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Difference between revisions of "2009 AIME II Problems/Problem 15"

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== Solutions ==
 
== Solutions ==
  
−
===Solution 1===
+
===Solution 1 (Quick Calculus)===
 +
Let <math>V = \overline{NM} \cap \overline{AC}</math> and <math>W = \overline{NM} \cap \overline{BC}</math>. Further more let <math>\angle NMC = \alpha</math> and <math>\angle MNC = 90^\circ - \alpha</math>. Angle chasing reveals <math>\angle NBC = \angle NAC = \alpha</math> and <math>\angle MBC = \angle MAC = 90^\circ - \alpha</math>. Additionally <math>NB = \frac{4}{5}</math> and <math>AN = AM</math> by the Pythagorean Theorem.
 +
 
 +
By the Angle Bisector Formula,
 +
<cmath>\frac{NV}{MV} = \frac{\sin (\alpha)}{\sin (90^\circ - \alpha)} = \tan (\alpha)</cmath>
 +
<cmath>\frac{MW}{NW} = \frac{3\sin (90^\circ - \alpha)}{4\sin (\alpha)} = \frac{3}{4} \cot (\alpha)</cmath>
 +
 
 +
As <math>NV + MV =MW + NW = 1</math> we compute <math>NW = \frac{1}{1+\frac{3}{4}\cot(\alpha)}</math> and <math>MV = \frac{1}{1+\tan (\alpha)}</math>, and finally <math>VW = NW + MV - 1 =  \frac{1}{1+\frac{3}{4}\cot(\alpha)} + \frac{1}{1+\tan (\alpha)} - 1</math>. Taking the derivative of <math>VW</math> with respect to <math>\alpha</math>, we arrive at
 +
<cmath>VW' = \frac{7\cos^2 (\alpha) - 4}{(\sin(\alpha) + \cos(\alpha))^2(4\sin(\alpha)+3\cos(\alpha))^2}</cmath>
 +
Clearly the maximum occurs when <math>\alpha = \cos^{-1}\left(\frac{2}{\sqrt{7}}\right)</math>. Plugging this back in, using the fact that <math>\tan(\cos^{-1}(x)) = \frac{\sqrt{1-x^2}}{x}</math> and <math>\cot(\cos^{-1}(x)) = \frac{x}{\sqrt{1-x^2}}</math>, we get
 +
 
 +
<math>VW = 7 - 4\sqrt{3}</math>
 +
with <math>7 + 4 + 3 = \boxed{014}</math>
 +
 
 +
~always_correct
 +
 
 +
===Solution 2 (Projective)===
 +
Since <math>MA = \frac{\sqrt{2}}{2} \approx 0.707 > \frac{3}{5}</math>, point <math>B</math> lies between <math>M</math> and <math>A</math> on the semicircular arc. We will first compute the length of <math>\overline{AB}</math>. By the law of cosines, <math>\cos \angle MOB = \frac{-(3/5)^2 + 2(1/2)^2}{2(1/2)^2} = \frac{7}{25}</math>, so <math>\cos \angle AOB = \sin \angle MOB = \frac{24}{25}</math>. Then <math>AB^2 = 2\left(\frac{1}{2}\right)^2 - 2\left(\frac{1}{2}\right)^2 \cdot \frac{24}{25} = \frac{1}{50}</math>, so <math>AB = \frac{1}{5\sqrt{2}}</math>.
 +
 
 +
Let <math>P = AC \cap MN</math> and <math>Q = BC \cap MN</math>, and let <math>MQ = x</math>, <math>PQ = d</math>, <math>PN = y</math>. Note that<cmath>(M, P; Q, N) \stackrel{C}{=} (M, A; B, N),</cmath>that is,<cmath>\frac{QP}{QM} \div \frac{NP}{NM} = \frac{BA}{BM} \div \frac{NA}{NM}</cmath>or<cmath>\frac{d}{xy} = \frac{1/(5\sqrt{2})}{(3/5) \cdot (\sqrt{2}/2)} = \frac{1}{3}.</cmath>Hence <math>d = \frac{1}{3}xy</math>, and we also know <math>d+x+y=1</math>. Now AM-GM gives<cmath>\frac{x+y}{2} \ge \sqrt{xy} \implies \frac{1-d}{2} \ge \sqrt{3d}.</cmath>This gives the quadratic inequality <math>d^2 - 14d + 1 \ge 0</math>, which solves as<cmath>d \in \left(-\infty, 7-4\sqrt3\right] \cup \left[7+4\sqrt3, \infty\right).</cmath>But <math>d \le 1</math>, so the greatest possible value of <math>d</math> is <math>7-4\sqrt3</math>. The answer is <math>7+4+3=\boxed{014}</math>.
 +
 
 +
~MSTang
 +
 
 +
===Solution 3 (Calculus)===
 
Let <math>O</math> be the center of the circle. Define <math>\angle{MOC}=t</math>, <math>\angle{BOA}=2a</math>, and let <math>BC</math> and <math>AC</math> intersect <math>MN</math> at points <math>X</math> and <math>Y</math>, respectively. We will express the length of <math>XY</math> as a function of <math>t</math> and maximize that function in the interval <math>[0, \pi]</math>.
 
Let <math>O</math> be the center of the circle. Define <math>\angle{MOC}=t</math>, <math>\angle{BOA}=2a</math>, and let <math>BC</math> and <math>AC</math> intersect <math>MN</math> at points <math>X</math> and <math>Y</math>, respectively. We will express the length of <math>XY</math> as a function of <math>t</math> and maximize that function in the interval <math>[0, \pi]</math>.
  
Line 86: Line 109:
 
so the maximum length of <math>XY</math> is <math>\sin(a)\times\frac{\sin(t')}{\sin(t' + a) + \cos(a)} = 7 - 4\sqrt{3}</math>, and the answer is <math>7 + 4 + 3 = \boxed{014}</math>.
 
so the maximum length of <math>XY</math> is <math>\sin(a)\times\frac{\sin(t')}{\sin(t' + a) + \cos(a)} = 7 - 4\sqrt{3}</math>, and the answer is <math>7 + 4 + 3 = \boxed{014}</math>.
  
−
===Solution 2===
+
===Solution 4===
 
<asy>
 
<asy>
 
unitsize(144);
 
unitsize(144);
Line 104: Line 127:
 
<cmath>\frac{MC}{MD} = \frac{NC}{ND}\Rightarrow MD = \frac{x}{x + y}.</cmath>
 
<cmath>\frac{MC}{MD} = \frac{NC}{ND}\Rightarrow MD = \frac{x}{x + y}.</cmath>
 
To find <math>ME</math>, we note that <math>\triangle BNE\sim\triangle MCE</math> and <math>\triangle BME\sim\triangle NCE</math>, so
 
To find <math>ME</math>, we note that <math>\triangle BNE\sim\triangle MCE</math> and <math>\triangle BME\sim\triangle NCE</math>, so
 +
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
 
\frac{BN}{NE} &= \frac{MC}{CE} \\
 
\frac{BN}{NE} &= \frac{MC}{CE} \\
 
\frac{ME}{BM} &= \frac{CE}{NC}.
 
\frac{ME}{BM} &= \frac{CE}{NC}.
 
\end{align*}</cmath>
 
\end{align*}</cmath>
 +
 
Writing <math>NE = 1 - ME</math>, we can substitute known values and multiply the equations to get
 
Writing <math>NE = 1 - ME</math>, we can substitute known values and multiply the equations to get
 +
 
<cmath>\frac{4(ME)}{3 - 3(ME)} = \frac{x}{y}\Rightarrow ME = \frac{3x}{3x + 4y}.</cmath>
 
<cmath>\frac{4(ME)}{3 - 3(ME)} = \frac{x}{y}\Rightarrow ME = \frac{3x}{3x + 4y}.</cmath>
 +
 
The value we wish to maximize is
 
The value we wish to maximize is
 +
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
 
DE &= MD - ME \\
 
DE &= MD - ME \\
Line 117: Line 145:
 
&= \frac{1}{3(x/y) + 4(y/x) + 7}.
 
&= \frac{1}{3(x/y) + 4(y/x) + 7}.
 
\end{align*}</cmath>
 
\end{align*}</cmath>
 +
 
By the AM-GM inequality, <math>3(x/y) + 4(y/x)\geq 2\sqrt{12} = 4\sqrt{3}</math>, so
 
By the AM-GM inequality, <math>3(x/y) + 4(y/x)\geq 2\sqrt{12} = 4\sqrt{3}</math>, so
 
<cmath>DE\leq \frac{1}{4\sqrt{3} + 7} = 7 - 4\sqrt{3},</cmath>
 
<cmath>DE\leq \frac{1}{4\sqrt{3} + 7} = 7 - 4\sqrt{3},</cmath>
 
giving the answer of <math>7 + 4 + 3 = \boxed{014}</math>. Equality is achieved when <math>3(x/y) = 4(y/x)</math> subject to the condition <math>x^2 + y^2 = 1</math>, which occurs for <math>x = \frac{2\sqrt{7}}{7}</math> and <math>y = \frac{\sqrt{21}}{7}</math>.
 
giving the answer of <math>7 + 4 + 3 = \boxed{014}</math>. Equality is achieved when <math>3(x/y) = 4(y/x)</math> subject to the condition <math>x^2 + y^2 = 1</math>, which occurs for <math>x = \frac{2\sqrt{7}}{7}</math> and <math>y = \frac{\sqrt{21}}{7}</math>.
 +
 +
===Solution 5 (Projective)===
 +
By Pythagoras in <math>\triangle BMN,</math> we get <math>BN=\dfrac{4}{5}.</math>
 +
 +
Since cross ratios are preserved upon projecting, note that <math>(M,Y;X,N)\stackrel{C}{=}(M,B;A,N).</math> By definition of a cross ratio, this becomes <cmath>\dfrac{XM}{NY}:\dfrac{NM}{NY}=\dfrac{AM}{AB}:\dfrac{MN}{NB}.</cmath> Let <math>MY=a,YX=b,XN=c</math> such that <math>a+b+c=1.</math> We know that <math>XM=a+b,XY=b,NM=1,NY=b+c,</math> so the LHS becomes <math>\dfrac{(a+b)(b+c)}{b}.</math>
 +
 +
In the RHS, we are given every value except for <math>AB.</math> However, Ptolemy's Theorem on <math>MBAN</math> gives <math>AB\cdot MN+AN\cdot BM=AM\cdot BN\implies AB+\dfrac{3}{5\sqrt{2}}=\dfrac{4}{5\sqrt{2}}\implies AB=\dfrac{1}{5\sqrt{2}}.</math> Substituting, we get <math>\dfrac{(a+b)(b+c)}{b}=4\implies b(a+b+c)+ac=4b, b=\dfrac{ac}{3}</math> where we use <math>a+b+c=1.</math>
 +
 +
Again using <math>a+b+c=1,</math> we have <math>a+b+c=1\implies a+\dfrac{ac}{3}+c=1\implies a=3\dfrac{1-c}{c+3}.</math> Then <math>b=\dfrac{ac}{3}=\dfrac{c-c^2}{c+3}.</math> Since this is a function in <math>c,</math> we differentiate WRT <math>c</math> to find its maximum. By quotient rule, it suffices to solve <cmath>(-2c+1)(c+3)-(c-c^2)=0 \implies c^2+6c-3,c=-3+2\sqrt{3}.</cmath> Substituting back yields <math>b=7-4\sqrt{3},</math> so <math>7+4+3=\boxed{014}</math> is the answer.
 +
 +
~Generic_Username
 +
 +
 +
==Video Solution==
 +
https://youtu.be/4OZyKVD05Zg?si=yg1ndnP_GperfUx6
 +
 +
~MathProblemSolvingSkills.com
 +
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2009|n=II|num-b=14|after=Last Problem}}
 
{{AIME box|year=2009|n=II|num-b=14|after=Last Problem}}
 +
[[Category: Intermediate Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 20:23, 23 May 2024

Problem

Let $\overline{MN}$ be a diameter of a circle with diameter 1. Let $A$ and $B$ be points on one of the semicircular arcs determined by $\overline{MN}$ such that $A$ is the midpoint of the semicircle and $MB=\frac{3}5$. Point $C$ lies on the other semicircular arc. Let $d$ be the length of the line segment whose endpoints are the intersections of diameter $\overline{MN}$ with chords $\overline{AC}$ and $\overline{BC}$. The largest possible value of $d$ can be written in the form $r-s\sqrt{t}$, where $r, s$ and $t$ are positive integers and $t$ is not divisible by the square of any prime. Find $r+s+t$.

Solutions

Solution 1 (Quick Calculus)

Let $V = \overline{NM} \cap \overline{AC}$ and $W = \overline{NM} \cap \overline{BC}$. Further more let $\angle NMC = \alpha$ and $\angle MNC = 90^\circ - \alpha$. Angle chasing reveals $\angle NBC = \angle NAC = \alpha$ and $\angle MBC = \angle MAC = 90^\circ - \alpha$. Additionally $NB = \frac{4}{5}$ and $AN = AM$ by the Pythagorean Theorem.

By the Angle Bisector Formula, \[\frac{NV}{MV} = \frac{\sin (\alpha)}{\sin (90^\circ - \alpha)} = \tan (\alpha)\] \[\frac{MW}{NW} = \frac{3\sin (90^\circ - \alpha)}{4\sin (\alpha)} = \frac{3}{4} \cot (\alpha)\]

As $NV + MV =MW + NW = 1$ we compute $NW = \frac{1}{1+\frac{3}{4}\cot(\alpha)}$ and $MV = \frac{1}{1+\tan (\alpha)}$, and finally $VW = NW + MV - 1 = \frac{1}{1+\frac{3}{4}\cot(\alpha)} + \frac{1}{1+\tan (\alpha)} - 1$. Taking the derivative of $VW$ with respect to $\alpha$, we arrive at \[VW' = \frac{7\cos^2 (\alpha) - 4}{(\sin(\alpha) + \cos(\alpha))^2(4\sin(\alpha)+3\cos(\alpha))^2}\] Clearly the maximum occurs when $\alpha = \cos^{-1}\left(\frac{2}{\sqrt{7}}\right)$. Plugging this back in, using the fact that $\tan(\cos^{-1}(x)) = \frac{\sqrt{1-x^2}}{x}$ and $\cot(\cos^{-1}(x)) = \frac{x}{\sqrt{1-x^2}}$, we get

$VW = 7 - 4\sqrt{3}$ with $7 + 4 + 3 = \boxed{014}$

~always_correct

Solution 2 (Projective)

Since $MA = \frac{\sqrt{2}}{2} \approx 0.707 > \frac{3}{5}$, point $B$ lies between $M$ and $A$ on the semicircular arc. We will first compute the length of $\overline{AB}$. By the law of cosines, $\cos \angle MOB = \frac{-(3/5)^2 + 2(1/2)^2}{2(1/2)^2} = \frac{7}{25}$, so $\cos \angle AOB = \sin \angle MOB = \frac{24}{25}$. Then $AB^2 = 2\left(\frac{1}{2}\right)^2 - 2\left(\frac{1}{2}\right)^2 \cdot \frac{24}{25} = \frac{1}{50}$, so $AB = \frac{1}{5\sqrt{2}}$.

Let $P = AC \cap MN$ and $Q = BC \cap MN$, and let $MQ = x$, $PQ = d$, $PN = y$. Note that\[(M, P; Q, N) \stackrel{C}{=} (M, A; B, N),\]that is,\[\frac{QP}{QM} \div \frac{NP}{NM} = \frac{BA}{BM} \div \frac{NA}{NM}\]or\[\frac{d}{xy} = \frac{1/(5\sqrt{2})}{(3/5) \cdot (\sqrt{2}/2)} = \frac{1}{3}.\]Hence $d = \frac{1}{3}xy$, and we also know $d+x+y=1$. Now AM-GM gives\[\frac{x+y}{2} \ge \sqrt{xy} \implies \frac{1-d}{2} \ge \sqrt{3d}.\]This gives the quadratic inequality $d^2 - 14d + 1 \ge 0$, which solves as\[d \in \left(-\infty, 7-4\sqrt3\right] \cup \left[7+4\sqrt3, \infty\right).\]But $d \le 1$, so the greatest possible value of $d$ is $7-4\sqrt3$. The answer is $7+4+3=\boxed{014}$.

~MSTang

Solution 3 (Calculus)

Let $O$ be the center of the circle. Define $\angle{MOC}=t$, $\angle{BOA}=2a$, and let $BC$ and $AC$ intersect $MN$ at points $X$ and $Y$, respectively. We will express the length of $XY$ as a function of $t$ and maximize that function in the interval $[0, \pi]$.

Let $C'$ be the foot of the perpendicular from $C$ to $MN$. We compute $XY$ as follows.

(a) By the Extended Law of Sines in triangle $ABC$, we have

\[CA\]

\[= \sin\angle{ABC}\]

\[= \sin\left(\frac{\widehat{AN} + \widehat{NC}}{2}\right)\]

\[= \sin\left(\frac{\frac{\pi}{2} + (\pi-t)}{2}\right)\]

\[= \sin\left(\frac{3\pi}{4} - \frac{t}{2}\right)\]

\[= \sin\left(\frac{\pi}{4} + \frac{t}{2}\right)\]

(b) Note that $CC' = CO\sin(t) = \left(\frac{1}{2}\right)\sin(t)$ and $AO = \frac{1}{2}$. Since $CC'Y$ and $AOY$ are similar right triangles, we have $CY/AY = CC'/AO = \sin(t)$, and hence,

\[CY/CA\]

\[= \frac{CY}{CY + AY}\]

\[= \frac{\sin(t)}{1 + \sin(t)}\]

\[= \frac{\sin(t)}{\sin\left(\frac{\pi}{2}\right) + \sin(t)}\]

\[= \frac{\sin(t)}{2\sin\left(\frac{\pi}{4} + \frac{t}{2}\right)\cos\left(\frac{\pi}{4} - \frac{t}{2}\right)}\]

(c) We have $\angle{XCY} = \frac{\widehat{AB}}{2}=a$ and $\angle{CXY} = \frac{\widehat{MB}+\widehat{CN}}{2} = \frac{\left(\frac{\pi}{2} - 2a\right) + (\pi - t)}{2} = \frac{3\pi}{4} - a - \frac{t}{2}$, and hence by the Law of Sines,

\[XY/CY\]

\[= \frac{\sin\angle{XCY}}{\sin\angle{CXY}}\]

\[= \frac{\sin(a)}{\sin\left(\frac{3\pi}{4} - a - \frac{t}{2}\right)}\]

\[= \frac{\sin(a)}{\sin\left(\frac{\pi}{4} + a + \frac{t}{2}\right)}\]

(d) Multiplying (a), (b), and (c), we have

\[XY\]

\[= CA * (CY/CA) * (XY/CY)\]

\[= \frac{\sin(t)\sin(a)}{2\cos\left(\frac{\pi}{4} - \frac{t}{2}\right)\sin\left(\frac{\pi}{4} + a + \frac{t}{2}\right)}\]

\[= \frac{\sin(t)\sin(a)}{\sin\left(\frac{\pi}{2} + a\right) + \sin(a + t)}\]

\[= \sin(a)\times\frac{\sin(t)}{\sin(t + a) + \cos(a)}\],

which is a function of $t$ (and the constant $a$). Differentiating this with respect to $t$ yields

\[\sin(a)\times\frac{\cos(t)(\sin(t + a) + \cos(a)) - \sin(t)\cos(t + a)}{(\sin(t + a) + \cos(a))^2}\],

and the numerator of this is

\[\sin(a) \times(\sin(t + a)\cos(t) - \cos(t + a)\sin(t) + \cos(a)\cos(t))\] \[= \sin(a) \times (\sin(a) + \cos(a)\cos(t))\],

which vanishes when $\sin(a) + \cos(a)\cos(t) = 0$. Therefore, the length of $XY$ is maximized when $t=t'$, where $t'$ is the value in $[0, \pi]$ that satisfies $\cos(t') = -\tan(a)$.

Note that

\[\frac{1 - \tan(a)}{1 + \tan(a)} = \tan\left(\frac{\pi}{4} - a\right) = \tan((\widehat{MB})/2) = \tan\angle{MNB} = \frac{3}{4}\],

so $\tan(a) = \frac{1}{7}$. We compute

\[\sin(a) = \frac{\sqrt{2}}{10}\]

\[\cos(a) = \frac{7\sqrt{2}}{10}\]

\[\cos(t') = -\tan(a) = -\frac{1}{7}\]

\[\sin(t') = \frac{4\sqrt{3}}{7}\]

\[\sin(t' + a)=\sin(t')\cos(a) + \cos(t')\sin(a) = \frac{28\sqrt{6} - \sqrt{2}}{70}\],

so the maximum length of $XY$ is $\sin(a)\times\frac{\sin(t')}{\sin(t' + a) + \cos(a)} = 7 - 4\sqrt{3}$, and the answer is $7 + 4 + 3 = \boxed{014}$.

Solution 4

[asy] unitsize(144); pair A, B, C, M, n; A = (0,1); B = (-7/25, 24/25); C=(1/7,-4*sqrt(3)/7); M = (-1,0); n = (1,0); pair [] D = intersectionpoints(A--C,M--n); pair [] e = intersectionpoints(B--C,M--n); draw(circle((0,0),1)); draw(M--n--B--M--A--n--C--A--B--C--cycle); label("$A$",A,N); label("$B$",B,NNW); label("$M$",M,W); label("$C$",C,SSE); label("$N$",n,E); label("$D$",D[0],SE); label("$E$",e[0],SW); label("$x$",(M+C)/2,SW); label("$y$",(n+C)/2,SE); [/asy]

Suppose $\overline{AC}$ and $\overline{BC}$ intersect $\overline{MN}$ at $D$ and $E$, respectively, and let $MC = x$ and $NC = y$. Since $A$ is the midpoint of arc $MN$, $\overline{CA}$ bisects $\angle MCN$, and we get \[\frac{MC}{MD} = \frac{NC}{ND}\Rightarrow MD = \frac{x}{x + y}.\] To find $ME$, we note that $\triangle BNE\sim\triangle MCE$ and $\triangle BME\sim\triangle NCE$, so

\begin{align*} \frac{BN}{NE} &= \frac{MC}{CE} \\ \frac{ME}{BM} &= \frac{CE}{NC}. \end{align*}

Writing $NE = 1 - ME$, we can substitute known values and multiply the equations to get

\[\frac{4(ME)}{3 - 3(ME)} = \frac{x}{y}\Rightarrow ME = \frac{3x}{3x + 4y}.\]

The value we wish to maximize is

\begin{align*} DE &= MD - ME \\ &= \frac{x}{x + y} - \frac{3x}{3x + 4y} \\ &= \frac{xy}{3x^2 + 7xy + 4y^2} \\ &= \frac{1}{3(x/y) + 4(y/x) + 7}. \end{align*}

By the AM-GM inequality, $3(x/y) + 4(y/x)\geq 2\sqrt{12} = 4\sqrt{3}$, so \[DE\leq \frac{1}{4\sqrt{3} + 7} = 7 - 4\sqrt{3},\] giving the answer of $7 + 4 + 3 = \boxed{014}$. Equality is achieved when $3(x/y) = 4(y/x)$ subject to the condition $x^2 + y^2 = 1$, which occurs for $x = \frac{2\sqrt{7}}{7}$ and $y = \frac{\sqrt{21}}{7}$.

Solution 5 (Projective)

By Pythagoras in $\triangle BMN,$ we get $BN=\dfrac{4}{5}.$

Since cross ratios are preserved upon projecting, note that $(M,Y;X,N)\stackrel{C}{=}(M,B;A,N).$ By definition of a cross ratio, this becomes \[\dfrac{XM}{NY}:\dfrac{NM}{NY}=\dfrac{AM}{AB}:\dfrac{MN}{NB}.\] Let $MY=a,YX=b,XN=c$ such that $a+b+c=1.$ We know that $XM=a+b,XY=b,NM=1,NY=b+c,$ so the LHS becomes $\dfrac{(a+b)(b+c)}{b}.$

In the RHS, we are given every value except for $AB.$ However, Ptolemy's Theorem on $MBAN$ gives $AB\cdot MN+AN\cdot BM=AM\cdot BN\implies AB+\dfrac{3}{5\sqrt{2}}=\dfrac{4}{5\sqrt{2}}\implies AB=\dfrac{1}{5\sqrt{2}}.$ Substituting, we get $\dfrac{(a+b)(b+c)}{b}=4\implies b(a+b+c)+ac=4b, b=\dfrac{ac}{3}$ where we use $a+b+c=1.$

Again using $a+b+c=1,$ we have $a+b+c=1\implies a+\dfrac{ac}{3}+c=1\implies a=3\dfrac{1-c}{c+3}.$ Then $b=\dfrac{ac}{3}=\dfrac{c-c^2}{c+3}.$ Since this is a function in $c,$ we differentiate WRT $c$ to find its maximum. By quotient rule, it suffices to solve \[(-2c+1)(c+3)-(c-c^2)=0 \implies c^2+6c-3,c=-3+2\sqrt{3}.\] Substituting back yields $b=7-4\sqrt{3},$ so $7+4+3=\boxed{014}$ is the answer.

~Generic_Username


Video Solution

https://youtu.be/4OZyKVD05Zg?si=yg1ndnP_GperfUx6

~MathProblemSolvingSkills.com


See Also

2009 AIME II (Problems • Answer Key • Resources)
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