<math>\asymp</math>||\asymp<br />
|-<br />
| <math>\subseteq</math>||\subseteq||<math>\supseteq</math>||\supseteq||<math>\cong</math>||\cong<br />
|-<br />
| <math>\smile</math>||\smile||<math>\sqsubset</math>||\sqsubset||<math>\sqsupset</math>||\sqsupset<br />
|-<br />
| <math>\equiv</math>||\equiv||<math>\frown</math>||\frown||<math>\sqsubseteq</math>||\sqsubseteq<br />
|-<br />
| <math>\sqsupseteq</math>||\sqsupseteq||<math>\propto</math>||\propto||<math>\bowtie</math>||\bowtie<br />
|-<br />
| <math>\in</math>||\in||<math>\ni</math>||\ni||<math>\prec</math>||\prec<br />
|-<br />
| <math>\succ</math>||\succ||<math>\vdash</math>||\vdash||<math>\dashv</math>||\dashv<br />
|-<br />
| <math>\preceq</math>||\preceq||<math>\succeq</math>||\succeq||<math>\models</math>||\models<br />
|-<br />
| <math>\perp</math>||\perp||<math>\parallel</math>||\parallel||<br />
|-<br />
| <math>\mid</math>||\mid||<math>\bumpeq</math>||\bumpeq||<br />
|}<br />
Negations of many of these relations can be formed by just putting \not before the symbol, or by slipping an n between the \ and the word. Here are a few examples, plus a few other negations; it works for many of the others as well.<br />
{| class="latextable"<br />
!Symbol!!Command!!Symbol!!Command!!Symbol!!Command<br />
|-<br />
|<math>\nmid</math>||\nmid||<math>\nleq</math>||\nleq||<math>\ngeq</math>||\ngeq<br />
|-<br />
| <math>\nsim</math>||\nsim||<math>\ncong</math>||\ncong||<math>\nparallel</math>||\nparallel<br />
|-<br />
| <math>\not<</math>||\not<||<math>\not></math>||\not>||<math>\not=</math>||\not=<br />
|-<br />
| <math>\not\le</math>||\not\le||<math>\not\ge</math>||\not\ge||<math>\not\sim</math>||\not\sim<br />
|-<br />
|<math>\not \approx</math>||\not\approx||<math>\not\cong</math>||\not\cong||<math>\not\equiv</math>||\not\equiv<br />
|-<br />
| <math>\not\parallel</math>||\not\parallel||<math>\nless</math>||\nless||<math>\ngtr</math>||\ngtr<br />
|-<br />
| <math>\lneq</math>||\lneq||<math>\gneq</math>||\gneq||<math>\lnsim</math>||\lnsim<br />
|-<br />
| <math>\lneqq</math>||\lneqq||<math>\gneqq</math>||\gneqq<br />
|}<br />
<br />
To use other relations not listed here, such as =, >, and <, in LaTeX, you may just use the symbols on your keyboard.<br />
<br />
==Greek Letters==<br />
{| class="latextable"<br />
|+ Lowercase Letters<br />
!Symbol!!Command!!Symbol!!Command!!Symbol!!Command!!Symbol!!Command<br />
|-<br />
|<math>\alpha</math>||\alpha||<math>\beta</math>||\beta||<math>\gamma</math>||\gamma||<math>\delta</math>||\delta<br />
|-<br />
| <math>\epsilon</math>||\epsilon||<math>\varepsilon</math>||\varepsilon||<math>\zeta</math>||\zeta||<math>\eta</math>||\eta<br />
|-<br />
| <math>\theta</math>||\theta||<math>\vartheta</math>||\vartheta||<math>\iota</math>||\iota||<math>\kappa</math>||\kappa<br />
|-<br />
| <math>\lambda</math>||\lambda||<math>\mu</math>||\mu||<math>\nu</math>||\nu||<math>\xi</math>||\xi<br />
|-<br />
|<math>\pi</math>||\pi||<math>\varpi</math>||\varpi||<math>\rho</math>||\rho||<math>\varrho</math>||\varrho<br />
|-<br />
| <math>\sigma</math>||\sigma||<math>\varsigma</math>||\varsigma||<math>\tau</math>||\tau||<math>\upsilon</math>||\upsilon<br />
|-<br />
| <math>\phi</math>||\phi||<math>\varphi</math>||\varphi||<math>\chi</math>||\chi||<math>\psi</math>||\psi<br />
|-<br />
| <math>\omega</math>||\omega<br />
|}<br />
<br />
<br />
{| class="latextable"<br />
|+ Capital Letters<br />
!Symbol!!Command!!Symbol!!Command!!Symbol!!Command!!Symbol!!Command<br />
|-<br />
|<math>\Gamma</math>||\Gamma||<math>\Delta</math>||\Delta||<math>\Theta</math>||\Theta||<math>\Lambda</math>||\Lambda<br />
|-<br />
| <math>\Xi</math>||\Xi||<math>\Pi</math>||\Pi||<math>\Sigma</math>||\Sigma||<math>\Upsilon</math>||\Upsilon<br />
|-<br />
| <math>\Phi</math>||\Phi||<math>\Psi</math>||\Psi||<math>\Omega</math>||\Omega<br />
|}<br />
<br />
<br />
== Arrows ==<br />
<br />
{| class="latextable"<br />
!Symbol !! Command!!Symbol !! Command<br />
|-<br />
|<math>\gets</math>||\gets||<math>\to</math>||\to<br />
|-<br />
|<math>\leftarrow</math>||\leftarrow||<math>\Leftarrow</math>||\Leftarrow<br />
|-<br />
|<math>\rightarrow</math>||\rightarrow||<math>\Rightarrow</math>||\Rightarrow<br />
|-<br />
|<math>\leftrightarrow</math>||\leftrightarrow||<math>\Leftrightarrow</math>||\Leftrightarrow<br />
|-<br />
|<math>\mapsto</math>||\mapsto||<math>\hookleftarrow</math>||\hookleftarrow<br />
|-<br />
|<math>\leftharpoonup</math>||\leftharpoonup||<math>\leftharpoondown</math>||\leftharpoondown<br />
|-<br />
|<math>\rightleftharpoons</math>||\rightleftharpoons||<math>\longleftarrow</math>||\longleftarrow<br />
|-<br />
|<math>\Longleftarrow</math>||\Longleftarrow||<math>\longrightarrow</math>||\longrightarrow<br />
|-<br />
|<math>\Longrightarrow</math>||\Longrightarrow||<math>\longleftrightarrow</math>||\longleftrightarrow<br />
|-<br />
|<math>\Longleftrightarrow</math>||\Longleftrightarrow||<math>\longmapsto</math>||\longmapsto<br />
|-<br />
|<math>\hookrightarrow</math>||\hookrightarrow||<math>\rightharpoonup</math>||\rightharpoonup<br />
|-<br />
|<math>\rightharpoondown</math>||\rightharpoondown||<math>\leadsto</math>||\leadsto<br />
|-<br />
|<math>\uparrow</math>||\uparrow||<math>\Uparrow</math>||\Uparrow<br />
|-<br />
|<math>\downarrow</math>||\downarrow||<math>\Downarrow</math>||\Downarrow<br />
|-<br />
|<math>\updownarrow</math>||\updownarrow||<math>\Updownarrow</math>||\Updownarrow<br />
|-<br />
|<math>\nearrow</math>||\nearrow||<math>\searrow</math>||\searrow<br />
|-<br />
|<math>\swarrow</math>||\swarrow||<math>\nwarrow</math>||\nwarrow<br />
|}<br />
(For those of you who hate typing long strings of letters, \iff and \implies can be used in place of \Longleftrightarrow and \Longrightarrow respectively.)<br />
<br />
==Dots==<br />
{| class="latextable"<br />
!Symbol!!Command!!Symbol!!Command!!<br />
|- <br />
|<math>\cdot</math>||\cdot|| |<math>\vdots</math>||\vdots|| <br />
|- <br />
|<math>\dots</math>||\dots|| |<math>\ddots</math>||\ddots||<br />
|-<br />
|<math>\cdots</math>||\cdots|| |<math>\iddots</math>||\iddots||<br />
<br />
|}<br />
<br />
==Accents==<br />
{| class="latextable"<br />
!Symbol !! Command!!Symbol !! Command!!Symbol !! Command<br />
|-<br />
|<math>\hat{x}</math>||\hat{x}||<math>\check{x}</math>||\check{x}||<math>\dot{x}</math>||\dot{x}<br />
|-<br />
|<math>\breve{x}</math>||\breve{x}||<math>\acute{x}</math>||\acute{x}||<math>\ddot{x}</math>||\ddot{x}<br />
|-<br />
|<math>\grave{x}</math>||\grave{x}||<math>\tilde{x}</math>||\tilde{x}||<math>\mathring{x}</math>||\mathring{x}<br />
|-<br />
|<math>\bar{x}</math>||\bar{x}||<math>\vec{x}</math>||\vec{x}<br />
|}<br />
When applying accents to i and j, you can use \imath and \jmath to keep the dots from interfering with the accents:<br />
{| class="latextable"<br />
!Symbol !! Command!!Symbol !! Command<br />
|-<br />
|<math>\vec{\jmath}</math>||\vec{\jmath}||<math>\tilde{\imath}</math>||\tilde{\imath}<br />
|}<br />
\tilde and \hat have wide versions that allow you to accent an expression:<br />
{| class="latextable"<br />
!Symbol !! Command!!Symbol !! Command<br />
|-<br />
|<math>\widehat{7+x}</math>||\widehat{7+x}||<math>\widetilde{abc}</math>||\widetilde{abc}<br />
|}<br />
<br />
==Others==<br />
{| class="latextable"<br />
!Symbol!!Command!!Symbol!!Command!!Symbol!!Command <br />
|-<br />
|<math>\infty</math>||\infty||<math>\triangle</math>||\triangle||<math>\angle</math>||\angle<br />
|-<br />
|<math>\aleph</math>||\aleph||<math>\hbar</math>||\hbar||<math>\imath</math>||\imath<br />
|-<br />
|<math>\jmath</math>||\jmath||<math>\ell</math>||\ell||<math>\wp</math>||\wp<br />
|-<br />
|<math>\Re</math>||\Re||<math>\Im</math>||\Im||<math>\mho</math>||\mho<br />
|-<br />
|<math>\prime</math>||\prime||<math>\emptyset</math>||\emptyset||<math>\nabla</math>||\nabla<br />
|-<br />
|<math>\surd</math>||\surd||<math>\partial</math>||\partial||<math>\top</math>||\top<br />
|-<br />
|<math>\bot</math>||\bot||<math>\vdash</math>||\vdash||<math>\dashv</math>||\dashv<br />
|-<br />
|<math>\forall</math>||\forall||<math>\exists</math>||\exists||<math>\neg</math>||\neg<br />
|-<br />
|<math>\flat</math>||\flat||<math>\natural</math>||\natural||<math>\sharp</math>||\sharp<br />
|-<br />
|<math>\backslash</math>||\backslash||<math>\Box</math>||\Box||<math>\Diamond</math>||\Diamond<br />
|-<br />
|<math>\clubsuit</math>||\clubsuit||<math>\diamondsuit</math>||\diamondsuit||<math>\heartsuit</math>||\heartsuit<br />
|-<br />
|[[Image:Spadesuit.gif]]||\spadesuit||<math>\Join</math>||\Join||<math>\blacksquare</math>||\blacksquare<br />
|-<br />
|<math>\S</math>||\S||<math>\P</math>||\P||<math>\copyright</math>||\copyright<br />
|-<br />
|<math>\pounds</math>||\pounds||<math>\overarc{ABC}</math>||\overarc{ABC}||<math>\underarc{XYZ}</math>||\underarc{XYZ}<br />
|-<br />
|<math>\bigstar</math>||\bigstar||<math>\in</math>||\in||<math>\cup</math>||\cup<br />
|-<br />
|<math>\square</math>||\square||<br />
|-<br />
|<math>\smiley</math>||\smiley||<br />
|-<br />
|<math>\mathbb{R}</math>||\mathbb{R} (represents all real numbers)||<br />
|-<br />
|<math>\checkmark</math>||\checkmark||<br />
|}<br />
<br />
==Command Symbols==<br />
Some symbols are used in commands so they need to be treated in a special way.<br />
{| class="latextable"<br />
!Symbol!!Command!!Symbol!!Command!!Symbol!!Command!!Symbol!!Command<br />
|-<br />
|<math>\textdollar</math>||\textdollar or \$||<math>\&</math>||\&||<math>\%</math>||\%||<math>\#</math>||\#<br />
|-<br />
|<math>\_</math>||\_||<math>\{</math>||\{||<math>\}</math>||\}||<math>\backslash</math>||\backslash<br />
|}<br />
<br />
(Warning: Using <nowiki>$</nowiki> for <math>\textdollar</math> will result in <math>\$</math>. This is a bug as far as we know. Depending on the version of <math>\LaTeX</math> this is not always a problem.)<br />
<br />
==European Language Symbols==<br />
{| class="latextable"<br />
!Symbol!!Command!!Symbol!!Command!!Symbol!!Command!!Symbol!!Command<br />
|-<br />
|<math>{\oe}</math>||{\oe}||<math>{\ae}</math>||{\ae}||<math>{\o}</math>||{\o}<br />
|-<br />
|<math>{\OE}</math>||{\OE}||<math>{\AE}</math>||{\AE}||<math>{\AA}</math>||{\AA}||<math>{\O}</math>||{\O}<br />
|-<br />
|<math>{\l}</math>||{\l}||<math>{\ss}</math>||{\ss}||<math>\text{!`}</math>||!`<br />
|-<br />
|<math>{\L}</math>||{\L}||<math>{\SS}</math>||{\SS}||<br />
|}<br />
<br />
==Bracketing Symbols==<br />
In mathematics, sometimes we need to enclose expressions in brackets or braces or parentheses. Some of these work just as you'd imagine in LaTeX; type ( and ) for parentheses, [ and ] for brackets, and | and | for absolute value. However, other symbols have special commands:<br />
{| class="latextable"<br />
!Symbol !! Command!!Symbol !! Command!!Symbol !! Command<br />
|-<br />
|<math>\{</math>||\{||<math>\}</math>||\}||<math>\|</math>||<nowiki>\|</nowiki><br />
|-<br />
| <math>\backslash</math>||\backslash||<math>\lfloor</math>||\lfloor||<math>\rfloor</math>||\rfloor<br />
|-<br />
| <math>\lceil</math>||\lceil||<math>\rceil</math>||\rceil||<math>\langle</math>||\langle<br />
|-<br />
| <math>\rangle</math>||\rangle<br />
|}<br />
You might notice that if you use any of these to typeset an expression that is vertically large, like<br />
<br />
:<tt>(\frac{a}{x} )^2</tt><br />
<br />
the parentheses don't come out the right size:<br />
<br />
:<math>(\frac{a}{x})^2</math><br />
<br />
If we put \left and \right before the relevant parentheses, we get a prettier expression:<br />
<br />
:<tt>\left(\frac{a}{x} \right)^2</tt><br />
<br />
gives<br />
<br />
:<math>\left(\frac{a}{x} \right)^2</math><br />
<br />
And with system of equations:<br />
<br />
<tt>\left\{\begin{array}{l}x+y=3\\2x+y=5\end{array}\right.</tt><br />
<br />
Gives<br />
<br />
<math>\left\{\begin{array}{l}x+y=3\\2x+y=5\end{array}\right.</math><br />
<br />
See that there's a dot after <tt>\right</tt>. You must put that dot or the code won't work.<br />
<br />
<br />
And, if you type this<br />
<br />
<tt>\underbrace{a_0+a_1+a_2+\cdots+a_n}_{x}</tt><br />
<br />
Gives<br />
<br />
<math>\underbrace{a_0+a_1+a_2+\cdots+a_n}_{x}</math><br />
<br />
Or<br />
<br />
<tt>\overbrace{a_0+a_1+a_2+\cdots+a_n}^{x}</tt><br />
<br />
Gives<br />
<br />
<math>\overbrace{a_0+a_1+a_2+\cdots+a_n}^{x}</math><br />
<br />
<br />
\left and \right can also be used to resize the following symbols:<br />
{| class="latextable"<br />
!Symbol !! Command!!Symbol !! Command!!Symbol !! Command<br />
|-<br />
|<math>\uparrow</math>||\uparrow||<math>\downarrow</math>||\downarrow||<math>\updownarrow</math>||\updownarrow<br />
|-<br />
| <math>\Uparrow</math>||\Uparrow||<math>\Downarrow</math>||\Downarrow||<math>\Updownarrow</math>||\Updownarrow<br />
|}<br />
<br />
==Multi-Size Symbols==<br />
Some symbols render differently in inline math mode and in display mode. Display mode occurs when you use <nowiki>\[...\]</nowiki> or <nowiki>$$...$$</nowiki>, or environments like \begin{equation}...\end{equation}, \begin{align}...\end{align}. Read more in the [[LaTeX:Commands|commands]] section of the guide about how symbols which take arguments above and below the symbols, such as a summation symbol, behave in the two modes.<br />
<br />
In each of the following, the two images show the symbol in display mode, then in inline mode.<br />
<br />
{| class="latextable"<br />
!Symbol !! Command!!Symbol !! Command!!Symbol !! Command<br />
|-<br />
|<math>\sum \textstyle\sum</math>||\sum||<math>\int \textstyle\int</math>||\int||<math>\oint \textstyle\oint</math>||\oint<br />
|-<br />
|<math>\prod \textstyle\prod</math>||\prod||<math>\coprod \textstyle\coprod</math>||\coprod||<math>\bigcap \textstyle\bigcap</math>||\bigcap<br />
|-<br />
|<math>\bigcup \textstyle\bigcup</math>||\bigcup||<math>\bigsqcup \textstyle\bigsqcup</math>||\bigsqcup||<math>\bigvee \textstyle\bigvee</math>||\bigvee<br />
|-<br />
|<math>\bigwedge \textstyle\bigwedge</math>||\bigwedge||<math>\bigodot \textstyle\bigodot</math>||\bigodot||<math>\bigotimes \textstyle\bigotimes</math>||\bigotimes<br />
|-<br />
|<math>\bigoplus \textstyle\bigoplus</math>||\bigoplus||<math>\biguplus \textstyle\biguplus</math>||\biguplus<br />
|}<br />
<br />
==See Also==<br />
*[[LaTeX:Commands | Next: Commands]]<br />
*[[LaTeX:Layout | Previous: Layout]]</div>Giraffefunhttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_24&diff=916382018 AMC 10B Problems/Problem 242018-02-16T22:13:37Z<p>Giraffefun: </p>
<hr />
<div>==Problem==<br />
<br />
Let <math>ABCDEF</math> be a regular hexagon with side length <math>1</math>. Denote <math>X</math>, <math>Y</math>, and <math>Z</math> the midpoints of sides <math>\overline {AB}</math>, <math>\overline{CD}</math>, and <math>\overline{EF}</math>, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of <math>\triangle ACE</math> and <math>\triangle XYZ</math>?<br />
<br />
<math>\textbf{(A)} \frac {3}{8}\sqrt{3} \qquad \textbf{(B)} \frac {7}{16}\sqrt{3} \qquad \textbf{(C)} \frac {15}{32}\sqrt{3} \qquad \textbf{(D)} \frac {1}{2}\sqrt{3} \qquad \textbf{(E)} \frac {9}{16}\sqrt{3} \qquad </math><br />
<br />
<br />
Answer: <math>\frac {15}{32}\sqrt{3}</math><br />
<br />
==Solution==<br />
<br />
<asy><br />
pair A,B,C,D,E,F,W,X,Y,Z,M,N,O;<br />
A=(0,sqrt(3));<br />
B=(1,sqrt(3));<br />
C=(3/2,sqrt(3)/2);<br />
D=(1,0);<br />
E=(0,0);<br />
F=(-1/2,sqrt(3)/2);<br />
X=(1/2, sqrt(3));<br />
Y=(5/4, sqrt(3)/4);<br />
Z=(-1/4, sqrt(3)/4); <br />
M=(0,sqrt(3)/2);<br />
N=(3/4,3sqrt(3)/4);<br />
O=(3/4,sqrt(3)/4);<br />
<br />
draw(A--B--C--D--E--F--cycle);<br />
<br />
draw(A--C--E--cycle);<br />
draw(X--Y--Z--cycle);<br />
draw(M--N--O);<br />
<br />
label("$A$",A,NW);<br />
label("$B$",B,NE);<br />
label("$C$",C,ESE);<br />
label("$D$",D,SE);<br />
label("$E$",E,SW);<br />
label("$F$",F,WSW);<br />
label("$X$", X, N);<br />
label("$Y$", Y, ESE);<br />
label("$Z$", Z, WSW);<br />
label("$M$", M, W);<br />
label("$N$", N, NE);<br />
label("$O$", O, SE);<br />
<br />
</asy><br />
<br />
Now, if we look at the figure, we can see that the complement of the hexagon we are trying to find is composed of 3 isosceles trapezoids (AXFZ, XBCY, and ZYED), and 3 right triangles (With one vertice on each of X, Y, and Z). Now we know that one base of each trapezoid is just the side length of the hexagon which is 1, and the other base is 3/2 (It is halfway in between the side and the longest diagonal) with a height of <math>sqrt(3)/4</math> (by using the Pythagorean theorem and the fact that it is an isosceles trapezoid) to give each trapezoid having an area of <math>5sqrt(3)/16</math> for a total area of <math>15sqrt(3)/16</math> (Alternatively, we could have calculated the area of hexagon ABCDEF and subtracted the area of triangle XYZ, which, as we showed before, had a side length of 3/2). Now, we need to find the area of each of the small triangles, which, if we look at the triangle that has a vertice on X, is similar to the triangle with a base of YC = 1/2. Using similar triangles we calculate the base to be 1/4 and the height to be <math>sqrt(3)/4</math> giving us an area of <math>sqrt(3)/32</math> per triangle, and a total area of <math>3sqrt(3)/32</math>. Adding the two areas together, we get <math>15sqrt(3)/16+3sqrt(3)/32=33sqrt(3)/32</math>. Finding the total area, we get <math>6*1^2*sqrt(3)/4=3sqrt(3)/2</math>. Taking the complement, we get <math>3sqrt(3)/2-33sqrt(3)/32=15sqrt(3)/32</math>, so the answer is C --- Arpitr20<br />
<br />
==See Also==<br />
{{AMC10 box|year=2018|ab=B|num-b=23|num-a=25}}<br />
{{AMC12 box|year=2018|ab=B|num-b=19|num-a=21}}<br />
{{MAA Notice}}<br />
<br />
==Problem==<br />
<br />
Let <math>ABCDEF</math> be a regular hexagon with side length <math>1</math>. Denote <math>X</math>, <math>Y</math>, and <math>Z</math> the midpoints of sides <math>\overline {AB}</math>, <math>\overline{CD}</math>, and <math>\overline{EF}</math>, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of <math>\triangle ACE</math> and <math>\triangle XYZ</math>?<br />
<br />
<math>\textbf{(A)} \frac {3}{8}\sqrt{3} \qquad \textbf{(B)} \frac {7}{16}\sqrt{3} \qquad \textbf{(C)} \frac {15}{32}\sqrt{3} \qquad \textbf{(D)} \frac {1}{2}\sqrt{3} \qquad \textbf{(E)} \frac {9}{16}\sqrt{3} \qquad </math><br />
<br />
<br />
Answer: <math>\frac {15}{32}\sqrt{3}</math><br />
<br />
==Solution==<br />
<br />
<asy><br />
pair A,B,C,D,E,F,W,X,Y,Z,M,N,O;<br />
A=(0,sqrt(3));<br />
B=(1,sqrt(3));<br />
C=(3/2,sqrt(3)/2);<br />
D=(1,0);<br />
E=(0,0);<br />
F=(-1/2,sqrt(3)/2);<br />
X=(1/2, sqrt(3));<br />
Y=(5/4, sqrt(3)/4);<br />
Z=(-1/4, sqrt(3)/4); <br />
M=(0,sqrt(3)/2);<br />
N=(1,3/2);<br />
O=(1,1/2);<br />
<br />
draw(A--B--C--D--E--F--cycle);<br />
<br />
draw(A--C--E--cycle);<br />
draw(X--Y--Z--cycle);<br />
draw(M--N--O);<br />
<br />
label("$A$",A,NW);<br />
label("$B$",B,NE);<br />
label("$C$",C,ESE);<br />
label("$D$",D,SE);<br />
label("$E$",E,SW);<br />
label("$F$",F,WSW);<br />
label("$X$", X, N);<br />
label("$Y$", Y, ESE);<br />
label("$Z$", Z, WSW);<br />
label("$M$", M, W);<br />
label("$N$", N, NE);<br />
label("$O$", O, SE);<br />
<br />
</asy><br />
<br />
Now, if we look at the figure, we can see that the complement of the hexagon we are trying to find is composed of 3 isosceles trapezoids (AXFZ, XBCY, and ZYED), and 3 right triangles (With one vertice on each of X, Y, and Z). Now we know that one base of each trapezoid is just the side length of the hexagon which is 1, and the other base is 3/2 (It is halfway in between the side and the longest diagonal) with a height of <math>sqrt(3)/4</math> (by using the Pythagorean theorem and the fact that it is an isosceles trapezoid) to give each trapezoid having an area of <math>5sqrt(3)/16</math> for a total area of <math>15sqrt(3)/16</math> (Alternatively, we could have calculated the area of hexagon ABCDEF and subtracted the area of triangle XYZ, which, as we showed before, had a side length of 3/2). Now, we need to find the area of each of the small triangles, which, if we look at the triangle that has a vertice on X, is similar to the triangle with a base of YC = 1/2. Using similar triangles we calculate the base to be 1/4 and the height to be <math>sqrt(3)/4</math> giving us an area of <math>sqrt(3)/32</math> per triangle, and a total area of <math>3sqrt(3)/32</math>. Adding the two areas together, we get <math>15sqrt(3)/16+3sqrt(3)/32=33sqrt(3)/32</math>. Finding the total area, we get <math>6*1^2*sqrt(3)/4=3sqrt(3)/2</math>. Taking the complement, we get <math>3sqrt(3)/2-33sqrt(3)/32=15sqrt(3)/32</math>, so the answer is C --- Arpitr20<br />
<br />
<br />
==See Also==<br />
{{AMC10 box|year=2018|ab=B|num-b=23|num-a=25}}<br />
{{AMC12 box|year=2018|ab=B|num-b=19|num-a=21}}<br />
{{MAA Notice}}<br />
<br />
==See Also==<br />
{{AMC10 box|year=2018|ab=B|num-b=23|num-a=25}}<br />
{{AMC12 box|year=2018|ab=B|num-b=19|num-a=21}}<br />
{{MAA Notice}}</div>Giraffefunhttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12B_Problems/Problem_8&diff=916232018 AMC 12B Problems/Problem 82018-02-16T22:00:05Z<p>Giraffefun: /* Problem */</p>
<hr />
<div>==Problem ==<br />
<br />
Line Segment <math>\overline{AB}</math> is a diameter of a circle with <math>AB = 24</math>. Point <math>C</math>, not equal to <math>A</math> or <math>B</math>, lies on the circle. As point <math>C</math> moves around the circle, the centroid (center of mass) of (insert triangle symbol)<math>ABC</math> traces out a closed curve missing two points. To the nearest positive integer, what is the area of the region bounded by this curve?<br />
<br />
<math>\textbf{(A)} \indent 25 \qquad \textbf{(B)} \indent 32 \qquad \textbf{(C)} \indent 50 \qquad \textbf{(D)} \indent 63 \qquad \textbf{(E)} \indent 75 </math><br />
<br />
==Solution==<br />
Draw the Median connecting C to the center O of the circle. Note that the centroid is <math>\frac{1}{3}</math> of the distance from O to C.<br />
Thus, as C traces a circle of radius 12, the Centroid will trace a circle of radius <math>\frac{12}{3}=4</math>.<br />
<br />
The area of this circle is <math>\pi\cdot4^2=16\pi \approx 50</math>.<br />
<br />
==See Also==<br />
{{AMC12 box|year=2018|ab=B|num-a=9|num-b=7}}<br />
{{MAA Notice}}</div>Giraffefunhttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12B_Problems/Problem_8&diff=916112018 AMC 12B Problems/Problem 82018-02-16T21:49:41Z<p>Giraffefun: </p>
<hr />
<div>==Problem ==<br />
<br />
Line Segment <math>\overline{AB}</math> is a diameter of a circle with <math>AB = 24</math>. Point <math>C</math>, not equal to <math>A</math> or <math>B</math>, lies on the circle. As point <math>C</math> moves around the circle, the centroid (center of mass) of (insert triangle symbol)<math>ABC</math> traces out a closed curve missing two points. To the nearest positive integer, what is the area of the region bounded by this curve?<br />
<br />
==Solution==<br />
<br />
==See Also==<br />
{{AMC12 box|year=2018|ab=B|num-a=9|num-b=7}}<br />
{{MAA Notice}}</div>Giraffefunhttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12B_Problems/Problem_6&diff=915432018 AMC 12B Problems/Problem 62018-02-16T21:09:01Z<p>Giraffefun: /* Solution 1 */</p>
<hr />
<div>==Problem==<br />
<br />
Suppose <math>S</math> cans of soda can be purchased from a vending machine for <math>Q</math> quarters. Which of the following expressions describes the number of cans of soda that can be purchased for <math>D</math> dollars, where 1 dollar is worth 4 quarters?<br />
<br />
<math>\textbf{(A)} \frac{4DQ}{S} \qquad \textbf{(B)} \frac{4DS}{Q} \qquad \textbf{(C)} \frac{4Q}{DS} \qquad \textbf{(D)} \frac{DQ}{4S} \qquad \textbf{(E)} \frac{DS}{4Q}</math><br />
<br />
==Solution 1==<br />
The unit price for a can of soda (in quarters) is <math>\frac{S}{Q}</math>. Thus, the number of cans which can be bought for <math>D</math> dollars (<math>4D</math> quarters) is<math> \boxed {\textbf{(B)} \frac{4DS}{Q}}</math> (Giraffefun)<br />
<br />
==See Also==<br />
{{AMC12 box|year=2018|ab=B|num-b=5|num-a=7}}<br />
{{MAA Notice}}</div>Giraffefunhttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12B_Problems/Problem_6&diff=915372018 AMC 12B Problems/Problem 62018-02-16T21:06:58Z<p>Giraffefun: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
Suppose <math>S</math> cans of soda can be purchased from a vending machine for <math>Q</math> quarters. Which of the following expressions describes the number of cans of soda that can be purchased for <math>D</math> dollars, where 1 dollar is worth 4 quarters?<br />
<br />
<math>\textbf{(A)} \frac{4DQ}{S} \qquad \textbf{(B)} \frac{4DS}{Q} \qquad \textbf{(C)} \frac{4Q}{DS} \qquad \textbf{(D)} \frac{DQ}{4S} \qquad \textbf{(E)} \frac{DS}{4Q}</math><br />
<br />
==Solution 1==<br />
The unit price for a can of soda (in quarters) is <math>\frac{S}{Q}</math>. Thus, the number of cans which can be bought for <math>D</math> dollars (<math>4D</math> quarters) is<math> \boxed {\textbf{(B)} \frac{4DS}{Q}}</math><br />
<br />
==See Also==<br />
{{AMC12 box|year=2018|ab=B|num-b=5|num-a=7}}<br />
{{MAA Notice}}</div>Giraffefunhttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12B_Problems/Problem_6&diff=915272018 AMC 12B Problems/Problem 62018-02-16T21:02:15Z<p>Giraffefun: /* Problem */</p>
<hr />
<div>==Problem==<br />
<br />
Suppose <math>S</math> cans of soda can be purchased from a vending machine for <math>Q</math> quarters. Which of the following expressions describes the number of cans of soda that can be purchased for <math>D</math> dollars, where 1 dollar is worth 4 quarters?<br />
<br />
<math>\textbf{(A)} \frac{4DQ}{S} \qquad \textbf{(B)} \frac{4DS}{Q} \qquad \textbf{(C)} \frac{4Q}{DS} \qquad \textbf{(D)} \frac{DQ}{4S} \qquad \textbf{(E)} \frac{DS}{4Q}</math><br />
<br />
==Solution==<br />
<br />
==See Also==<br />
{{AMC12 box|year=2018|ab=B|num-b=5|num-a=7}}<br />
{{MAA Notice}}</div>Giraffefunhttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12B_Problems/Problem_6&diff=915202018 AMC 12B Problems/Problem 62018-02-16T20:58:04Z<p>Giraffefun: /* Problem */</p>
<hr />
<div>==Problem==<br />
<br />
Suppose <math>S</math> cans of soda can be purchased from a vending machine for <math>Q</math> quarters. Which of the following expressions describes the number of cans of soda that can be purchased for <math>D</math> dollars, where 1 dollar is worth 4 quarters?<br />
<br />
==Solution==<br />
<br />
==See Also==<br />
{{AMC12 box|year=2018|ab=B|num-b=5|num-a=7}}<br />
{{MAA Notice}}</div>Giraffefunhttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12B_Problems/Problem_6&diff=915112018 AMC 12B Problems/Problem 62018-02-16T20:55:12Z<p>Giraffefun: /* Also See */</p>
<hr />
<div>==Problem==<br />
<br />
==Solution==<br />
<br />
==See Also==<br />
{{AMC12 box|year=2018|ab=B|num-b=5|num-a=7}}<br />
{{MAA Notice}}</div>Giraffefunhttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12B_Problems/Problem_6&diff=915042018 AMC 12B Problems/Problem 62018-02-16T20:52:44Z<p>Giraffefun: Created page with "==Problem== ==Solution== ==Also See=="</p>
<hr />
<div>==Problem==<br />
<br />
==Solution==<br />
<br />
==Also See==</div>Giraffefunhttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_24&diff=915002018 AMC 10B Problems/Problem 242018-02-16T20:50:54Z<p>Giraffefun: /* Problem */</p>
<hr />
<div>==Problem==<br />
<br />
Let <math>ABCDEFG</math> be a regular hexagon with side length <math>1</math>. Denote <math>X</math>, <math>Y</math>, and <math>Z</math> the midpoints of sides <math>\overline {AB}</math>, <math>\overline{CD}</math>, and <math>\overline{EF}</math>, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of <math>\triangle ACE</math> and <math>\triangle XYZ</math>?<br />
<br />
<math>\textbf{(A)} \frac {3}{8}\sqrt{3} \qquad \textbf{(B)} \frac {7}{16}\sqrt{3} \qquad \textbf{(C)} \frac {15}{32}\sqrt{3} \qquad \textbf{(D)} \frac {1}{2}\sqrt{3} \qquad \textbf{(E)} \frac {9}{16}\sqrt{3} \qquad </math><br />
<br />
<br />
Answer: <math>\frac {15}{32}\sqrt{3}</math><br />
<br />
==Solution==<br />
<br />
==See Also==<br />
{{AMC10 box|year=2018|ab=B|num-b=23|num-a=25}}<br />
{{AMC12 box|year=2018|ab=B|num-b=19|num-a=21}}<br />
{{MAA Notice}}</div>Giraffefunhttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_24&diff=914352018 AMC 10B Problems/Problem 242018-02-16T20:09:04Z<p>Giraffefun: </p>
<hr />
<div>==Problem==<br />
<br />
Let ABCDEFG be a regular hexagon with side length 1. Denote X, Y, and Z the midpoints of sides (segment) AB, (segment) CD, and (segment) EF, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of (insert) triangle symbol) ACE and (insert triangle symbol) XYZ?<br />
<br />
<math>\textbf{(A)} \frac {3}{8}\sqrt{3} \qquad \textbf{(B)} \frac {7}{16}\sqrt{3} \qquad \textbf{(C)} \frac {15}{32}\sqrt{3} \qquad \textbf{(D)} \frac {1}{2}\sqrt{3} \qquad \textbf{(E)} \frac {9}{16}\sqrt{3} \qquad </math><br />
<br />
<br />
<br />
Answer: 15sqrt(3)/32</div>Giraffefunhttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_24&diff=914292018 AMC 10B Problems/Problem 242018-02-16T20:06:20Z<p>Giraffefun: </p>
<hr />
<div>==Problem==<br />
<br />
Let ABCDEFG be a regular hexagon with side length 1. Denote X, Y, and Z the midpoints of sides (segment) AB, (segment) CD, and (segment) EF, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of (insert) triangle symbol) ACE and (insert triangle symbol) XYZ?<br />
<br />
<math>\textbf{(A)} \frac {3}{8}\sqrt{3}</math><br />
<br />
<br />
<br />
Answer: 15sqrt(3)/32</div>Giraffefunhttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_24&diff=914212018 AMC 10B Problems/Problem 242018-02-16T20:03:54Z<p>Giraffefun: </p>
<hr />
<div>==Problem==<br />
<br />
Let ABCDEFG be a regular hexagon with side length 1. Denote X, Y, and Z the midpoints of sides (segment) AB, (segment) CD, and (segment) EF, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of (insert) triangle symbol) ACE and (insert triangle symbol) XYZ?<br />
<br />
<br />
<br />
Answer: 15sqrt(3)/32</div>Giraffefunhttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_24&diff=914202018 AMC 10B Problems/Problem 242018-02-16T20:03:26Z<p>Giraffefun: /* Problem */</p>
<hr />
<div>(This is not the exact wording of the question)<br />
==Problem==<br />
<br />
Let ABCDEFG be a regular hexagon with side length 1. Denote X, Y, and Z the midpoints of sides (segment) AB, (segment) CD, and (segment) EF, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of (insert) triangle symbol) ACE and (insert triangle symbol) XYZ?<br />
<br />
<br />
<br />
Answer: 15sqrt(3)/32</div>Giraffefunhttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_24&diff=914142018 AMC 10B Problems/Problem 242018-02-16T19:57:51Z<p>Giraffefun: </p>
<hr />
<div>(This is not the exact wording of the question)<br />
==Problem==<br />
<br />
Given regular hexagon ABCDEF with side length 1, define X to be the midpoint of side AB, Y to be the midpoint of side CD, and Z to be the midpoint of side EF, as shown. What is the area of the hexagon formed by the intersection of triangles ACE and XYZ<br />
<br />
choices?<br />
<br />
Answer: 15sqrt(3)/32</div>Giraffefunhttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12B_Problems/Problem_3&diff=914082018 AMC 12B Problems/Problem 32018-02-16T19:53:24Z<p>Giraffefun: </p>
<hr />
<div>==Problem==<br />
<br />
A line with slope 2 intersects a line with slope 6 at the point <math>(40,30)</math>. What is the distance between the <math>x</math>-intercepts of these two lines? <br />
<br />
<math>(\text{A}) 5 \qquad (\text{B}) 10 \qquad (\text{C}) 20 \qquad (\text{D}) 25 \qquad (\text{E}) 50</math><br />
<br />
<br />
==Solution 1==<br />
<br />
<br />
Using the slope-intercept form, we get the equations <math>y-30 = 6(x-40)</math> and <math>y-30 = 2(x-40)</math>. Simplifying, we get <math>6x-y=210</math> and <math>2x-y=50</math>. Letting <math>y=0</math> in both equations and solving for <math>x</math> gives the <math>x</math>-intercepts: <math>x=35</math> and <math>x=25</math>, respectively. Thus the distance between them is <math>35-25 = 10 \Rightarrow \boxed{(\text{B}) 10}<br />
\indent</math> (Giraffefun)<br />
<br />
==See Also==<br />
<br />
{{AMC12 box|year=2018|ab=B|num-b=2|num-a=4}}<br />
{{MAA Notice}}</div>Giraffefunhttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12B_Problems/Problem_5&diff=914062018 AMC 12B Problems/Problem 52018-02-16T19:52:41Z<p>Giraffefun: </p>
<hr />
<div>==Problem==<br />
<br />
How many subsets of <math>\{2,3,4,5,6,7,8,9\}</math> contain at least one prime number?<br />
<math>(\text{A}) \indent 128 \qquad (\text{B}) \indent 192 \qquad (\text{C}) \indent 224 \qquad (\text{D}) \indent 240 \qquad (\text{E}) \indent 256 </math><br />
<br />
==Solution 1 ==<br />
Since an element of a subset is either in or out, the total number of subsets of the 8 element set is <math>2^8 = 256</math>. However, since we are only concerned about the subsets with at least 1 prime in it, we can use complementary counting to count the subsets without a prime and subtract that from the total. Because there are 4 non-primes, there are <math>2^8 -2^4 = 240</math> subsets with at least 1 prime so the answer is <math>\Rightarrow \boxed { (\textbf{D}) 240 }\indent</math> (Giraffefun)<br />
<br />
==See Also==<br />
{{AMC12 box|year=2018|ab=B|num-a=6|num-b=4}}<br />
{{MAA Notice}}</div>Giraffefunhttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12B_Problems/Problem_3&diff=914012018 AMC 12B Problems/Problem 32018-02-16T19:45:12Z<p>Giraffefun: /* Solution 1 */</p>
<hr />
<div>A line with slope 2 intersects a line with slope 6 at the point <math>(40,30)</math>. What is the distance between the <math>x</math>-intercepts of these two lines? <br />
<br />
<math>(\text{A}) 5 \qquad (\text{B}) 10 \qquad (\text{C}) 20 \qquad (\text{D}) 25 \qquad (\text{E}) 50</math><br />
<br />
<br />
==Solution 1==<br />
<br />
<br />
Using the slope-intercept form, we get the equations <math>y-30 = 6(x-40)</math> and <math>y-30 = 2(x-40)</math>. Simplifying, we get <math>6x-y=210</math> and <math>2x-y=50</math>. Letting <math>y=0</math> in both equations and solving for <math>x</math> gives the <math>x</math>-intercepts: <math>x=35</math> and <math>x=25</math>, respectively. Thus the distance between them is <math>35-25 = 10 \Rightarrow \boxed{(\text{B}) 10}<br />
\indent</math> (Giraffefun)<br />
<br />
==See Also==<br />
<br />
{{AMC12 box|year=2018|ab=B|num-b=2|num-a=4}}<br />
{{MAA Notice}}</div>Giraffefunhttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12B_Problems/Problem_5&diff=913982018 AMC 12B Problems/Problem 52018-02-16T19:37:54Z<p>Giraffefun: /* Solution 1 */</p>
<hr />
<div>How many subsets of <math>\{2,3,4,5,6,7,8,9\}</math> contain at least one prime number?<br />
<math>(\text{A}) \indent 128 \qquad (\text{B}) \indent 192 \qquad (\text{C}) \indent 224 \qquad (\text{D}) \indent 240 \qquad (\text{E}) \indent 256 </math><br />
<br />
==Solution 1 ==<br />
Since an element of a subset is either in or out, the total number of subsets of the 8 element set is <math>2^8 = 256</math>. However, since we are only concerned about the subsets with at least 1 prime in it, we can use complementary counting to count the subsets without a prime and subtract that from the total. Because there are 4 non-primes, there are <math>2^8 -2^4 = 240</math> subsets with at least 1 prime so the answer is <math>\Rightarrow \boxed { (\textbf{D}) 240 }\indent</math> (Giraffefun)<br />
<br />
==See Also==<br />
{{AMC12 box|year=2018|ab=B|num-a=6|num-b=4}}<br />
{{MAA Notice}}</div>Giraffefunhttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12B_Problems/Problem_3&diff=913972018 AMC 12B Problems/Problem 32018-02-16T19:34:58Z<p>Giraffefun: /* \Large Solution 1 */</p>
<hr />
<div>A line with slope 2 intersects a line with slope 6 at the point <math>(40,30)</math>. What is the distance between the <math>x</math>-intercepts of these two lines? <br />
<br />
<math>(\text{A}) 5 \qquad (\text{B}) 10 \qquad (\text{C}) 20 \qquad (\text{D}) 25 \qquad (\text{E}) 50</math><br />
<br />
<br />
==Solution 1==<br />
<br />
<br />
Using the slope-intercept form, we get the equations <math>y-30 = 6(x-40)</math> and <math>y-30 = 2(x-40)</math>. Simplifying, we get <math>6x-y=210</math> and <math>2x-y=50</math>. Letting <math>y=0</math> in both equations gives the <math>x</math>-intercepts: <math>x=35</math> and <math>x=25</math>, respectively. Thus the distance between them is <math>35-25 = 10 \Rightarrow (\text{B})<br />
\indent</math> (Giraffefun)<br />
<br />
==See Also==<br />
<br />
{{AMC12 box|year=2018|ab=B|num-b=2|num-a=4}}<br />
{{MAA Notice}}</div>Giraffefunhttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12B_Problems/Problem_5&diff=913962018 AMC 12B Problems/Problem 52018-02-16T19:34:11Z<p>Giraffefun: /* \Large See Also */</p>
<hr />
<div>How many subsets of <math>\{2,3,4,5,6,7,8,9\}</math> contain at least one prime number?<br />
<math>(\text{A}) \indent 128 \qquad (\text{B}) \indent 192 \qquad (\text{C}) \indent 224 \qquad (\text{D}) \indent 240 \qquad (\text{E}) \indent 256 </math><br />
<br />
==Solution 1 ==<br />
Since an element of a subset is either in or out, the total number of subsets of the 8 element set is <math>2^8 = 256</math>. However, since we are only concerned about the subsets with at least 1 prime in it, we can use complementary counting to count the subsets without a prime and subtract that from the total. Because there are 4 non-primes, there are <math>2^8 -2^4 = 240</math> subsets with at least 1 prime so the answer is <math>\Rightarrow (\text{D}) \indent</math> (Giraffefun)<br />
<br />
==See Also==<br />
{{AMC12 box|year=2018|ab=B|num-a=6|num-b=4}}<br />
{{MAA Notice}}</div>Giraffefunhttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12B_Problems/Problem_5&diff=913952018 AMC 12B Problems/Problem 52018-02-16T19:33:34Z<p>Giraffefun: /* \Large See Also */</p>
<hr />
<div>How many subsets of <math>\{2,3,4,5,6,7,8,9\}</math> contain at least one prime number?<br />
<math>(\text{A}) \indent 128 \qquad (\text{B}) \indent 192 \qquad (\text{C}) \indent 224 \qquad (\text{D}) \indent 240 \qquad (\text{E}) \indent 256 </math><br />
<br />
==Solution 1 ==<br />
Since an element of a subset is either in or out, the total number of subsets of the 8 element set is <math>2^8 = 256</math>. However, since we are only concerned about the subsets with at least 1 prime in it, we can use complementary counting to count the subsets without a prime and subtract that from the total. Because there are 4 non-primes, there are <math>2^8 -2^4 = 240</math> subsets with at least 1 prime so the answer is <math>\Rightarrow (\text{D}) \indent</math> (Giraffefun)<br />
<br />
==<math>\Large</math> See Also==<br />
{{AMC12 box|year=2018|ab=B|num-a=6|num-b=4}}<br />
{{MAA Notice}}</div>Giraffefunhttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12B_Problems/Problem_5&diff=913902018 AMC 12B Problems/Problem 52018-02-16T19:26:46Z<p>Giraffefun: /* \Large See Also */</p>
<hr />
<div>How many subsets of <math>\{2,3,4,5,6,7,8,9\}</math> contain at least one prime number?<br />
<math>(\text{A}) \indent 128 \qquad (\text{B}) \indent 192 \qquad (\text{C}) \indent 224 \qquad (\text{D}) \indent 240 \qquad (\text{E}) \indent 256 </math><br />
<br />
==<math>\Large</math> See Also==<br />
{{AMC12 box|year=2018|ab=B|num-a=6|num-b=4}}<br />
{{MAA Notice}}</div>Giraffefunhttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12B_Problems/Problem_5&diff=913892018 AMC 12B Problems/Problem 52018-02-16T19:26:13Z<p>Giraffefun: Created page with "How many subsets of <math>\{2,3,4,5,6,7,8,9\}</math> contain at least one prime number? <math>(\text{A}) \indent 128 \qquad (\text{B}) \indent 192 \qquad (\text{C}) \indent..."</p>
<hr />
<div>How many subsets of <math>\{2,3,4,5,6,7,8,9\}</math> contain at least one prime number?<br />
<math>(\text{A}) \indent 128 \qquad (\text{B}) \indent 192 \qquad (\text{C}) \indent 224 \qquad (\text{D}) \indent 240 \qquad (\text{E}) \indent 256 </math><br />
<br />
==<math>\Large</math> See Also==<br />
{{AMC12 box|year=2015|ab=B|num-a=6|num-b=4}}<br />
{{MAA Notice}}</div>Giraffefunhttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12B_Problems/Problem_3&diff=913832018 AMC 12B Problems/Problem 32018-02-16T19:16:43Z<p>Giraffefun: /* Solution 1 */</p>
<hr />
<div>A line with slope 2 intersects a line with slope 6 at the point <math>(40,30)</math>. What is the distance between the <math>x</math>-intercepts of these two lines? <br />
<br />
<math>(\text{A}) 5 \qquad (\text{B}) 10 \qquad (\text{C}) 20 \qquad (\text{D}) 25 \qquad (\text{E}) 50</math><br />
<br />
<br />
===<math>\Large </math>Solution 1===<br />
<br />
<br />
Using the slope-intercept form, we get the equations <math>y-30 = 6(x-40)</math> and <math>y-30 = 2(x-40)</math>. Simplifying, we get <math>6x-y=210</math> and <math>2x-y=50</math>. Letting <math>y=0</math> in both equations gives the <math>x</math>-intercepts: <math>x=35</math> and <math>x=25</math>, respectively. Thus the distance between them is <math>35-25 = 10 \rightarrow (\text{B})</math><br />
<math>\indent</math> (Giraffefun)<br />
<br />
==See Also==<br />
<br />
{{AMC12 box|year=2018|ab=B|num-b=2|num-a=4}}<br />
{{MAA Notice}}</div>Giraffefunhttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12B_Problems/Problem_3&diff=913772018 AMC 12B Problems/Problem 32018-02-16T19:13:09Z<p>Giraffefun: /* Solution 1 */</p>
<hr />
<div>A line with slope 2 intersects a line with slope 6 at the point <math>(40,30)</math>. What is the distance between the <math>x</math>-intercepts of these two lines? <br />
<br />
<math>(\text{A}) 5 \qquad (\text{B}) 10 \qquad (\text{C}) 20 \qquad (\text{D}) 25 \qquad (\text{E}) 50</math><br />
<br />
<br />
===Solution 1===<br />
<br />
Using the slope-intercept form, we get the equations <math>y-30 = 6(x-40)</math> and <math>y-30 = 2(x-40)</math>. Simplifying, we get <math>6x-y=210</math> and <math>2x-y=50</math>. Letting <math>y=0</math> in both equations gives the <math>x</math>-intercepts: <math>x=35</math> and <math>x=25</math>, respectively. Thus the distance between them is <math>35-25 = 10 \rightarrow (\text{B})</math><br />
(Giraffefun)<br />
<br />
==See Also==<br />
<br />
{{AMC12 box|year=2018|ab=B|num-b=2|num-a=4}}<br />
{{MAA Notice}}</div>Giraffefunhttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12B_Problems/Problem_3&diff=913672018 AMC 12B Problems/Problem 32018-02-16T19:06:30Z<p>Giraffefun: /* Solution 1 */</p>
<hr />
<div>A line with slope 2 intersects a line with slope 6 at the point <math>(40,30)</math>. What is the distance between the <math>x</math>-intercepts of these two lines? <br />
<br />
<math>(\text{A}) 5 \qquad (\text{B}) 10 \qquad (\text{C}) 20 \qquad (\text{D}) 25 \qquad (\text{E}) 50</math><br />
<br />
<br />
===Solution 1===<br />
<br />
Using the slope-intercept form, we get the equations <math>y-30 = 6(x-40)</math> and <math>y-30 = 2(x-40)</math>. Simplifying, we get <math>6x-y=210</math> and <math>2x-y=50</math>. Letting <math>y=0</math> in both equations gives the <math>x</math>-intercepts: <math>x=35</math> and <math>x=25</math>, respectively. Thus the distance between them is <math>35-25 = 10 \rightarrow (\text{B})</math><br />
(Giraffefun)</div>Giraffefunhttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12B_Problems/Problem_3&diff=913662018 AMC 12B Problems/Problem 32018-02-16T19:05:28Z<p>Giraffefun: /* Solution 1 */</p>
<hr />
<div>A line with slope 2 intersects a line with slope 6 at the point <math>(40,30)</math>. What is the distance between the <math>x</math>-intercepts of these two lines? <br />
<br />
<math>(\text{A}) 5 \qquad (\text{B}) 10 \qquad (\text{C}) 20 \qquad (\text{D}) 25 \qquad (\text{E}) 50</math><br />
<br />
<br />
===Solution 1===<br />
<br />
Using the slope-intercept form, we get the equations <math>y-30 = 6(x-40)</math> and <math>y-30 = 2(x-40)</math>. Simplifying, we get <math>6x-y=210</math> and <math>2x-y=50</math>. Letting <math>y=0</math> in both equations gives the <math>x</math>-intercepts: <math>x=35</math> and <math>x=25</math>, respectively. Thus the distance between them is <math>35-25 = 10 \rightarrow (\text{B})</math></div>Giraffefunhttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12B_Problems/Problem_3&diff=913602018 AMC 12B Problems/Problem 32018-02-16T19:03:12Z<p>Giraffefun: </p>
<hr />
<div>A line with slope 2 intersects a line with slope 6 at the point <math>(40,30)</math>. What is the distance between the <math>x</math>-intercepts of these two lines? <br />
<br />
<math>(\text{A}) 5 \qquad (\text{B}) 10 \qquad (\text{C}) 20 \qquad (\text{D}) 25 \qquad (\text{E}) 50</math><br />
<br />
<br />
===Solution 1===<br />
<br />
Using the slope-intercept form, we get the equations <math>y-30 = 6(x-40)</math> and <math>y-30 = 2(x-40)</math>. Simplifying, we get <math>6x-y=210</math> and <math>2x-y=50</math>. Letting <math>y=0</math> in both equations gives the <math>x</math>-intercepts</div>Giraffefunhttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12B_Problems/Problem_3&diff=913522018 AMC 12B Problems/Problem 32018-02-16T18:57:58Z<p>Giraffefun: </p>
<hr />
<div>A line with slope 2 intersects a line with slope 6 at the point <math>(40,30)</math>. What is the distance between the x-intercepts of these two lines? <br />
<br />
<math>(\text{A}) 5 \qquad (\text{B}) 10 \qquad (\text{C}) 20 \qquad (\text{D}) 25 \qquad (\text{E}) 50</math></div>Giraffefunhttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12B_Problems/Problem_3&diff=913472018 AMC 12B Problems/Problem 32018-02-16T18:54:54Z<p>Giraffefun: Created page with "A line with slope 2 intersects a line with slope 6 at the point <math>(40,30)</math>. What is the distance between the x-intercepts of these two lines?"</p>
<hr />
<div>A line with slope 2 intersects a line with slope 6 at the point <math>(40,30)</math>. What is the distance between the x-intercepts of these two lines?</div>Giraffefunhttps://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_5&diff=910902011 AMC 12A Problems/Problem 52018-02-12T03:45:19Z<p>Giraffefun: /* Solution */</p>
<hr />
<div>== Problem ==<br />
Last summer <math>30\%</math> of the birds living on Town Lake were geese, <math>25\%</math> were swans, <math>10\%</math> were herons, and <math>35\%</math> were ducks. What percent of the birds that were not swans were geese?<br />
<br />
<math><br />
\textbf{(A)}\ 20 \qquad<br />
\textbf{(B)}\ 30 \qquad<br />
\textbf{(C)}\ 40 \qquad<br />
\textbf{(D)}\ 50 \qquad<br />
\textbf{(E)}\ 60</math><br />
<br />
== Solution ==<br />
To simplify the problem, WLOG, let us say that there were a total of <math>100</math> birds. The number of birds that are not swans is <math>75</math>. The number of geese is <math>30</math>. Therefore the percentage is just <math>\frac{30}{75} \times 100 = 40 \Rightarrow \boxed{C}</math><br />
<br />
== See also ==<br />
{{AMC12 box|year=2011|num-b=4|num-a=6|ab=A}}<br />
{{MAA Notice}}</div>Giraffefunhttps://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12A_Problems/Problem_2&diff=897792017 AMC 12A Problems/Problem 22018-01-15T09:07:32Z<p>Giraffefun: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
The sum of two nonzero real numbers is 4 times their product. What is the sum of the reciprocals of the two numbers?<br />
<br />
<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 12</math><br />
<br />
==Solution==<br />
<br />
Let <math>x, y</math> be our two numbers. Then <math>x+y = 4xy</math>. Thus, <br />
<br />
<math> \frac{1}{x} + \frac{1}{y} = \frac{x+y}{xy} = 4</math>. <br />
<br />
<math>\boxed{ \textbf{C}}</math>.<br />
<br />
==See Also==<br />
{{AMC10 box|year=2017|ab=A|num-b=4|num-a=6}}<br />
{{AMC12 box|year=2017|ab=A|num-b=1|num-a=3}}<br />
{{MAA Notice}}</div>Giraffefunhttps://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10B_Problems/Problem_15&diff=891852013 AMC 10B Problems/Problem 152017-12-26T01:50:58Z<p>Giraffefun: /* Solution 1 */</p>
<hr />
<div>==Problem==<br />
<br />
A wire is cut into two pieces, one of length <math>a</math> and the other of length <math>b</math>. The piece of length <math>a</math> is bent to form an equilateral triangle, and the piece of length <math>b</math> is bent to form a regular hexagon. The triangle and the hexagon have equal area. What is <math>\frac{a}{b}</math>?<br />
<br />
<math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ \frac{\sqrt{6}}{2}\qquad\textbf{(C)}\ \sqrt{3} \qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \frac{3\sqrt{2}}{2} </math><br />
<br />
==Solution 1==<br />
<br />
Using the area formulas for an equilateral triangle <math>\left(\frac{{s}^{2}\sqrt{3}}{4}\right)</math> and regular hexagon <math>\left(\frac{3{s}^{2}\sqrt{3}}{2}\right)</math>, with side length <math>s</math> and plugging <math>\frac{a}{3}</math> and <math>\frac{b}{6}</math> into each equation, we find that <math>\frac{{a}^{2}\sqrt{3}}{36}=\frac{{b}^{2}\sqrt{3}}{24}</math>. Simplifying this, we get <math>\frac{a}{b}=\boxed{\textbf{(B)} \frac{\sqrt{6}}{2}}</math><br />
<br />
==Solution 2==<br />
The regular hexagon can be broken into 6 small equilateral triangles, each of which is similar to the big equilateral triangle. The big triangle's area is 6 times the area of one of the little triangles. Therefore each side of the big triangle is <math>\sqrt{6}</math> times the side of the small triangle. The desired ratio is <math>\frac{3\sqrt{6}}{6}=\frac{\sqrt{6}}{2}\Rightarrow(B).</math><br />
<br />
== See also ==<br />
{{AMC10 box|year=2013|ab=B|num-b=14|num-a=16}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
[[Category:Area Problems]]<br />
{{MAA Notice}}</div>Giraffefunhttps://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10B_Problems/Problem_15&diff=891842013 AMC 10B Problems/Problem 152017-12-26T01:50:25Z<p>Giraffefun: /* Solution 1 */</p>
<hr />
<div>==Problem==<br />
<br />
A wire is cut into two pieces, one of length <math>a</math> and the other of length <math>b</math>. The piece of length <math>a</math> is bent to form an equilateral triangle, and the piece of length <math>b</math> is bent to form a regular hexagon. The triangle and the hexagon have equal area. What is <math>\frac{a}{b}</math>?<br />
<br />
<math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ \frac{\sqrt{6}}{2}\qquad\textbf{(C)}\ \sqrt{3} \qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \frac{3\sqrt{2}}{2} </math><br />
<br />
==Solution 1==<br />
<br />
Using the area formulas for an equilateral triangle <math>\left(\frac{{s}^{2}\sqrt{3}}{4}\right)</math> and regular hexagon <math>\left(\frac{3{s}^{2}\sqrt{3}}{2}\right)</math>, with side length <cmath>s</cmath> and plugging <math>\frac{a}{3}</math> and <math>\frac{b}{6}</math> into each equation, we find that <math>\frac{{a}^{2}\sqrt{3}}{36}=\frac{{b}^{2}\sqrt{3}}{24}</math>. Simplifying this, we get <math>\frac{a}{b}=\boxed{\textbf{(B)} \frac{\sqrt{6}}{2}}</math><br />
<br />
==Solution 2==<br />
The regular hexagon can be broken into 6 small equilateral triangles, each of which is similar to the big equilateral triangle. The big triangle's area is 6 times the area of one of the little triangles. Therefore each side of the big triangle is <math>\sqrt{6}</math> times the side of the small triangle. The desired ratio is <math>\frac{3\sqrt{6}}{6}=\frac{\sqrt{6}}{2}\Rightarrow(B).</math><br />
<br />
== See also ==<br />
{{AMC10 box|year=2013|ab=B|num-b=14|num-a=16}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
[[Category:Area Problems]]<br />
{{MAA Notice}}</div>Giraffefunhttps://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10B_Problems/Problem_10&diff=891832013 AMC 10B Problems/Problem 102017-12-26T01:44:41Z<p>Giraffefun: /* Solution */</p>
<hr />
<div>==Problem==<br />
A basketball team's players were successful on 50% of their two-point shots and 40% of their three-point shots, which resulted in 54 points. They attempted 50% more two-point shots than three-point shots. How many three-point shots did they attempt?<br />
<br />
<math> \textbf{(A) }10\qquad\textbf{(B) }15\qquad\textbf{(C) }20\qquad\textbf{(D) }25\qquad\textbf{(E) }30 </math><br />
<br />
==Solution==<br />
Let <math>x</math> be the number of two point shots attempted and <math>y</math> the number of three point shots attempted. Because each two point shot is worth two points and the team made 50% and each three point shot is worth 3 points and the team made 40%, <math>0.5(2x)+0.4(3y)=54</math> or <math>x+1.2y=54</math>. Because the team attempted 50% more two point shots then threes, <math>x=1.5y</math>. Substituting <math>1.5y</math> for <math>x</math> in the first equation gives <math>1.5y+1.2y=54</math>, which equals <math>2.7y=54</math> so <math>y=</math> <math>\boxed{\textbf{(C) }20}</math><br />
<br />
== See also ==<br />
{{AMC10 box|year=2013|ab=B|num-b=9|num-a=11}}<br />
{{MAA Notice}}</div>Giraffefunhttps://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10B_Problems/Problem_7&diff=891822013 AMC 10B Problems/Problem 72017-12-26T01:43:47Z<p>Giraffefun: /* Solution */</p>
<hr />
<div>==Problem==<br />
Six points are equally spaced around a circle of radius 1. Three of these points are the vertices of a triangle that is neither equilateral nor isosceles. What is the area of this triangle?<br />
<br />
<math> \textbf{(A)}\ \frac{\sqrt{3}}{3}\qquad\textbf{(B)}\ \frac{\sqrt{3}}{2}\qquad\textbf{(C)}\ \textbf{1}\qquad\textbf{(D)}\ \sqrt{2}\qquad\textbf{(E)}\ \text{2} </math><br />
[[Category: Introductory Geometry Problems]]<br />
<br />
==Solution==<br />
If there are no two points on the circle that are adjacent, then the triangle would be equilateral. If the three points are all adjacent, it would be isosceles. Thus, the only possibility is two adjacent points and one point two away. Because one of the sides of this triangle is the diameter, the opposite angle is a right angle. Also, because the two adjacent angles are one sixth of the circle apart, the angle opposite them is thirty degrees. This is a <math>30-60-90</math> triangle. <br />
If the original six points are connected, a regular hexagon is created. This hexagon consists of six equilateral triangles, so the radius is equal to one of its side lengths. The radius is <math>1</math>, so the side opposite the thirty degree angle in the triangle is also <math>1</math>. From the properties of <math>30-60-90</math> triangles, the area is <math>1\cdot\sqrt3/2</math>=<math>\boxed{\textbf{(B) } \frac{\sqrt3}{2}}</math><br />
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== See also ==<br />
{{AMC10 box|year=2013|ab=B|num-b=6|num-a=8}}<br />
{{MAA Notice}}</div>Giraffefunhttps://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12B_Problems/Problem_4&diff=891812013 AMC 12B Problems/Problem 42017-12-26T01:43:02Z<p>Giraffefun: /* Solution */</p>
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<div>{{duplicate|[[2013 AMC 12B Problems|2013 AMC 12B #4]] and [[2013 AMC 10B Problems|2013 AMC 10B #8]]}}<br />
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==Problem==<br />
Ray's car averages <math>40</math> miles per gallon of gasoline, and Tom's car averages <math>10</math> miles per gallon of gasoline. Ray and Tom each drive the same number of miles. What is the cars' combined rate of miles per gallon of gasoline?<br \><br />
<math>\textbf{(A)}\ 10 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 30 \qquad \textbf{(E)}\ 40</math><br />
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==Solution==<br />
Suppose both Ray and Tom drive 40 miles. Ray's car would require <math>\frac{40}{40}=1</math> gallon of gas and Tom's car would require <math>\frac{40}{10}=4</math> gallons of gas. They would have driven a total of <math>40+40=80</math> miles, on <math>1+4=5</math> gallons of gas, for a combined rate of <math>\frac{80}{5}=</math> <math>\boxed{\textbf{(B) }16}</math><br />
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== See also ==<br />
{{AMC10 box|year=2013|ab=B|num-b=7|num-a=9}}<br />
{{AMC12 box|year=2013|ab=B|num-b=3|num-a=5}}<br />
{{MAA Notice}}</div>Giraffefun