https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Krishkhushi09&feedformat=atom AoPS Wiki - User contributions [en] 2022-01-17T20:41:53Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10B_Problems/Problem_11&diff=141168 2017 AMC 10B Problems/Problem 11 2020-12-31T18:51:14Z <p>Krishkhushi09: /* Solution */</p> <hr /> <div>==Problem==<br /> At Typico High School, &lt;math&gt;60\%&lt;/math&gt; of the students like dancing, and the rest dislike it. Of those who like dancing, &lt;math&gt;80\%&lt;/math&gt; say that they like it, and the rest say that they dislike it. Of those who dislike dancing, &lt;math&gt;90\%&lt;/math&gt; say that they dislike it, and the rest say that they like it. What fraction of students who say they dislike dancing actually like it?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 10\%\qquad\textbf{(B)}\ 12\%\qquad\textbf{(C)}\ 20\%\qquad\textbf{(D)}\ 25\%\qquad\textbf{(E)}\ 33\frac{1}{3}\%&lt;/math&gt;<br /> ==Solution==<br /> &lt;math&gt;60\% \cdot 20\% = 12\%&lt;/math&gt; of the people that claim that they like dancing say they dislike it, and &lt;math&gt;40\% \cdot 90\% = 36\%&lt;/math&gt; of the people that claim that they dislike dancing actually dislike it. Therefore, the answer is &lt;math&gt;\frac{12\%}{12\%+36\%} = \boxed{\textbf{(D) } 25\%}&lt;/math&gt;.<br /> <br /> ==Video Solution==<br /> https://youtu.be/93lThricxLE<br /> <br /> ~savannahsolver<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=B|num-b=10|num-a=12}}<br /> {{MAA Notice}}</div> Krishkhushi09 https://artofproblemsolving.com/wiki/index.php?title=1988_AIME_Problems/Problem_13&diff=126219 1988 AIME Problems/Problem 13 2020-06-22T18:29:03Z <p>Krishkhushi09: Another more easier solution for beginners to AIME problems, utilizing the fibonacci pattern.</p> <hr /> <div>== Problem ==<br /> Find &lt;math&gt;a&lt;/math&gt; if &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are [[integer]]s such that &lt;math&gt;x^2 - x - 1&lt;/math&gt; is a factor of &lt;math&gt;ax^{17} + bx^{16} + 1&lt;/math&gt;.<br /> <br /> __TOC__<br /> === Solution 1 (Not rigorous)===<br /> Let's work backwards! Let &lt;math&gt;F(x) = ax^{17} + bx^{16} + 1&lt;/math&gt; and let &lt;math&gt;P(x)&lt;/math&gt; be the [[polynomial]] such that &lt;math&gt;P(x)(x^2 - x - 1) = F(x)&lt;/math&gt;.<br /> <br /> Clearly, the [[constant]] term of &lt;math&gt;P(x)&lt;/math&gt; must be &lt;math&gt;- 1&lt;/math&gt;. Now, we have &lt;math&gt;(x^2 - x - 1)(c_1x^{15} + c_2x^{14} + \cdots + c_{15}x - 1)&lt;/math&gt;, where &lt;math&gt;c_{i}&lt;/math&gt; is some [[coefficient]]. However, since &lt;math&gt;F(x)&lt;/math&gt; has no &lt;math&gt;x&lt;/math&gt; term, it must be true that &lt;math&gt;c_{15} = 1&lt;/math&gt;. <br /> <br /> Let's find &lt;math&gt;c_{14}&lt;/math&gt; now. Notice that all we care about in finding &lt;math&gt;c_{14}&lt;/math&gt; is that &lt;math&gt;(x^2 - x - 1)(\cdots + c_{14}x^2 + x - 1) = \text{something} + 0x^2 + \text{something}&lt;/math&gt;. Therefore, &lt;math&gt;c_{14} = - 2&lt;/math&gt;. Undergoing a similar process, &lt;math&gt;c_{13} = 3&lt;/math&gt;, &lt;math&gt;c_{12} = - 5&lt;/math&gt;, &lt;math&gt;c_{11} = 8&lt;/math&gt;, and we see a nice pattern. The coefficients of &lt;math&gt;P(x)&lt;/math&gt; are just the [[Fibonacci sequence]] with alternating signs! Therefore, &lt;math&gt;a = c_1 = F_{16}&lt;/math&gt;, where &lt;math&gt;F_{16}&lt;/math&gt; denotes the 16th Fibonnaci number and &lt;math&gt;a = \boxed{987}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> Let &lt;math&gt;F_n&lt;/math&gt; represent the &lt;math&gt;n&lt;/math&gt;th number in the Fibonacci sequence. Therefore,<br /> <br /> &lt;math&gt;x^2 - x - 1 = 0\Longrightarrow x^n = F_n(x),\ n\in N\Longrightarrow x^{n + 2} = F_{n + 1}\cdot x + F_n,\ n\in N\ .&lt;/math&gt;<br /> <br /> The above uses the similarity between the Fibonacci [[recursion|recursive]] definition, &lt;math&gt;F_{n+2} - F_{n+1} - F_n = 0&lt;/math&gt;, and the polynomial &lt;math&gt;x^2 - x - 1 = 0&lt;/math&gt;. <br /> <br /> &lt;math&gt;0 = ax^{17} + bx^{16} + 1 = a(F_{17}\cdot x + F_{16}) + b(F_{16}\cdot x + F_{15}) + 1\Longrightarrow&lt;/math&gt;<br /> <br /> &lt;math&gt;(aF_{17} + bF_{16})\cdot x + (aF_{16} + bF_{15} + 1) = 0,\ x\not\in Q\Longrightarrow&lt;/math&gt;<br /> <br /> &lt;math&gt;aF_{17} + bF_{16} = 0&lt;/math&gt; and &lt;math&gt;aF_{16} + bF_{15} + 1 = 0\Longrightarrow&lt;/math&gt;<br /> <br /> &lt;math&gt;a = F_{16},\ b = - F_{17}\Longrightarrow a = \boxed {987}\ .&lt;/math&gt;<br /> <br /> === Solution 3 ===<br /> We can long divide and search for a pattern; then the remainder would be set to zero to solve for &lt;math&gt;a&lt;/math&gt;. Writing out a few examples quickly shows us that the remainders after each subtraction follow the Fibonacci sequence. Carrying out this pattern, we find that the remainder is &lt;math&gt;(F_{16}b + F_{17}a)x + F_{15}b + F_{16}a + 1 = 0&lt;/math&gt;. Since the coefficient of &lt;math&gt;x&lt;/math&gt; must be zero, this gives us two equations, &lt;math&gt;F_{16}b + F_{17}a = 0&lt;/math&gt; and &lt;math&gt;F_{15}b + F_{16}a + 1 = 0&lt;/math&gt;. Solving these two as above, we get that &lt;math&gt;a = \boxed{987}&lt;/math&gt;.<br /> <br /> There are various similar solutions which yield the same pattern, such as repeated substitution of &lt;math&gt;x^2 = x + 1&lt;/math&gt; into the larger polynomial.<br /> <br /> === Solution 4 ===<br /> The roots of &lt;math&gt;x^2-x-1&lt;/math&gt; are &lt;math&gt;\phi&lt;/math&gt; (the [[Golden Ratio]]) and &lt;math&gt;1-\phi&lt;/math&gt;. These two must also be roots of &lt;math&gt;ax^{17}+bx^{16}+1&lt;/math&gt;. Thus, we have two equations: &lt;math&gt;a\phi^{17}+b\phi^{16}+1=0&lt;/math&gt; and &lt;math&gt;a(1-\phi)^{17}+b(1-\phi)^{16}+1=0&lt;/math&gt;. Subtract these two and divide by &lt;math&gt;\sqrt{5}&lt;/math&gt; to get &lt;math&gt;\frac{a(\phi^{17}-(1-\phi)^{17})}{\sqrt{5}}+\frac{b(\phi^{16}-(1-\phi)^{16})}{\sqrt{5}}=0&lt;/math&gt;. Noting that the formula for the &lt;math&gt;n&lt;/math&gt;th [[Fibonacci number]] is &lt;math&gt;\frac{\phi^n-(1-\phi)^n}{\sqrt{5}}&lt;/math&gt;, we have &lt;math&gt;1597a+987b=0&lt;/math&gt;. Since &lt;math&gt;1597&lt;/math&gt; and &lt;math&gt;987&lt;/math&gt; are coprime, the solutions to this equation under the integers are of the form &lt;math&gt;a=987k&lt;/math&gt; and &lt;math&gt;b=-1597k&lt;/math&gt;, of which the only integral solutions for &lt;math&gt;a&lt;/math&gt; on &lt;math&gt;[0,999]&lt;/math&gt; are &lt;math&gt;0&lt;/math&gt; and &lt;math&gt;987&lt;/math&gt;. &lt;math&gt;(a,b)=(0,0)&lt;/math&gt; cannot work since &lt;math&gt;x^2-x-1&lt;/math&gt; does not divide &lt;math&gt;1&lt;/math&gt;, so the answer must be &lt;math&gt;\boxed{987}&lt;/math&gt;. (Note that this solution would not be valid on an Olympiad or any test that does not restrict answers as integers between &lt;math&gt;000&lt;/math&gt; and &lt;math&gt;999&lt;/math&gt;).<br /> <br /> === Solution 5: For Beginners (less technical) ===<br /> Trying to divide &lt;math&gt;ax^17 + bx^16 + 1&lt;/math&gt; by &lt;math&gt;x^2-x-1&lt;/math&gt; would be very tough, so let's try to divide using smaller degrees of x. Doing &lt;math&gt;\frac{ax^3+bx^2+1}{x^2-x-1}&lt;/math&gt;, we get the following systems of equations:<br /> \begin{align*}<br /> a+b &amp; = -1\\2a+b &amp; = 0<br /> \end{align*}<br /> Continuing with &lt;math&gt;\frac{ax^4+bx^3+1}{x^2-x-1}&lt;/math&gt;:<br /> \begin{align*}<br /> 2a+b &amp; = -1\\3a+2b &amp; = 0<br /> \end{align*}<br /> There is somewhat of a pattern showing up, so let's try &lt;math&gt;\frac{ax^5+bx^4+1}{x^2-x-1}&lt;/math&gt; to verify. We get:<br /> \begin{align*}<br /> 3a+2b &amp; = -1\\5a+3b &amp; = 0<br /> \end{align*}<br /> Now we begin to see that our pattern is actually the Fibonacci Number's! Using the previous equations, we can make a general statement about &lt;math&gt;\frac{ax^n+bx^{n-1}+1}{x^2-x-1}&lt;/math&gt;<br /> \begin{align*}<br /> af_{n-1}+bf_{n-2} &amp; = -1\\af_n+bf_{n-1} &amp; = 0<br /> \end{align*}<br /> Also, noticing our solutions from the previous systems of equations, we can create the following statement:<br /> <br /> [b]If &lt;math&gt;{ax^n+bx^{n-1}+1}&lt;/math&gt; has &lt;math&gt;x^2-x-1&lt;/math&gt; as a factor, then &lt;math&gt;a=f_{n-1}&lt;/math&gt; and &lt;math&gt;b = f_n&lt;/math&gt;[/b]<br /> <br /> Thus, if &lt;math&gt;{ax^17+bx^16+1}&lt;/math&gt; has &lt;math&gt;{x^2-x-1}&lt;/math&gt; as a factor, we get that a = 987 and b = 1597, so a = &lt;math&gt;\boxed {987}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=1988|num-b=12|num-a=14}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Krishkhushi09 https://artofproblemsolving.com/wiki/index.php?title=1999_AIME_Problems/Problem_3&diff=121543 1999 AIME Problems/Problem 3 2020-04-22T23:36:12Z <p>Krishkhushi09: </p> <hr /> <div>== Problem ==<br /> Find the sum of all [[positive integer]]s &lt;math&gt;n&lt;/math&gt; for which &lt;math&gt;n^2-19n+99&lt;/math&gt; is a [[perfect square]].<br /> <br /> <br /> == Solution 1==<br /> If &lt;math&gt;n^2-19n+99=x^2&lt;/math&gt; for some positive integer &lt;math&gt;x&lt;/math&gt;, then rearranging we get &lt;math&gt;n^2-19n+99-x^2=0&lt;/math&gt;. Now from the quadratic formula, <br /> <br /> :&lt;math&gt;n=\frac{19\pm \sqrt{4x^2-35}}{2}&lt;/math&gt;<br /> <br /> Because &lt;math&gt;n&lt;/math&gt; is an integer, this means &lt;math&gt;4x^2-35=q^2&lt;/math&gt; for some nonnegative integer &lt;math&gt;q&lt;/math&gt;. Rearranging gives &lt;math&gt;(2x+q)(2x-q)=35&lt;/math&gt;. Thus &lt;math&gt;(2x+q, 2x-q)=(35, 1)&lt;/math&gt; or &lt;math&gt;(7,5)&lt;/math&gt;, giving &lt;math&gt;x=3&lt;/math&gt; or &lt;math&gt;9&lt;/math&gt;. This gives &lt;math&gt;n=1, 9, 10,&lt;/math&gt; or &lt;math&gt;18&lt;/math&gt;, and the sum is &lt;math&gt;1+9+10+18=\boxed{38}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> Suppose there is some &lt;math&gt;k&lt;/math&gt; such that &lt;math&gt;x^2 - 19x + 99 = k^2&lt;/math&gt;. Completing the square, we have that &lt;math&gt;(x - 19/2)^2 + 99 - (19/2)^2 = k^2&lt;/math&gt;, that is, &lt;math&gt;(x - 19/2)^2 + 35/4 = k^2&lt;/math&gt;. Multiplying both sides by 4 and rearranging, we see that &lt;math&gt;(2k)^2 - (2x - 19)^2 = 35&lt;/math&gt;. Thus, &lt;math&gt;(2k - 2x + 19)(2k + 2x - 19) = 35&lt;/math&gt;. We then proceed as we did in the previous solution.<br /> <br /> <br /> ==Solution 3==<br /> <br /> When &lt;math&gt; n \geq 12 &lt;/math&gt;, we have <br /> &lt;cmath&gt; (n-10)^2 &lt; n^2 -19n + 99 &lt; (n-8)^2. &lt;/cmath&gt;<br /> <br /> So if &lt;math&gt; n \geq 12&lt;/math&gt; and &lt;math&gt; n^2 -19n + 99 &lt;/math&gt; is a perfect square, then <br /> &lt;cmath&gt; n^2 -19n + 99 = (n-9)^2 &lt;/cmath&gt;<br /> <br /> or &lt;math&gt; n = 18 &lt;/math&gt;.<br /> <br /> For &lt;math&gt; 1 \leq n &lt; 12 &lt;/math&gt;, it is easy to check that &lt;math&gt; n^2 -19n + 99 &lt;/math&gt; is a perfect square when &lt;math&gt; n = 1, 9 &lt;/math&gt; and &lt;math&gt; 10 &lt;/math&gt; ( using the identity &lt;math&gt; n^2 -19n + 99 = (n-10)^2 + n - 1.) &lt;/math&gt;<br /> <br /> We conclude that the answer is &lt;math&gt;1 + 9 + 10 + 18 = \boxed{38}.&lt;/math&gt;<br /> <br /> ==Solution 4: Graphing==<br /> <br /> If we graphed &lt;cmath&gt; y = \sqrt{x^2-19x+99} &lt;/cmath&gt; we would see that only four values of x return integer values of y: 10, 9, 1, 18. Thus, the answer if &lt;math&gt;\boxed{38}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=1999|num-b=2|num-a=4}}<br /> [[Category:Intermediate Number Theory Problems]]<br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Krishkhushi09 https://artofproblemsolving.com/wiki/index.php?title=1992_AIME_Problems/Problem_5&diff=121542 1992 AIME Problems/Problem 5 2020-04-22T23:27:44Z <p>Krishkhushi09: Another solution</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;S^{}_{}&lt;/math&gt; be the set of all rational numbers &lt;math&gt;r^{}_{}&lt;/math&gt;, &lt;math&gt;0^{}_{}&lt;r&lt;1&lt;/math&gt;, that have a repeating decimal expansion in the form &lt;math&gt;0.abcabcabc\ldots=0.\overline{abc}&lt;/math&gt;, where the digits &lt;math&gt;a^{}_{}&lt;/math&gt;, &lt;math&gt;b^{}_{}&lt;/math&gt;, and &lt;math&gt;c^{}_{}&lt;/math&gt; are not necessarily distinct. To write the elements of &lt;math&gt;S^{}_{}&lt;/math&gt; as fractions in lowest terms, how many different numerators are required?<br /> <br /> == Solution ==<br /> We consider the method in which repeating decimals are normally converted to fractions with an example:<br /> <br /> &lt;math&gt;x=0.\overline{176}&lt;/math&gt;<br /> <br /> &lt;math&gt;\Rightarrow 1000x=176.\overline{176}&lt;/math&gt;<br /> <br /> &lt;math&gt;\Rightarrow 999x=1000x-x=176&lt;/math&gt;<br /> <br /> &lt;math&gt;\Rightarrow x=\frac{176}{999}&lt;/math&gt;<br /> <br /> Thus, let &lt;math&gt;x=0.\overline{abc}&lt;/math&gt;<br /> <br /> &lt;math&gt;\Rightarrow 1000x=abc.\overline{abc}&lt;/math&gt;<br /> <br /> &lt;math&gt;\Rightarrow 999x=1000x-x=abc&lt;/math&gt;<br /> <br /> &lt;math&gt;\Rightarrow x=\frac{abc}{999}&lt;/math&gt;<br /> <br /> If &lt;math&gt;abc&lt;/math&gt; is not divisible by &lt;math&gt;3&lt;/math&gt; or &lt;math&gt;37&lt;/math&gt;, then this is in lowest terms. Let us consider the other multiples: &lt;math&gt;333&lt;/math&gt; multiples of &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;27&lt;/math&gt; of &lt;math&gt;37&lt;/math&gt;, and &lt;math&gt;9&lt;/math&gt; of &lt;math&gt;3&lt;/math&gt; and &lt;math&gt;37&lt;/math&gt;, so &lt;math&gt;999-333-27+9 = 648&lt;/math&gt;, which is the amount that are neither. The &lt;math&gt;12&lt;/math&gt; numbers that are multiples of &lt;math&gt;81&lt;/math&gt; reduce to multiples of &lt;math&gt;3&lt;/math&gt;. We have to count these since it will reduce to a multiple of &lt;math&gt;3&lt;/math&gt; which we have removed from &lt;math&gt;999&lt;/math&gt;, but, this cannot be removed since the numerator cannot cancel the &lt;math&gt;3&lt;/math&gt;.There aren't any numbers which are multiples of &lt;math&gt;37^2&lt;/math&gt;, so we can't get numerators which are multiples of &lt;math&gt;37&lt;/math&gt;. Therefore &lt;math&gt;648 + 12 = \boxed{660}&lt;/math&gt;.<br /> <br /> == Solution 2==<br /> We can see that the denominator of the fraction will be 999b, where b is either 1, 1/3, 1/111, 1/9, or 1/333. This means that the numerator stays abc unless abc divides 999. Finding the numbers relatively prime to 999 gives us the numerators when the fraction cannot be simplified past 999.<br /> <br /> &lt;math&gt;\Rightarrow 999(1 - \frac{1}{3})(1- \frac{1}{111}) = 660&lt;/math&gt;<br /> <br /> Now, look at a number that does divide 999:<br /> <br /> &lt;math&gt;\Rightarrow 555/999 = 5/9&lt;/math&gt;<br /> <br /> We can see that if a number does divide 999, it will be simplified into a number which is relatively prime to 999 (since that is what simplifying is). There are 660 numbers less than or relatively prime to 999, so the answer is &lt;math&gt;\boxed{660}&lt;/math&gt;.<br /> - krishkhushi09<br /> <br /> {{AIME box|year=1992|num-b=4|num-a=6}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Krishkhushi09 https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_I_Problems/Problem_1&diff=121105 2013 AIME I Problems/Problem 1 2020-04-18T17:00:36Z <p>Krishkhushi09: /* Solution */</p> <hr /> <div>== Problem 1 ==<br /> The AIME Triathlon consists of a half-mile swim, a 30-mile bicycle ride, and an eight-mile run. Tom swims, bicycles, and runs at constant rates. He runs fives times as fast as he swims, and he bicycles twice as fast as he runs. Tom completes the AIME Triathlon in four and a quarter hours. How many minutes does he spend bicycling?<br /> <br /> == Solution ==<br /> <br /> Let &lt;math&gt;r&lt;/math&gt; represent the rate Tom swims in miles per minute. Then we have<br /> <br /> &lt;math&gt;\frac{1/2}{r} + \frac{8}{5r} + \frac{30}{10r} = 255&lt;/math&gt;<br /> <br /> Solving for &lt;math&gt;r&lt;/math&gt;, we find &lt;math&gt;r = 1/50&lt;/math&gt;, so the time Tom spends biking is &lt;math&gt;\frac{30}{(10)(1/50)} = \boxed{150}&lt;/math&gt; minutes.<br /> <br /> == See also ==<br /> {{AIME box|year=2013|n=I|before=First Problem|num-a=2}}<br /> {{MAA Notice}}</div> Krishkhushi09 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_24&diff=119990 2018 AMC 10A Problems/Problem 24 2020-03-23T16:50:59Z <p>Krishkhushi09: This is a new solution using similar areas and giving areas different variables and solving.</p> <hr /> <div>== Problem ==<br /> <br /> Triangle &lt;math&gt;ABC&lt;/math&gt; with &lt;math&gt;AB=50&lt;/math&gt; and &lt;math&gt;AC=10&lt;/math&gt; has area &lt;math&gt;120&lt;/math&gt;. Let &lt;math&gt;D&lt;/math&gt; be the midpoint of &lt;math&gt;\overline{AB}&lt;/math&gt;, and let &lt;math&gt;E&lt;/math&gt; be the midpoint of &lt;math&gt;\overline{AC}&lt;/math&gt;. The angle bisector of &lt;math&gt;\angle BAC&lt;/math&gt; intersects &lt;math&gt;\overline{DE}&lt;/math&gt; and &lt;math&gt;\overline{BC}&lt;/math&gt; at &lt;math&gt;F&lt;/math&gt; and &lt;math&gt;G&lt;/math&gt;, respectively. What is the area of quadrilateral &lt;math&gt;FDBG&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) }60 \qquad<br /> \textbf{(B) }65 \qquad<br /> \textbf{(C) }70 \qquad<br /> \textbf{(D) }75 \qquad<br /> \textbf{(E) }80 \qquad<br /> &lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> <br /> Let &lt;math&gt;BC = a&lt;/math&gt;, &lt;math&gt;BG = x&lt;/math&gt;, &lt;math&gt;GC = y&lt;/math&gt;, and the length of the perpendicular to &lt;math&gt;BC&lt;/math&gt; through &lt;math&gt;A&lt;/math&gt; be &lt;math&gt;h&lt;/math&gt;. By angle bisector theorem, we have that &lt;cmath&gt;\frac{50}{x} = \frac{10}{y},&lt;/cmath&gt; where &lt;math&gt;y = -x+a&lt;/math&gt;. Therefore substituting we have that &lt;math&gt;BG=\frac{5a}{6}&lt;/math&gt;. By similar triangles, we have that &lt;math&gt;DF=\frac{5a}{12}&lt;/math&gt;, and the height of this trapezoid is &lt;math&gt;\frac{h}{2}&lt;/math&gt;. Then, we have that &lt;math&gt;\frac{ah}{2}=120&lt;/math&gt;. We wish to compute &lt;math&gt;\frac{5a}{8}\cdot\frac{h}{2}&lt;/math&gt;, and we have that it is &lt;math&gt;\boxed{75}&lt;/math&gt; by substituting.<br /> <br /> == Solution 2 ==<br /> <br /> For this problem, we have &lt;math&gt;\triangle{ADE}\sim\triangle{ABC}&lt;/math&gt; because of SAS and &lt;math&gt;DE = \frac{BC}{2}&lt;/math&gt;. Therefore, &lt;math&gt;\bigtriangleup ADE&lt;/math&gt; is a quarter of the area of &lt;math&gt;\bigtriangleup ABC&lt;/math&gt;, which is &lt;math&gt;30&lt;/math&gt;. Subsequently, we can compute the area of quadrilateral &lt;math&gt;BDEC&lt;/math&gt; to be &lt;math&gt;120 - 30 = 90&lt;/math&gt;. Using the angle bisector theorem in the same fashion as the previous problem, we get that &lt;math&gt;\overline{BG}&lt;/math&gt; is &lt;math&gt;5&lt;/math&gt; times the length of &lt;math&gt;\overline{GC}&lt;/math&gt;. We want the larger piece, as described by the problem. Because the heights are identical, one area is &lt;math&gt;5&lt;/math&gt; times the other, and &lt;math&gt;\frac{5}{6} \cdot 90 = \boxed{75}&lt;/math&gt;.<br /> <br /> == Solution 3 ==<br /> The area of &lt;math&gt;\bigtriangleup ABG&lt;/math&gt; to the area of &lt;math&gt;\bigtriangleup ACG&lt;/math&gt; is &lt;math&gt;5:1&lt;/math&gt; by Law of Sines. So the area of &lt;math&gt;\bigtriangleup ABG&lt;/math&gt; is &lt;math&gt;100&lt;/math&gt;. Since &lt;math&gt;\overline{DE}&lt;/math&gt; is the midsegment of &lt;math&gt;\bigtriangleup ABC&lt;/math&gt;, so &lt;math&gt;\overline{DF}&lt;/math&gt; is the midsegment of &lt;math&gt;\bigtriangleup ABG&lt;/math&gt; . So the area of &lt;math&gt;\bigtriangleup ADF&lt;/math&gt; to the area of &lt;math&gt;\bigtriangleup ABG&lt;/math&gt; is &lt;math&gt;1:4&lt;/math&gt; , so the area of &lt;math&gt;\bigtriangleup ACG&lt;/math&gt; is &lt;math&gt;25&lt;/math&gt;, by similar triangles. Therefore the area of quad &lt;math&gt;FDBG&lt;/math&gt; is &lt;math&gt;100-25=\boxed{75}&lt;/math&gt;<br /> <br /> ==Solution 4 ==<br /> The area of quadrilateral &lt;math&gt;FDBG&lt;/math&gt; is the area of &lt;math&gt;\bigtriangleup ABG&lt;/math&gt; minus the area of &lt;math&gt;\bigtriangleup ADF&lt;/math&gt;. Notice, &lt;math&gt;\overline{DE} || \overline{BC}&lt;/math&gt;, so &lt;math&gt;\bigtriangleup ABG \sim \bigtriangleup ADF&lt;/math&gt;, and since &lt;math&gt;\overline{AD}:\overline{AB}=1:2&lt;/math&gt;, the area of &lt;math&gt;\bigtriangleup ADF:\bigtriangleup ABG=(1:2)^2=1:4&lt;/math&gt;. Given that the area of &lt;math&gt;\bigtriangleup ABC&lt;/math&gt; is &lt;math&gt;120&lt;/math&gt;, using &lt;math&gt;\frac{bh}{2}&lt;/math&gt; on side &lt;math&gt;AB&lt;/math&gt; yields &lt;math&gt;\frac{50h}{2}=120\implies h=\frac{240}{50}=\frac{24}{5}&lt;/math&gt;. Using the Angle Bisector Theorem, &lt;math&gt;\overline{BG}:\overline{BC}=50:(10+50)=5:6&lt;/math&gt;, so the height of &lt;math&gt;\bigtriangleup ABG: \bigtriangleup ACB=5:6&lt;/math&gt;. Therefore our answer is &lt;math&gt;\big[ FDBG\big] = \big[ABG\big]-\big[ ADF\big] = \big[ ABG\big]\big(1-\frac{1}{4}\big)=\frac{3}{4}\cdot \frac{bh}{2}=\frac{3}{8}\cdot 50\cdot \frac{5}{6}\cdot \frac{24}{5}=\frac{3}{8}\cdot 200=\boxed{75}&lt;/math&gt;<br /> <br /> ==Solution 5: Trig ==<br /> We try to find the area of quadrilateral &lt;math&gt;FDBG&lt;/math&gt; by subtracting the area outside the quadrilateral but inside triangle &lt;math&gt;ABC&lt;/math&gt;. Note that the area of &lt;math&gt;\triangle ADE&lt;/math&gt; is equal to &lt;math&gt;\frac{1}{2} \cdot 25 \cdot 5 \cdot \sin{A}&lt;/math&gt; and the area of triangle &lt;math&gt;ABC&lt;/math&gt; is equal to &lt;math&gt;\frac{1}{2} \cdot 50 \cdot 10 \cdot \sin A&lt;/math&gt;. The ratio &lt;math&gt;\frac{[ADE]}{[ABC]}&lt;/math&gt; is thus equal to &lt;math&gt;\frac{1}{4}&lt;/math&gt; and the area of triangle &lt;math&gt;ADE&lt;/math&gt; is &lt;math&gt;\frac{1}{4} \cdot 120 = 30&lt;/math&gt;. Let side &lt;math&gt;BC&lt;/math&gt; be equal to &lt;math&gt;6x&lt;/math&gt;, then &lt;math&gt;BG = 5x, GC = x&lt;/math&gt; by the angle bisector theorem. Similarly, we find the area of triangle &lt;math&gt;AGC&lt;/math&gt; to be &lt;math&gt;\frac{1}{2} \cdot 10 \cdot x \cdot \sin C&lt;/math&gt; and the area of triangle &lt;math&gt;ABC&lt;/math&gt; to be &lt;math&gt;\frac{1}{2} \cdot 6x \cdot 10 \cdot \sin C&lt;/math&gt;. A ratio between these two triangles yields &lt;math&gt;\frac{[ACG]}{[ABC]} = \frac{x}{6x} = \frac{1}{6}&lt;/math&gt;, so &lt;math&gt;[AGC] = 20&lt;/math&gt;. Now we just need to find the area of triangle &lt;math&gt;AFE&lt;/math&gt; and subtract it from the combined areas of &lt;math&gt;[ADE]&lt;/math&gt; and &lt;math&gt;[ACG]&lt;/math&gt;, since we count it twice. Note that the angle bisector theorem also applies for &lt;math&gt;\triangle ADE&lt;/math&gt; and &lt;math&gt;\frac{AE}{AD} = \frac{1}{5}&lt;/math&gt;, so thus &lt;math&gt;\frac{EF}{ED} = \frac{1}{6}&lt;/math&gt; and we find &lt;math&gt;[AFE] = \frac{1}{6} \cdot 30 = 5&lt;/math&gt;, and the area outside &lt;math&gt;FDBG&lt;/math&gt; must be &lt;math&gt; [ADE] + [AGC] - [AFE] = 30 + 20 - 5 = 45&lt;/math&gt;, and we finally find &lt;math&gt;[FDBG] = [ABC] - 45 = 120 -45 = \boxed{75}&lt;/math&gt;, and we are done. <br /> <br /> ==Solution 6: Areas ==<br /> &lt;asy&gt;<br /> draw((0,0)--(1,3)--(5,0)--cycle);<br /> draw((0,0)--(2,2.25));<br /> draw((0.5,1.5)--(2.5,0));<br /> label(&quot;A&quot;,(0,0),SW);<br /> label(&quot;B&quot;,(5,0),SE);<br /> label(&quot;C&quot;,(1,3),N);<br /> label(&quot;G&quot;,(2,2.25),NE);<br /> label(&quot;D&quot;,(2.5,0),S);<br /> label(&quot;E&quot;,(0.5,1.5),NW);<br /> label(&quot;3Y&quot;,(2.5,0.75),N);<br /> label(&quot;Y&quot;,(1,0.2),N);<br /> label(&quot;X&quot;,(0.5,0.5),N);<br /> label(&quot;3X&quot;,(1.25,1.75),N);<br /> &lt;/asy&gt;<br /> Give triangle &lt;math&gt;AEF&lt;/math&gt; area X. Then, by similarity, since &lt;math&gt;\frac{AC}{AE} = \frac{2}{1}&lt;/math&gt;, &lt;math&gt;ACG&lt;/math&gt; has area 4X. Thus, &lt;math&gt;FGCE&lt;/math&gt; has area 3X.<br /> Doing the same for triangle &lt;math&gt;AGB&lt;/math&gt;, we get that triangle &lt;math&gt;AFD&lt;/math&gt; has area Y and quadrilateral &lt;math&gt;GFDB&lt;/math&gt; has area 3Y. Since &lt;math&gt;AEF&lt;/math&gt; has the same height as &lt;math&gt;AFD&lt;/math&gt;, the ratios of the areas is equal to the ratios of the bases. Because of the Angle Bisector Theorem, &lt;math&gt;\frac{CG}{GD} = \frac{1}{5}&lt;/math&gt;. So, &lt;math&gt;\frac{[AEF]}{[AFD]} = \frac{1}{5}&lt;/math&gt;. Since &lt;math&gt;AEF&lt;/math&gt; has area X, we can write the equation 5X = Y and substitute 5X for Y.<br /> &lt;asy&gt;<br /> draw((0,0)--(1,3)--(5,0)--cycle);<br /> draw((0,0)--(2,2.25));<br /> draw((0.5,1.5)--(2.5,0));<br /> label(&quot;A&quot;,(0,0),SW);<br /> label(&quot;B&quot;,(5,0),SE);<br /> label(&quot;C&quot;,(1,3),N);<br /> label(&quot;G&quot;,(2,2.25),NE);<br /> label(&quot;D&quot;,(2.5,0),S);<br /> label(&quot;E&quot;,(0.5,1.5),NW);<br /> label(&quot;15X&quot;,(2.5,0.75),N);<br /> label(&quot;5X&quot;,(1,0.2),N);<br /> label(&quot;X&quot;,(0.5,0.5),N);<br /> label(&quot;3X&quot;,(1.25,1.75),N);<br /> &lt;/asy&gt;<br /> Now we can solve for X by adding up all the sums. X + 3X + 5X + 15X = 120, so X = 5. Since we want to find &lt;math&gt;GFDB&lt;/math&gt;, we substitute 5 for 15X to get &lt;math&gt;\boxed{75}&lt;/math&gt;.<br /> &lt;math&gt;\sim&lt;/math&gt;krishkhushi09<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2018|ab=A|num-b=23|num-a=25}}<br /> {{AMC12 box|year=2018|ab=A|num-b=17|num-a=19}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Intermediate Geometry Problems]]</div> Krishkhushi09 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_24&diff=119975 2018 AMC 10A Problems/Problem 24 2020-03-23T05:01:42Z <p>Krishkhushi09: /* Solution 6: Areas */</p> <hr /> <div>== Problem ==<br /> <br /> Triangle &lt;math&gt;ABC&lt;/math&gt; with &lt;math&gt;AB=50&lt;/math&gt; and &lt;math&gt;AC=10&lt;/math&gt; has area &lt;math&gt;120&lt;/math&gt;. Let &lt;math&gt;D&lt;/math&gt; be the midpoint of &lt;math&gt;\overline{AB}&lt;/math&gt;, and let &lt;math&gt;E&lt;/math&gt; be the midpoint of &lt;math&gt;\overline{AC}&lt;/math&gt;. The angle bisector of &lt;math&gt;\angle BAC&lt;/math&gt; intersects &lt;math&gt;\overline{DE}&lt;/math&gt; and &lt;math&gt;\overline{BC}&lt;/math&gt; at &lt;math&gt;F&lt;/math&gt; and &lt;math&gt;G&lt;/math&gt;, respectively. What is the area of quadrilateral &lt;math&gt;FDBG&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) }60 \qquad<br /> \textbf{(B) }65 \qquad<br /> \textbf{(C) }70 \qquad<br /> \textbf{(D) }75 \qquad<br /> \textbf{(E) }80 \qquad<br /> &lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> <br /> Let &lt;math&gt;BC = a&lt;/math&gt;, &lt;math&gt;BG = x&lt;/math&gt;, &lt;math&gt;GC = y&lt;/math&gt;, and the length of the perpendicular to &lt;math&gt;BC&lt;/math&gt; through &lt;math&gt;A&lt;/math&gt; be &lt;math&gt;h&lt;/math&gt;. By angle bisector theorem, we have that &lt;cmath&gt;\frac{50}{x} = \frac{10}{y},&lt;/cmath&gt; where &lt;math&gt;y = -x+a&lt;/math&gt;. Therefore substituting we have that &lt;math&gt;BG=\frac{5a}{6}&lt;/math&gt;. By similar triangles, we have that &lt;math&gt;DF=\frac{5a}{12}&lt;/math&gt;, and the height of this trapezoid is &lt;math&gt;\frac{h}{2}&lt;/math&gt;. Then, we have that &lt;math&gt;\frac{ah}{2}=120&lt;/math&gt;. We wish to compute &lt;math&gt;\frac{5a}{8}\cdot\frac{h}{2}&lt;/math&gt;, and we have that it is &lt;math&gt;\boxed{75}&lt;/math&gt; by substituting.<br /> <br /> == Solution 2 ==<br /> <br /> For this problem, we have &lt;math&gt;\triangle{ADE}\sim\triangle{ABC}&lt;/math&gt; because of SAS and &lt;math&gt;DE = \frac{BC}{2}&lt;/math&gt;. Therefore, &lt;math&gt;\bigtriangleup ADE&lt;/math&gt; is a quarter of the area of &lt;math&gt;\bigtriangleup ABC&lt;/math&gt;, which is &lt;math&gt;30&lt;/math&gt;. Subsequently, we can compute the area of quadrilateral &lt;math&gt;BDEC&lt;/math&gt; to be &lt;math&gt;120 - 30 = 90&lt;/math&gt;. Using the angle bisector theorem in the same fashion as the previous problem, we get that &lt;math&gt;\overline{BG}&lt;/math&gt; is &lt;math&gt;5&lt;/math&gt; times the length of &lt;math&gt;\overline{GC}&lt;/math&gt;. We want the larger piece, as described by the problem. Because the heights are identical, one area is &lt;math&gt;5&lt;/math&gt; times the other, and &lt;math&gt;\frac{5}{6} \cdot 90 = \boxed{75}&lt;/math&gt;.<br /> <br /> == Solution 3 ==<br /> The area of &lt;math&gt;\bigtriangleup ABG&lt;/math&gt; to the area of &lt;math&gt;\bigtriangleup ACG&lt;/math&gt; is &lt;math&gt;5:1&lt;/math&gt; by Law of Sines. So the area of &lt;math&gt;\bigtriangleup ABG&lt;/math&gt; is &lt;math&gt;100&lt;/math&gt;. Since &lt;math&gt;\overline{DE}&lt;/math&gt; is the midsegment of &lt;math&gt;\bigtriangleup ABC&lt;/math&gt;, so &lt;math&gt;\overline{DF}&lt;/math&gt; is the midsegment of &lt;math&gt;\bigtriangleup ABG&lt;/math&gt; . So the area of &lt;math&gt;\bigtriangleup ADF&lt;/math&gt; to the area of &lt;math&gt;\bigtriangleup ABG&lt;/math&gt; is &lt;math&gt;1:4&lt;/math&gt; , so the area of &lt;math&gt;\bigtriangleup ACG&lt;/math&gt; is &lt;math&gt;25&lt;/math&gt;, by similar triangles. Therefore the area of quad &lt;math&gt;FDBG&lt;/math&gt; is &lt;math&gt;100-25=\boxed{75}&lt;/math&gt;<br /> <br /> ==Solution 4 ==<br /> The area of quadrilateral &lt;math&gt;FDBG&lt;/math&gt; is the area of &lt;math&gt;\bigtriangleup ABG&lt;/math&gt; minus the area of &lt;math&gt;\bigtriangleup ADF&lt;/math&gt;. Notice, &lt;math&gt;\overline{DE} || \overline{BC}&lt;/math&gt;, so &lt;math&gt;\bigtriangleup ABG \sim \bigtriangleup ADF&lt;/math&gt;, and since &lt;math&gt;\overline{AD}:\overline{AB}=1:2&lt;/math&gt;, the area of &lt;math&gt;\bigtriangleup ADF:\bigtriangleup ABG=(1:2)^2=1:4&lt;/math&gt;. Given that the area of &lt;math&gt;\bigtriangleup ABC&lt;/math&gt; is &lt;math&gt;120&lt;/math&gt;, using &lt;math&gt;\frac{bh}{2}&lt;/math&gt; on side &lt;math&gt;AB&lt;/math&gt; yields &lt;math&gt;\frac{50h}{2}=120\implies h=\frac{240}{50}=\frac{24}{5}&lt;/math&gt;. Using the Angle Bisector Theorem, &lt;math&gt;\overline{BG}:\overline{BC}=50:(10+50)=5:6&lt;/math&gt;, so the height of &lt;math&gt;\bigtriangleup ABG: \bigtriangleup ACB=5:6&lt;/math&gt;. Therefore our answer is &lt;math&gt;\big[ FDBG\big] = \big[ABG\big]-\big[ ADF\big] = \big[ ABG\big]\big(1-\frac{1}{4}\big)=\frac{3}{4}\cdot \frac{bh}{2}=\frac{3}{8}\cdot 50\cdot \frac{5}{6}\cdot \frac{24}{5}=\frac{3}{8}\cdot 200=\boxed{75}&lt;/math&gt;<br /> <br /> ==Solution 5: Trig ==<br /> We try to find the area of quadrilateral &lt;math&gt;FDBG&lt;/math&gt; by subtracting the area outside the quadrilateral but inside triangle &lt;math&gt;ABC&lt;/math&gt;. Note that the area of &lt;math&gt;\triangle ADE&lt;/math&gt; is equal to &lt;math&gt;\frac{1}{2} \cdot 25 \cdot 5 \cdot \sin{A}&lt;/math&gt; and the area of triangle &lt;math&gt;ABC&lt;/math&gt; is equal to &lt;math&gt;\frac{1}{2} \cdot 50 \cdot 10 \cdot \sin A&lt;/math&gt;. The ratio &lt;math&gt;\frac{[ADE]}{[ABC]}&lt;/math&gt; is thus equal to &lt;math&gt;\frac{1}{4}&lt;/math&gt; and the area of triangle &lt;math&gt;ADE&lt;/math&gt; is &lt;math&gt;\frac{1}{4} \cdot 120 = 30&lt;/math&gt;. Let side &lt;math&gt;BC&lt;/math&gt; be equal to &lt;math&gt;6x&lt;/math&gt;, then &lt;math&gt;BG = 5x, GC = x&lt;/math&gt; by the angle bisector theorem. Similarly, we find the area of triangle &lt;math&gt;AGC&lt;/math&gt; to be &lt;math&gt;\frac{1}{2} \cdot 10 \cdot x \cdot \sin C&lt;/math&gt; and the area of triangle &lt;math&gt;ABC&lt;/math&gt; to be &lt;math&gt;\frac{1}{2} \cdot 6x \cdot 10 \cdot \sin C&lt;/math&gt;. A ratio between these two triangles yields &lt;math&gt;\frac{[ACG]}{[ABC]} = \frac{x}{6x} = \frac{1}{6}&lt;/math&gt;, so &lt;math&gt;[AGC] = 20&lt;/math&gt;. Now we just need to find the area of triangle &lt;math&gt;AFE&lt;/math&gt; and subtract it from the combined areas of &lt;math&gt;[ADE]&lt;/math&gt; and &lt;math&gt;[ACG]&lt;/math&gt;, since we count it twice. Note that the angle bisector theorem also applies for &lt;math&gt;\triangle ADE&lt;/math&gt; and &lt;math&gt;\frac{AE}{AD} = \frac{1}{5}&lt;/math&gt;, so thus &lt;math&gt;\frac{EF}{ED} = \frac{1}{6}&lt;/math&gt; and we find &lt;math&gt;[AFE] = \frac{1}{6} \cdot 30 = 5&lt;/math&gt;, and the area outside &lt;math&gt;FDBG&lt;/math&gt; must be &lt;math&gt; [ADE] + [AGC] - [AFE] = 30 + 20 - 5 = 45&lt;/math&gt;, and we finally find &lt;math&gt;[FDBG] = [ABC] - 45 = 120 -45 = \boxed{75}&lt;/math&gt;, and we are done. <br /> <br /> ==See Also==<br /> {{AMC10 box|year=2018|ab=A|num-b=23|num-a=25}}<br /> {{AMC12 box|year=2018|ab=A|num-b=17|num-a=19}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Intermediate Geometry Problems]]</div> Krishkhushi09 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_24&diff=119974 2018 AMC 10A Problems/Problem 24 2020-03-23T04:48:27Z <p>Krishkhushi09: </p> <hr /> <div>== Problem ==<br /> <br /> Triangle &lt;math&gt;ABC&lt;/math&gt; with &lt;math&gt;AB=50&lt;/math&gt; and &lt;math&gt;AC=10&lt;/math&gt; has area &lt;math&gt;120&lt;/math&gt;. Let &lt;math&gt;D&lt;/math&gt; be the midpoint of &lt;math&gt;\overline{AB}&lt;/math&gt;, and let &lt;math&gt;E&lt;/math&gt; be the midpoint of &lt;math&gt;\overline{AC}&lt;/math&gt;. The angle bisector of &lt;math&gt;\angle BAC&lt;/math&gt; intersects &lt;math&gt;\overline{DE}&lt;/math&gt; and &lt;math&gt;\overline{BC}&lt;/math&gt; at &lt;math&gt;F&lt;/math&gt; and &lt;math&gt;G&lt;/math&gt;, respectively. What is the area of quadrilateral &lt;math&gt;FDBG&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) }60 \qquad<br /> \textbf{(B) }65 \qquad<br /> \textbf{(C) }70 \qquad<br /> \textbf{(D) }75 \qquad<br /> \textbf{(E) }80 \qquad<br /> &lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> <br /> Let &lt;math&gt;BC = a&lt;/math&gt;, &lt;math&gt;BG = x&lt;/math&gt;, &lt;math&gt;GC = y&lt;/math&gt;, and the length of the perpendicular to &lt;math&gt;BC&lt;/math&gt; through &lt;math&gt;A&lt;/math&gt; be &lt;math&gt;h&lt;/math&gt;. By angle bisector theorem, we have that &lt;cmath&gt;\frac{50}{x} = \frac{10}{y},&lt;/cmath&gt; where &lt;math&gt;y = -x+a&lt;/math&gt;. Therefore substituting we have that &lt;math&gt;BG=\frac{5a}{6}&lt;/math&gt;. By similar triangles, we have that &lt;math&gt;DF=\frac{5a}{12}&lt;/math&gt;, and the height of this trapezoid is &lt;math&gt;\frac{h}{2}&lt;/math&gt;. Then, we have that &lt;math&gt;\frac{ah}{2}=120&lt;/math&gt;. We wish to compute &lt;math&gt;\frac{5a}{8}\cdot\frac{h}{2}&lt;/math&gt;, and we have that it is &lt;math&gt;\boxed{75}&lt;/math&gt; by substituting.<br /> <br /> == Solution 2 ==<br /> <br /> For this problem, we have &lt;math&gt;\triangle{ADE}\sim\triangle{ABC}&lt;/math&gt; because of SAS and &lt;math&gt;DE = \frac{BC}{2}&lt;/math&gt;. Therefore, &lt;math&gt;\bigtriangleup ADE&lt;/math&gt; is a quarter of the area of &lt;math&gt;\bigtriangleup ABC&lt;/math&gt;, which is &lt;math&gt;30&lt;/math&gt;. Subsequently, we can compute the area of quadrilateral &lt;math&gt;BDEC&lt;/math&gt; to be &lt;math&gt;120 - 30 = 90&lt;/math&gt;. Using the angle bisector theorem in the same fashion as the previous problem, we get that &lt;math&gt;\overline{BG}&lt;/math&gt; is &lt;math&gt;5&lt;/math&gt; times the length of &lt;math&gt;\overline{GC}&lt;/math&gt;. We want the larger piece, as described by the problem. Because the heights are identical, one area is &lt;math&gt;5&lt;/math&gt; times the other, and &lt;math&gt;\frac{5}{6} \cdot 90 = \boxed{75}&lt;/math&gt;.<br /> <br /> == Solution 3 ==<br /> The area of &lt;math&gt;\bigtriangleup ABG&lt;/math&gt; to the area of &lt;math&gt;\bigtriangleup ACG&lt;/math&gt; is &lt;math&gt;5:1&lt;/math&gt; by Law of Sines. So the area of &lt;math&gt;\bigtriangleup ABG&lt;/math&gt; is &lt;math&gt;100&lt;/math&gt;. Since &lt;math&gt;\overline{DE}&lt;/math&gt; is the midsegment of &lt;math&gt;\bigtriangleup ABC&lt;/math&gt;, so &lt;math&gt;\overline{DF}&lt;/math&gt; is the midsegment of &lt;math&gt;\bigtriangleup ABG&lt;/math&gt; . So the area of &lt;math&gt;\bigtriangleup ADF&lt;/math&gt; to the area of &lt;math&gt;\bigtriangleup ABG&lt;/math&gt; is &lt;math&gt;1:4&lt;/math&gt; , so the area of &lt;math&gt;\bigtriangleup ACG&lt;/math&gt; is &lt;math&gt;25&lt;/math&gt;, by similar triangles. Therefore the area of quad &lt;math&gt;FDBG&lt;/math&gt; is &lt;math&gt;100-25=\boxed{75}&lt;/math&gt;<br /> <br /> ==Solution 4 ==<br /> The area of quadrilateral &lt;math&gt;FDBG&lt;/math&gt; is the area of &lt;math&gt;\bigtriangleup ABG&lt;/math&gt; minus the area of &lt;math&gt;\bigtriangleup ADF&lt;/math&gt;. Notice, &lt;math&gt;\overline{DE} || \overline{BC}&lt;/math&gt;, so &lt;math&gt;\bigtriangleup ABG \sim \bigtriangleup ADF&lt;/math&gt;, and since &lt;math&gt;\overline{AD}:\overline{AB}=1:2&lt;/math&gt;, the area of &lt;math&gt;\bigtriangleup ADF:\bigtriangleup ABG=(1:2)^2=1:4&lt;/math&gt;. Given that the area of &lt;math&gt;\bigtriangleup ABC&lt;/math&gt; is &lt;math&gt;120&lt;/math&gt;, using &lt;math&gt;\frac{bh}{2}&lt;/math&gt; on side &lt;math&gt;AB&lt;/math&gt; yields &lt;math&gt;\frac{50h}{2}=120\implies h=\frac{240}{50}=\frac{24}{5}&lt;/math&gt;. Using the Angle Bisector Theorem, &lt;math&gt;\overline{BG}:\overline{BC}=50:(10+50)=5:6&lt;/math&gt;, so the height of &lt;math&gt;\bigtriangleup ABG: \bigtriangleup ACB=5:6&lt;/math&gt;. Therefore our answer is &lt;math&gt;\big[ FDBG\big] = \big[ABG\big]-\big[ ADF\big] = \big[ ABG\big]\big(1-\frac{1}{4}\big)=\frac{3}{4}\cdot \frac{bh}{2}=\frac{3}{8}\cdot 50\cdot \frac{5}{6}\cdot \frac{24}{5}=\frac{3}{8}\cdot 200=\boxed{75}&lt;/math&gt;<br /> <br /> ==Solution 5: Trig ==<br /> We try to find the area of quadrilateral &lt;math&gt;FDBG&lt;/math&gt; by subtracting the area outside the quadrilateral but inside triangle &lt;math&gt;ABC&lt;/math&gt;. Note that the area of &lt;math&gt;\triangle ADE&lt;/math&gt; is equal to &lt;math&gt;\frac{1}{2} \cdot 25 \cdot 5 \cdot \sin{A}&lt;/math&gt; and the area of triangle &lt;math&gt;ABC&lt;/math&gt; is equal to &lt;math&gt;\frac{1}{2} \cdot 50 \cdot 10 \cdot \sin A&lt;/math&gt;. The ratio &lt;math&gt;\frac{[ADE]}{[ABC]}&lt;/math&gt; is thus equal to &lt;math&gt;\frac{1}{4}&lt;/math&gt; and the area of triangle &lt;math&gt;ADE&lt;/math&gt; is &lt;math&gt;\frac{1}{4} \cdot 120 = 30&lt;/math&gt;. Let side &lt;math&gt;BC&lt;/math&gt; be equal to &lt;math&gt;6x&lt;/math&gt;, then &lt;math&gt;BG = 5x, GC = x&lt;/math&gt; by the angle bisector theorem. Similarly, we find the area of triangle &lt;math&gt;AGC&lt;/math&gt; to be &lt;math&gt;\frac{1}{2} \cdot 10 \cdot x \cdot \sin C&lt;/math&gt; and the area of triangle &lt;math&gt;ABC&lt;/math&gt; to be &lt;math&gt;\frac{1}{2} \cdot 6x \cdot 10 \cdot \sin C&lt;/math&gt;. A ratio between these two triangles yields &lt;math&gt;\frac{[ACG]}{[ABC]} = \frac{x}{6x} = \frac{1}{6}&lt;/math&gt;, so &lt;math&gt;[AGC] = 20&lt;/math&gt;. Now we just need to find the area of triangle &lt;math&gt;AFE&lt;/math&gt; and subtract it from the combined areas of &lt;math&gt;[ADE]&lt;/math&gt; and &lt;math&gt;[ACG]&lt;/math&gt;, since we count it twice. Note that the angle bisector theorem also applies for &lt;math&gt;\triangle ADE&lt;/math&gt; and &lt;math&gt;\frac{AE}{AD} = \frac{1}{5}&lt;/math&gt;, so thus &lt;math&gt;\frac{EF}{ED} = \frac{1}{6}&lt;/math&gt; and we find &lt;math&gt;[AFE] = \frac{1}{6} \cdot 30 = 5&lt;/math&gt;, and the area outside &lt;math&gt;FDBG&lt;/math&gt; must be &lt;math&gt; [ADE] + [AGC] - [AFE] = 30 + 20 - 5 = 45&lt;/math&gt;, and we finally find &lt;math&gt;[FDBG] = [ABC] - 45 = 120 -45 = \boxed{75}&lt;/math&gt;, and we are done. <br /> <br /> ==Solution 6: Areas ==<br /> &lt;asy&gt;<br /> draw((0,0)--(1,3)--(5,0)--cycle);<br /> label($A$, (0,0), N);<br /> &lt;/asy&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2018|ab=A|num-b=23|num-a=25}}<br /> {{AMC12 box|year=2018|ab=A|num-b=17|num-a=19}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Intermediate Geometry Problems]]</div> Krishkhushi09 https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_II_Problems/Problem_4&diff=112386 2011 AIME II Problems/Problem 4 2019-11-30T15:10:51Z <p>Krishkhushi09: /* Solution 1 */</p> <hr /> <div>== Problem 4 ==<br /> In triangle &lt;math&gt;ABC&lt;/math&gt;, &lt;math&gt;AB=\frac{20}{11} AC&lt;/math&gt;. The angle bisector of &lt;math&gt;\angle A&lt;/math&gt; intersects &lt;math&gt;BC&lt;/math&gt; at point &lt;math&gt;D&lt;/math&gt;, and point &lt;math&gt;M&lt;/math&gt; is the midpoint of &lt;math&gt;AD&lt;/math&gt;. Let &lt;math&gt;P&lt;/math&gt; be the point of the intersection of &lt;math&gt;AC&lt;/math&gt; and &lt;math&gt;BM&lt;/math&gt;. The ratio of &lt;math&gt;CP&lt;/math&gt; to &lt;math&gt;PA&lt;/math&gt; can be expressed in the form &lt;math&gt;\dfrac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> __TOC__<br /> <br /> == Solutions ==<br /> === Solution 1 ===<br /> &lt;asy&gt;<br /> pointpen = black; pathpen = linewidth(0.7);<br /> <br /> pair A = (0,0), C= (11,0), B=IP(CR(A,20),CR(C,18)), D = IP(B--C,CR(B,20/31*abs(B-C))), M = (A+D)/2, P = IP(M--2*M-B, A--C), D2 = IP(D--D+P-B, A--C);<br /> <br /> D(MP(&quot;A&quot;,D(A))--MP(&quot;B&quot;,D(B),N)--MP(&quot;C&quot;,D(C))--cycle); D(A--MP(&quot;D&quot;,D(D),NE)--MP(&quot;D'&quot;,D(D2))); D(B--MP(&quot;P&quot;,D(P))); D(MP(&quot;M&quot;,M,NW)); MP(&quot;20&quot;,(B+D)/2,ENE); MP(&quot;11&quot;,(C+D)/2,ENE);<br /> <br /> &lt;/asy&gt; Let &lt;math&gt;D'&lt;/math&gt; be on &lt;math&gt;\overline{AC}&lt;/math&gt; such that &lt;math&gt;BP \parallel DD'&lt;/math&gt;. It follows that &lt;math&gt;\triangle BPC \sim \triangle DD'C&lt;/math&gt;, so &lt;cmath&gt;\frac{PC}{D'C} = 1 + \frac{BD}{DC} = 1 + \frac{AB}{AC} = \frac{31}{11}&lt;/cmath&gt; by the [[Angle Bisector Theorem]]. Similarly, we see by the midline theorem that &lt;math&gt;AP = PD'&lt;/math&gt;. Thus, &lt;cmath&gt;\frac{CP}{PA} = \frac{1}{\frac{PD'}{PC}} = \frac{1}{1 - \frac{D'C}{PC}} = \frac{31}{20},&lt;/cmath&gt; and &lt;math&gt;m+n = \boxed{51}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> Assign [[mass points]] as follows: by Angle-Bisector Theorem, &lt;math&gt;BD / DC = 20/11&lt;/math&gt;, so we assign &lt;math&gt;m(B) = 11, m(C) = 20, m(D) = 31&lt;/math&gt;. Since &lt;math&gt;AM = MD&lt;/math&gt;, then &lt;math&gt;m(A) = 31&lt;/math&gt;, and &lt;math&gt;\frac{CP}{PA} = \frac{m(A) }{ m(C)} = \frac{31}{20}&lt;/math&gt;, so &lt;math&gt;m+n = \boxed{51}&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> By [[Menelaus' Theorem]] on &lt;math&gt;\triangle ACD&lt;/math&gt; with [[transversal]] &lt;math&gt;PB&lt;/math&gt;, &lt;cmath&gt;1 = \frac{CP}{PA} \cdot \frac{AM}{MD} \cdot \frac{DB}{CB} = \frac{CP}{PA} \cdot \left(\frac{1}{1+\frac{AC}{AB}}\right) \quad \Longrightarrow \quad \frac{CP}{PA} = \frac{31}{20}. &lt;/cmath&gt; So &lt;math&gt;m+n = \boxed{051}&lt;/math&gt;.<br /> <br /> ===Solution 4===<br /> We will use barycentric coordinates. Let &lt;math&gt;A = (1, 0, 0)&lt;/math&gt;, &lt;math&gt;B = (0, 1, 0)&lt;/math&gt;, &lt;math&gt;C = (0, 0, 1)&lt;/math&gt;. By the [[Angle Bisector Theorem]], &lt;math&gt;D = [0:11:20] = \left(0, \frac{11}{31}, \frac{20}{31}\right)&lt;/math&gt;. Since &lt;math&gt;M&lt;/math&gt; is the midpoint of &lt;math&gt;AD&lt;/math&gt;, &lt;math&gt;M = \frac{A + D}{2} = \left(\frac{1}{2}, \frac{11}{62}, \frac{10}{31}\right)&lt;/math&gt;. Therefore, the equation for line BM is &lt;math&gt;20x = 31z&lt;/math&gt;. Let &lt;math&gt;P = (x, 0, 1-x)&lt;/math&gt;. Using the equation for &lt;math&gt;BM&lt;/math&gt;, we get &lt;cmath&gt;20x = 31(1-x)&lt;/cmath&gt;<br /> &lt;cmath&gt;x = \frac{31}{51}&lt;/cmath&gt; Therefore, &lt;math&gt;\frac{CP}{PA} = \frac{1-x}{x} = \frac{31}{20}&lt;/math&gt; so the answer is &lt;math&gt;\boxed{051}&lt;/math&gt;.<br /> <br /> === Solution 5 ===<br /> Let &lt;math&gt;DC=x&lt;/math&gt;. Then by the Angle Bisector Theorem, &lt;math&gt;BD=\frac{20}{11}x&lt;/math&gt;. By the Ratio Lemma, we have that &lt;math&gt;\frac{PC}{AP}=\frac{\frac{31}{11}x\sin\angle PBC}{20\sin\angle ABP}.&lt;/math&gt; Notice that &lt;math&gt;[\triangle BAM]=[\triangle BMD]&lt;/math&gt; since their bases have the same length and they share a height. By the sin area formula, we have that &lt;cmath&gt;\frac{1}{2}\cdot20\cdot BM\cdot \sin\angle ABP=\frac{1}{2}\cdot \frac{20}{11}x\cdot BM\cdot\sin\angle PBC.&lt;/cmath&gt; Simplifying, we get that &lt;math&gt;\frac{\sin\angle PBC}{\sin\angle ABP}=\frac{11}{x}.&lt;/math&gt; Plugging this into what we got from the Ratio Lemma, we have that &lt;math&gt;\frac{PC}{AP}=\frac{31}{20}\implies\boxed{051.}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year = 2011|n=II|num-b=3|num-a=5}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Krishkhushi09 https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_II_Problems/Problem_4&diff=112385 2011 AIME II Problems/Problem 4 2019-11-30T15:10:33Z <p>Krishkhushi09: /* Solution 2 */</p> <hr /> <div>== Problem 4 ==<br /> In triangle &lt;math&gt;ABC&lt;/math&gt;, &lt;math&gt;AB=\frac{20}{11} AC&lt;/math&gt;. The angle bisector of &lt;math&gt;\angle A&lt;/math&gt; intersects &lt;math&gt;BC&lt;/math&gt; at point &lt;math&gt;D&lt;/math&gt;, and point &lt;math&gt;M&lt;/math&gt; is the midpoint of &lt;math&gt;AD&lt;/math&gt;. Let &lt;math&gt;P&lt;/math&gt; be the point of the intersection of &lt;math&gt;AC&lt;/math&gt; and &lt;math&gt;BM&lt;/math&gt;. The ratio of &lt;math&gt;CP&lt;/math&gt; to &lt;math&gt;PA&lt;/math&gt; can be expressed in the form &lt;math&gt;\dfrac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> __TOC__<br /> <br /> == Solutions ==<br /> === Solution 1 ===<br /> &lt;asy&gt;<br /> pointpen = black; pathpen = linewidth(0.7);<br /> <br /> pair A = (0,0), C= (11,0), B=IP(CR(A,20),CR(C,18)), D = IP(B--C,CR(B,20/31*abs(B-C))), M = (A+D)/2, P = IP(M--2*M-B, A--C), D2 = IP(D--D+P-B, A--C);<br /> <br /> D(MP(&quot;A&quot;,D(A))--MP(&quot;B&quot;,D(B),N)--MP(&quot;C&quot;,D(C))--cycle); D(A--MP(&quot;D&quot;,D(D),NE)--MP(&quot;D'&quot;,D(D2))); D(B--MP(&quot;P&quot;,D(P))); D(MP(&quot;M&quot;,M,NW)); MP(&quot;20&quot;,(B+D)/2,ENE); MP(&quot;11&quot;,(C+D)/2,ENE);<br /> <br /> &lt;/asy&gt; Let &lt;math&gt;D'&lt;/math&gt; be on &lt;math&gt;\overline{AC}&lt;/math&gt; such that &lt;math&gt;BP \parallel DD'&lt;/math&gt;. It follows that &lt;math&gt;\triangle BPC \sim \triangle DD'C&lt;/math&gt;, so &lt;cmath&gt;\frac{PC}{D'C} = 1 + \frac{BD}{DC} = 1 + \frac{AB}{AC} = \frac{31}{11}&lt;/cmath&gt; by the [[Angle Bisector Theorem]]. Similarly, we see by the midline theorem that &lt;math&gt;AP = PD'&lt;/math&gt;. Thus, &lt;cmath&gt;\frac{CP}{PA} = \frac{1}{\frac{PD'}{PC}} = \frac{1}{1 - \frac{D'C}{PC}} = \frac{31}{20},&lt;/cmath&gt; and &lt;math&gt;m+n = \boxed{051}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> Assign [[mass points]] as follows: by Angle-Bisector Theorem, &lt;math&gt;BD / DC = 20/11&lt;/math&gt;, so we assign &lt;math&gt;m(B) = 11, m(C) = 20, m(D) = 31&lt;/math&gt;. Since &lt;math&gt;AM = MD&lt;/math&gt;, then &lt;math&gt;m(A) = 31&lt;/math&gt;, and &lt;math&gt;\frac{CP}{PA} = \frac{m(A) }{ m(C)} = \frac{31}{20}&lt;/math&gt;, so &lt;math&gt;m+n = \boxed{51}&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> By [[Menelaus' Theorem]] on &lt;math&gt;\triangle ACD&lt;/math&gt; with [[transversal]] &lt;math&gt;PB&lt;/math&gt;, &lt;cmath&gt;1 = \frac{CP}{PA} \cdot \frac{AM}{MD} \cdot \frac{DB}{CB} = \frac{CP}{PA} \cdot \left(\frac{1}{1+\frac{AC}{AB}}\right) \quad \Longrightarrow \quad \frac{CP}{PA} = \frac{31}{20}. &lt;/cmath&gt; So &lt;math&gt;m+n = \boxed{051}&lt;/math&gt;.<br /> <br /> ===Solution 4===<br /> We will use barycentric coordinates. Let &lt;math&gt;A = (1, 0, 0)&lt;/math&gt;, &lt;math&gt;B = (0, 1, 0)&lt;/math&gt;, &lt;math&gt;C = (0, 0, 1)&lt;/math&gt;. By the [[Angle Bisector Theorem]], &lt;math&gt;D = [0:11:20] = \left(0, \frac{11}{31}, \frac{20}{31}\right)&lt;/math&gt;. Since &lt;math&gt;M&lt;/math&gt; is the midpoint of &lt;math&gt;AD&lt;/math&gt;, &lt;math&gt;M = \frac{A + D}{2} = \left(\frac{1}{2}, \frac{11}{62}, \frac{10}{31}\right)&lt;/math&gt;. Therefore, the equation for line BM is &lt;math&gt;20x = 31z&lt;/math&gt;. Let &lt;math&gt;P = (x, 0, 1-x)&lt;/math&gt;. Using the equation for &lt;math&gt;BM&lt;/math&gt;, we get &lt;cmath&gt;20x = 31(1-x)&lt;/cmath&gt;<br /> &lt;cmath&gt;x = \frac{31}{51}&lt;/cmath&gt; Therefore, &lt;math&gt;\frac{CP}{PA} = \frac{1-x}{x} = \frac{31}{20}&lt;/math&gt; so the answer is &lt;math&gt;\boxed{051}&lt;/math&gt;.<br /> <br /> === Solution 5 ===<br /> Let &lt;math&gt;DC=x&lt;/math&gt;. Then by the Angle Bisector Theorem, &lt;math&gt;BD=\frac{20}{11}x&lt;/math&gt;. By the Ratio Lemma, we have that &lt;math&gt;\frac{PC}{AP}=\frac{\frac{31}{11}x\sin\angle PBC}{20\sin\angle ABP}.&lt;/math&gt; Notice that &lt;math&gt;[\triangle BAM]=[\triangle BMD]&lt;/math&gt; since their bases have the same length and they share a height. By the sin area formula, we have that &lt;cmath&gt;\frac{1}{2}\cdot20\cdot BM\cdot \sin\angle ABP=\frac{1}{2}\cdot \frac{20}{11}x\cdot BM\cdot\sin\angle PBC.&lt;/cmath&gt; Simplifying, we get that &lt;math&gt;\frac{\sin\angle PBC}{\sin\angle ABP}=\frac{11}{x}.&lt;/math&gt; Plugging this into what we got from the Ratio Lemma, we have that &lt;math&gt;\frac{PC}{AP}=\frac{31}{20}\implies\boxed{051.}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year = 2011|n=II|num-b=3|num-a=5}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Krishkhushi09 https://artofproblemsolving.com/wiki/index.php?title=Set_of_Complex_Numbers&diff=109588 Set of Complex Numbers 2019-09-01T15:53:21Z <p>Krishkhushi09: This page is the set of all complex numbers. Complex numbers are real and imaginary. Real numbers are irrational and rational. Rational numbers include the integers, which include the natural numbers.</p> <hr /> <div>The set of complex numbers is a set of all numbers ever existed, from pi to 1. <br /> <br /> == Complex Numbers == <br /> Complex numbers are numbers which have a real part and an imaginary part, usually written in the from a + b&lt;math&gt;i&lt;/math&gt;.<br /> === Real Numbers ===<br /> Real numbers is the set of all numbers which aren't imaginary, eg. don't have the b&lt;math&gt;i&lt;/math&gt; counterpart and can be squared into some nonnegative number.<br /> === Imaginary Numbers ===<br /> [[Imaginary numbers]] is the set of numbers which do have the b&lt;math&gt;i&lt;/math&gt; counterpart. They will always be squared into a negative number.<br /> == Set of all Real Numbers == <br /> The set of all real numbers can be classified into two major sub sets, the set of irrational and rational numbers.<br /> === Irrational Numbers ===<br /> Irrational numbers are all numbers which cannot be simplified into the from a/b where b isn't 0.<br /> === Rational Numbers ===<br /> Rational numbers are all numbers which can be simplified into a/b. The rational numbers can be further split into to more subsets, subsubsets, and so on.<br /> == Set of Real Numbers ==<br /> === Integers ===<br /> Integers are all numbers which aren't fractions but can be simplified into a/b. For example, 2 is an integer and 2/3 isn't. Usually integers are numbers we use for our daily counting. They are both negative and nonnegative.<br /> === Natural Numbers ===<br /> Natural numbers are all positive integers. Examples are 2 and 3. 0 is nonnegative, but isn't positive so is therefore excluded. Natural numbers are a subset of integers, which are a subset of rational numbers, which combine with irrational numbers to make real numbers, which combine with imaginary numbers to make the set of all complex numbers.</div> Krishkhushi09 https://artofproblemsolving.com/wiki/index.php?title=Coefficient&diff=109587 Coefficient 2019-09-01T15:26:16Z <p>Krishkhushi09: </p> <hr /> <div>In [[mathematics]], a '''coefficient''' is a [[constant]] multiplicative [[divisor | factor]] of a specified object. The object can be a [[variable]], a [[vector]], a [[function]], or anything else that might be subject to multiplication. <br /> <br /> For example, the coefficient of &lt;math&gt;a&lt;/math&gt; in the [[expression]] &lt;math&gt;5a + b&lt;/math&gt; is 5. Note that it is important that we specify what we are looking at the coefficients of: 5 also has a coefficient in &lt;math&gt;5a + b&lt;/math&gt;, namely &lt;math&gt;a&lt;/math&gt;.<br /> <br /> Coefficients come up most frequently in a discussion of [[polynomial]]s. <br /> <br /> == Leading Coefficient == <br /> The leading coefficient is the coefficient of the highest degree in a polynomial. It is used to write the polynomial in standard form, and is especially used to factor quadratics.<br /> <br /> == See Also ==<br /> * [[Linear algebra]]<br /> * [[Binomial coefficient]]s<br /> * [[Binomial theorem]]<br /> <br /> {{stub}}<br /> [[Category:Definition]]</div> Krishkhushi09