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2006 AMC 12B Problems/Problem 23

Revision as of 18:19, 21 October 2021 by Shihan (talk | contribs) (→‎Solution 2)

Problem

Isosceles $\triangle ABC$ has a right angle at $C$. Point $P$ is inside $\triangle ABC$, such that $PA=11$, $PB=7$, and $PC=6$. Legs $\overline{AC}$ and $\overline{BC}$ have length $s=\sqrt{a+b\sqrt{2}}$, where $a$ and $b$ are positive integers. What is $a+b$?

[asy] pathpen = linewidth(0.7); pen f = fontsize(10); size(5cm); pair B = (0,sqrt(85+42*sqrt(2))); pair A = (B.y,0); pair C = (0,0); pair P = IP(arc(B,7,180,360),arc(C,6,0,90)); D(A--B--C--cycle); D(P--A); D(P--B); D(P--C); MP("A",D(A),plain.E,f); MP("B",D(B),plain.N,f); MP("C",D(C),plain.SW,f); MP("P",D(P),plain.NE,f); [/asy]

$\mathrm{(A)}\ 85 \qquad \mathrm{(B)}\ 91 \qquad \mathrm{(C)}\ 108 \qquad \mathrm{(D)}\ 121 \qquad \mathrm{(E)}\ 127$

Solution

[asy] pathpen = linewidth(0.7); pen f = fontsize(10); size(5cm); pair B = (0,sqrt(85+42*sqrt(2))); pair A = (B.y,0); pair C = (0,0); pair P = IP(arc(B,7,180,360),arc(C,6,0,90)); D(A--B--C--cycle); D(P--A); D(P--B); D(P--C); MP("A",D(A),plain.E,f); MP("B",D(B),plain.N,f); MP("C",D(C),plain.SW,f); MP("P",D(P),plain.NE,f); MP("\alpha",C,5*dir(80),f); MP("90^\circ-\alpha",C,3*dir(30),f); MP("s",(A+C)/2,plain.S,f); MP("s",(B+C)/2,plain.W,f); [/asy] Using the Law of Cosines on $\triangle PBC$, we have:

\begin{align*} PB^2&=BC^2+PC^2-2\cdot BC\cdot PC\cdot \cos(\alpha) \Rightarrow 49 = 36 + s^2 - 12s\cos(\alpha) \Rightarrow \cos(\alpha) = \dfrac{s^2-13}{12s}. \end{align*}

Using the Law of Cosines on $\triangle PAC$, we have: \begin{align*} PA^2&=AC^2+PC^2-2\cdot AC\cdot PC\cdot \cos(90^\circ-\alpha) \Rightarrow 121 = 36 + s^2 - 12s\sin(\alpha) \Rightarrow \sin(\alpha) = \dfrac{s^2-85}{12s}. \end{align*}

Now we use $\sin^2(\alpha) + \cos^2(\alpha) = 1$. \begin{align*} \sin^2(\alpha)+\cos^2(\alpha) = 1 &\Rightarrow \frac{s^4-26s^2+169}{144s^2} + \frac{s^4-170s^2+7225}{144s^2} = 1 \\ &\Rightarrow 2s^4-340s^2+7394 = 0 \\ &\Rightarrow s^4-170s^2+3697 = 0 \\ &\Rightarrow s^2 = \dfrac{170 \pm \sqrt{170^2 - 4\cdot3697}}{2}\\ &\Rightarrow s^2 = \dfrac{170 \pm \sqrt{28900 - 14788}}{2}\\ &\Rightarrow s^2 = \dfrac{170 \pm \sqrt{14112}}{2}\\ &\Rightarrow s^2 = \dfrac{170 \pm \sqrt{2^5\cdot3^2\cdot7^2}}{2}\\ &\Rightarrow s^2 = \dfrac{170 \pm 84\sqrt{2}}{2} = 85 \pm 42\sqrt2 \end{align*}

Note that we know that we want the solution with $s^2 > 85$ since we know that $\sin(\alpha) > 0$. Thus, $a+b=85+42=\boxed{127}$.

Solution 2

Rotate triangle $PAC$ 90 degrees counterclockwise about $C$ so that the image of $A$ rests on $B$. Now let the image of $P$ be $P'$. [asy] pathpen = linewidth(0.7); pen f = fontsize(10); size(10cm); pair B = (0,sqrt(85+42*sqrt(2))); pair A = (B.y,0); pair C = (0,0); pair P = IP(arc(B,7,180,360),arc(C,6,0,90)); draw(A--B--C--cycle, black+0.8); D(P--A); D(P--B); D(P--C); MP("A",D(A),E,f); MP("B, \widetilde{A}",D(B),N,f); MP("C",D(C),S,f); MP("P",D(P),NE,f); pair Bp = dir(90)*(B-origin); pair Pp = dir(90)*(P-origin); D(B--Bp--C--Pp--B); MP("\widetilde{P}",D(Pp),SW,f); MP("\widetilde{B}",D(Bp),W,f); [/asy] Note that $P'C=6$, meaning triangle $PCP'$ is right isosceles, and $\angle PP'C=45^\circ$. Then $PP'=6\sqrt{2}$. Now because $PB=7$ and $P'B=11$, we observe that $\angle P'PB=90^\circ$, by the Pythagorean Theorem on $P'PB$. Now we have that $\angle APC=\angle BP'C=\angle BP'P + \angle PP'C$. So we take the cosine of the second equality, using that fact that $\angle PP'C=45^\circ$, to get $\cos(BP'C)=\frac{6\sqrt{2}-7}{11\sqrt{2}}$. Finally, we use the fact that $\cos(BP'C)=\cos(CPA)$ and use the Law of Cosines on triangle $CPA$ to arrive at the value of $s^2$.

Or notice that since $\angle P'PB=90^\circ$ and $\angle PP'C=45^\circ$, we have $\angle BPC=135^\circ$, and Law of Cosines on triangle $BPC$ gives the value of $s^2$.

Solution 3 (coordinate bash)

Let point $P$ have coordinates $(x,y)$ and $C$ have coordinates $(0,0).$ Then, $A$ has $(s,0)$ and $B$ has $(0,s)$.

By distance formula, we have \[x^2+y^2=36 \tag{1}.\] \[x^2+(y-s)^2=49 \tag{2}.\] \[(x-s)^2+y^2=121 \tag{3}\]

Expanding $(2)$ and $(3)$ gives \[x^2+y^2-2ys+s^2=49,\] and \[x^2+y^2-2sx+s^2=121,\] respectively. Then, using equation $(1)$, we have that \[-2ys+s^2=49-36=13,\] and \[-2sx+s^2=121-36=85.\]

Then, solving for $x$ and $y$ gives \[x=\frac{85-s^2}{-2s}=\frac{s^2-85}{2s},\] and \[y=\frac{13-s^2}{-2s}=\frac{s^2-13}{2s}.\] Plugging these values of $x$ and $y$ into equation $(1)$ yields \[\left(\frac{s^2-85}{2s}\right)^2+\left(\frac{s^2-13}{2s}\right)^2=36.\]

We multiply both sides by $4s^2$ and expand, yielding the equation \[s^4-170s^2+7225+s^4-26s^2+169=144s^2.\] Simplifying gives the equation \[2s^4-340s^2+7394=0.\] Solving this quadratic gives $s^2=85\pm 42\sqrt{2}.$ Now, if this were the actual test, we stop here, noting that the question tells us $a$ and $b$ are positive, so $s^2$ must be $85+42\sqrt{2}$, and our solution is $127$.

However, here is why $s^2$ cannot be $85-42\sqrt{2}$:

If $s^2=85-42\sqrt{2}$, using $1.4$ as an approximation for $\sqrt{2}$, $s^2\approx 85-42\cdot 1.4=85-58.8=26.2$, and $s$ is slightly greater than $5$. Also note that this implies that $AB\approx 7$.

Note that at least one of $\angle BPA$, $\angle APC$ and $\angle BPC$ must be obtuse, since they sum to $360^{\circ}$. Then, note the well known fact that if $\angle A$ is the largest angle in $\Delta ABC$, $BC$ must be the largest side. However, combined with the first fact, implies that either $AB$ is the largest side of $\Delta ABP$, $BC$ is the largest side of $\Delta BPC$, or $AC$ is the largest side of $\Delta APC$. By our approximations, this cannot possibly be true, even if we are generous with our margin of error, so $s^2$ cannot equal $85-42\sqrt{2}$, and $s^2=85+42\sqrt{2}$.

The answer is $85+42=\boxed{127}$

See also

2006 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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