Difference between revisions of "2006 AMC 12B Problems/Problem 23"
M1sterzer0 (talk | contribs) (→Problem) |
(→Solution) |
||
(14 intermediate revisions by 9 users not shown) | |||
Line 1: | Line 1: | ||
− | |||
− | |||
== Problem == | == Problem == | ||
− | Isosceles <math>\triangle ABC</math> has a right angle at <math>C</math>. Point <math>P</math> is inside <math>\triangle ABC</math>, such that <math>PA=11</math>, <math>PB=7</math>, and <math>PC=6</math>. Legs <math>\overline{AC}</math> and <math>\overline{BC}</math> have length <math>s=\sqrt{a+b\sqrt{2} | + | Isosceles <math>\triangle ABC</math> has a right angle at <math>C</math>. Point <math>P</math> is inside <math>\triangle ABC</math>, such that <math>PA=11</math>, <math>PB=7</math>, and <math>PC=6</math>. Legs <math>\overline{AC}</math> and <math>\overline{BC}</math> have length <math>s=\sqrt{a+b\sqrt{2}}</math>, where <math>a</math> and <math>b</math> are positive integers. What is <math>a+b</math>? |
<asy> | <asy> | ||
Line 35: | Line 33: | ||
== Solution == | == 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 <math>\triangle PBC</math>, we have: | ||
+ | |||
+ | <cmath> | ||
+ | \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*} | ||
+ | </cmath> | ||
+ | |||
+ | Using the Law of Cosines on <math>\triangle PAC</math>, we have: | ||
+ | <cmath> | ||
+ | \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*} | ||
+ | </cmath> | ||
+ | |||
+ | Now we use <math>\sin^2(\alpha) + \cos^2(\alpha) = 1</math>. | ||
+ | <cmath> | ||
+ | \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 s^4-170s^2+3697 = 0 \\ | ||
+ | &\Rightarrow s^2 = \dfrac{170 \pm 84\sqrt{2}}{2} = 85 \pm 42\sqrt2 | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | Note that we know that we want the solution with <math>s^2 > 85</math> since we know that <math>\sin(\alpha) > 0</math>. Thus, <math>a+b=85+42=\boxed{127}</math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | Rotate triangle <math>PAC</math> 90 degrees counterclockwise about <math>C</math> so that the image of <math>A</math> rests on <math>B</math>. Now let the image of <math>P</math> be <math>\widetilde{P}</math>. | ||
+ | <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 <math>\widetilde{P}C=6</math>, meaning triangle <math>PC\widetilde{P}</math> is right isosceles, and <math>\angle P\widetilde{P}C=45^\circ</math>. Then <math>P\widetilde{P}=6\sqrt{2}</math>. Now because <math>PB=7</math> and <math>\widetilde{P}B=11</math>, we observe that <math>\angle \widetilde{P}PB=90^\circ</math>, by the Pythagorean Theorem on <math>\widetilde{P}PB</math>. Now we have that <math>\angle APC=\angle B\widetilde{P}C=\angle B\widetilde{P}P + \angle P\widetilde{P}C</math>. So we take the cosine of the second equality, using that fact that <math>\angle P\widetilde{P}C=45^\circ</math>, to get <math>\cos(B\widetilde{P}C)=\frac{6\sqrt{2}-7}{11\sqrt{2}}</math>. Finally, we use the fact that <math>\cos(B\widetilde{P}C)=\cos(CPA)</math> and use the Law of Cosines on triangle <math>CPA</math> to arrive at the value of <math>s^2</math>. | ||
+ | |||
+ | Or notice that since <math>\angle \widetilde{P}PB=90^\circ</math> and <math>\angle P\widetilde{P}C=45^\circ</math>, we have <math>\angle BPC=135^\circ</math>, and Law of Cosines on triangle <math>BPC</math> gives the value of <math>s^2</math>. | ||
+ | |||
+ | ==Solution 3 (coordinate bash)== | ||
+ | |||
+ | Let point <math>P</math> have coordinates <math>(x,y)</math> and <math>C</math> have coordinates <math>(0,0).</math> Then, <math>A</math> has <math>(s,0)</math> and <math>B</math> has <math>(0,s)</math>. | ||
+ | |||
+ | By distance formula, we have | ||
+ | <cmath>x^2+y^2=36 \tag{1}.</cmath> | ||
+ | <cmath>x^2+(y-s)^2=49 \tag{2}.</cmath> | ||
+ | <cmath>(x-s)^2+y^2=121 \tag{3}</cmath> | ||
+ | |||
+ | Expanding <math>(2)</math> and <math>(3)</math> gives <cmath>x^2+y^2-2ys+s^2=49,</cmath> and | ||
+ | <cmath>x^2+y^2-2sx+s^2=121,</cmath> respectively. Then, using equation <math>(1)</math>, we have that <cmath>-2ys+s^2=49-36=13,</cmath> and <cmath>-2sx+s^2=121-36=85.</cmath> | ||
+ | |||
+ | Then, solving for <math>x</math> and <math>y</math> gives | ||
+ | <cmath>x=\frac{85-s^2}{-2s}=\frac{s^2-85}{2s},</cmath> and | ||
+ | <cmath>y=\frac{13-s^2}{-2s}=\frac{s^2-13}{2s}.</cmath> | ||
+ | Plugging these values of <math>x</math> and <math>y</math> into equation <math>(1)</math> yields <cmath>\left(\frac{s^2-85}{2s}\right)^2+\left(\frac{s^2-13}{2s}\right)^2=36.</cmath> | ||
+ | |||
+ | We multiply both sides by <math>4s^2</math> and expand, yielding the equation <cmath>s^4-170s^2+7225+s^4-26s^2+169=144s^2.</cmath> Simplifying gives the equation <cmath>2s^4-340s^2+7394=0.</cmath> Solving this quadratic gives <math>s^2=85\pm 42\sqrt{2}.</math> Now, if this were the actual test, we stop here, noting that the question tells us <math>a</math> and <math>b</math> are positive, so <math>s^2</math> must be <math>85+42\sqrt{2}</math>, and our solution is <math>127</math>. | ||
+ | |||
+ | However, here is why <math>s^2</math> cannot be <math>85-42\sqrt{2}</math>: | ||
+ | |||
+ | If <math>s^2=85-42\sqrt{2}</math>, using <math>1.4</math> as an approximation for <math>\sqrt{2}</math>, <math>s^2\approx 85-42\cdot 1.4=85-58.8=26.2</math>, and <math>s</math> is slightly greater than <math>5</math>. Also note that this implies that <math>AB\approx 7</math>. | ||
+ | |||
+ | Note that at least one of <math>\angle BPA</math>, <math>\angle APC</math> and <math>\angle BPC</math> must be obtuse, since they sum to <math>360^{\circ}</math>. Then, note the well known fact that if <math>\angle A</math> is the largest angle in <math>\Delta ABC</math>, <math>BC</math> must be the largest side. However, combined with the first fact, implies that either <math>AB</math> is the largest side of <math>\Delta ABP</math>, <math>BC</math> is the largest side of <math>\Delta BPC</math>, or <math>AC</math> is the largest side of <math>\Delta APC</math>. By our approximations, this cannot possibly be true, even if we are generous with our margin of error, so <math>s^2</math> cannot equal <math>85-42\sqrt{2}</math>, and <math>s^2=85+42\sqrt{2}</math>. | ||
+ | |||
+ | The answer is <math>85+42=\boxed{127}</math> | ||
== See also == | == See also == | ||
{{AMC12 box|year=2006|ab=B|num-b=22|num-a=24}} | {{AMC12 box|year=2006|ab=B|num-b=22|num-a=24}} | ||
+ | {{MAA Notice}} |
Latest revision as of 14:04, 10 June 2022
Problem
Isosceles has a right angle at
. Point
is inside
, such that
,
, and
. Legs
and
have length
, where
and
are positive integers. What is
?
Solution
Using the Law of Cosines on
, we have:
Using the Law of Cosines on , we have:
Now we use .
Note that we know that we want the solution with since we know that
. Thus,
.
Solution 2
Rotate triangle 90 degrees counterclockwise about
so that the image of
rests on
. Now let the image of
be
.
Note that
, meaning triangle
is right isosceles, and
. Then
. Now because
and
, we observe that
, by the Pythagorean Theorem on
. Now we have that
. So we take the cosine of the second equality, using that fact that
, to get
. Finally, we use the fact that
and use the Law of Cosines on triangle
to arrive at the value of
.
Or notice that since and
, we have
, and Law of Cosines on triangle
gives the value of
.
Solution 3 (coordinate bash)
Let point have coordinates
and
have coordinates
Then,
has
and
has
.
By distance formula, we have
Expanding and
gives
and
respectively. Then, using equation
, we have that
and
Then, solving for and
gives
and
Plugging these values of
and
into equation
yields
We multiply both sides by and expand, yielding the equation
Simplifying gives the equation
Solving this quadratic gives
Now, if this were the actual test, we stop here, noting that the question tells us
and
are positive, so
must be
, and our solution is
.
However, here is why cannot be
:
If , using
as an approximation for
,
, and
is slightly greater than
. Also note that this implies that
.
Note that at least one of ,
and
must be obtuse, since they sum to
. Then, note the well known fact that if
is the largest angle in
,
must be the largest side. However, combined with the first fact, implies that either
is the largest side of
,
is the largest side of
, or
is the largest side of
. By our approximations, this cannot possibly be true, even if we are generous with our margin of error, so
cannot equal
, and
.
The answer is
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.