AIME 2009 II · 第 15 题

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

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

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

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

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Video Solution

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

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