Difference between revisions of "2011 AIME I Problems/Problem 4"
(→Solution 2) |
(soln 5) |
||
(34 intermediate revisions by 16 users not shown) | |||
Line 1: | Line 1: | ||
− | == Problem | + | == Problem == |
In triangle <math>ABC</math>, <math>AB=125</math>, <math>AC=117</math> and <math>BC=120</math>. The angle bisector of angle <math>A</math> intersects <math> \overline{BC} </math> at point <math>L</math>, and the angle bisector of angle <math>B</math> intersects <math> \overline{AC} </math> at point <math>K</math>. Let <math>M</math> and <math>N</math> be the feet of the perpendiculars from <math>C</math> to <math> \overline{BK}</math> and <math> \overline{AL}</math>, respectively. Find <math>MN</math>. | In triangle <math>ABC</math>, <math>AB=125</math>, <math>AC=117</math> and <math>BC=120</math>. The angle bisector of angle <math>A</math> intersects <math> \overline{BC} </math> at point <math>L</math>, and the angle bisector of angle <math>B</math> intersects <math> \overline{AC} </math> at point <math>K</math>. Let <math>M</math> and <math>N</math> be the feet of the perpendiculars from <math>C</math> to <math> \overline{BK}</math> and <math> \overline{AL}</math>, respectively. Find <math>MN</math>. | ||
− | |||
== Solution 1 == | == Solution 1 == | ||
− | Extend <math>{CM}</math> and <math>{CN}</math> such that they | + | Extend <math>{CM}</math> and <math>{CN}</math> such that they intersect line <math>{AB}</math> at points <math>P</math> and <math>Q</math>, respectively. |
− | Since <math>{BM}</math> is the angle bisector of angle B | + | <asy> |
− | Hence<math>MN=\ | + | defaultpen(fontsize(10)+0.8); size(200); |
+ | pen p=fontsize(9)+linewidth(3); | ||
+ | pair A,B,C,D,K,L,M,N,P,Q; | ||
+ | A=origin; B=(125,0); C=IP(CR(A,117),CR(B,120)); L=extension(B,C,A,bisectorpoint(B,A,C)); K=extension(A,C,B,bisectorpoint(C,B,A)); M=foot(C,B,K); N=foot(C,A,L); | ||
+ | draw(A--B--C--A); draw(A--L^^B--K, gray+dashed+0.5); draw(M--C--N^^N--extension(A,B,C,N)^^M--extension(A,B,C,M), gray+0.5); | ||
+ | dot("$A$",A,dir(200),p); dot("$B$",B,right,p); dot("$C$",C,up,p); dot("$L$",L,2*dir(70),p); dot("$N$",N,2*dir(-90),p); dot("$M$",M,2*dir(-90),p); dot("$P$",extension(A,B,C,M),2*down,p); dot("$Q$",extension(A,B,C,N),2*down,p); | ||
+ | label("$125$",A--B,down,fontsize(10)); label("$117$",A--C,2*dir(130),fontsize(10)); label("$120$",B--C,1.5*dir(30),fontsize(10)); | ||
+ | </asy> | ||
+ | Since <math>{BM}</math> is the angle bisector of angle <math>B</math> and <math>{CM}</math> is perpendicular to <math>{BM}</math>, <math>\triangle BCP</math> must be an isoceles triangle, so <math>BP=BC=120</math>, and <math>M</math> is the midpoint of <math>{CP}</math>. For the same reason, <math>AQ=AC=117</math>, and <math>N</math> is the midpoint of <math>{CQ}</math>. | ||
+ | Hence <math>MN=\tfrac 12 PQ</math>. Since <cmath>PQ=BP+AQ-AB=120+117-125=112,</cmath> so <math>MN=\boxed{056}</math>. | ||
== Solution 2 == | == Solution 2 == | ||
− | + | Let <math>I</math> be the incenter of <math>ABC</math>. Since <math>I</math> lies on <math>BM</math> and <math>AN</math>, <math>IM \perp MC</math> and <math>IN \perp NC</math>, so <math>\angle IMC + \angle INC = 180^\circ</math>. This means that <math>CMIN</math> is a cyclic quadrilateral. By the Law of Sines, <math>\frac{MN}{\sin \angle MIN} = \frac{2R}{\sin \angle CMI} = 2R = CI</math>, where <math>R</math> is the radius of the circumcircle of <math>CMIN</math>. Since <math>\sin \angle MIN = \sin \angle BIA = \sin (90^\circ + \tfrac 12 \angle BCA) = \cos \tfrac 12 \angle BCA = \cos \angle BCI</math>, we have that <math>MN = CI \cdot \sin \angle MIN = CI \cdot \cos \angle BCI</math>. Letting <math>H</math> be the point of contact of the incircle of <math>ABC</math> with side <math>BC</math>, we have <math>MN = CI \cdot \cos \angle BCI = CI \cdot \frac{CH}{CI} = CH</math>. Thus, <math>MN = s - AB = \frac{117+120-125}{2}=\boxed{056}</math>. | |
− | Let <math>I</math> be the incenter of <math>ABC</math>. | ||
== Solution 3 (Bash) == | == Solution 3 (Bash) == | ||
− | Project <math>I</math> onto <math>AC</math> and <math>BC</math> as <math>D</math> and <math>E</math>. <math>ID</math> and <math>IE</math> are both in-radii of <math>\triangle ABC</math> so we get right triangles with legs <math>r</math> (the in-radius length) and <math>s - c = 56</math>. Since <math>IC</math> is the hypotenuse for the 4 triangles (<math>\triangle INC, \triangle IMC, \triangle IDC,</math> and <math>\triangle IEC</math>), <math>C, D, M, I, N, E</math> are con-cyclic on a circle we shall denote as <math>\omega</math> which is also the circumcircle of <math>\triangle CMN</math> and <math>\triangle CDE</math>. To find <math>MN</math>, we can use the Law of Cosines on <math>\angle MON \implies MN^2 = 2R^2(1 - \cos{2\angle MCN})</math> where <math>O</math> is the center of <math>\omega</math>. Now, the circumradius <math>R</math> can be found with Pythagorean Theorem with <math>\triangle CDI</math> or <math>\triangle CEI</math>: <math>r^2 + 56^2 = (2R)^2</math>. To find <math>r</math>, we can use the formula <math>rs = [ABC]</math> and by Heron's, <math>[ABC] = \sqrt{181 \cdot 61 \cdot 56 \cdot 64} \implies r = \sqrt{\frac{61 \cdot 56 \cdot 64}{181}} \implies 2R^2 = \frac{393120}{181}</math>. To find <math>\angle MCN</math>, we can find <math>\angle MIN</math> since <math>\angle MCN = 180 - \angle MIN</math>. <math>\angle MIN = \angle MIC + \angle NIC = 180 - \angle BIC + 180 - \angle AIC = 180 - (180 - \frac{\angle A + \angle C}{2}) + 180 - (180 - \frac{\angle B + \angle C}{2}) = \frac{\angle A + \angle B + \angle C}{2} + \frac{\angle C}{2}</math>. Thus, <math>\angle MCN = 180 - \frac{\angle A + \angle B + \angle C}{2} - \frac{\angle C}{2}</math> and since <math>\angle A + \angle B + \angle C = 180</math>, we have <math>\angle A + \angle B + \angle C - \frac{\angle A + \angle B + \angle C}{2} - \frac{\angle C}{2} = \frac{\angle A + \angle B}{2}</math>. Plugging this into our Law of Cosines formula gives <math>MN^2 = 2R^2(1 - \cos{\angle A + \angle B}) = 2R^2(1 + \cos{\angle C})</math>. To find <math>\cos{\angle C}</math>, we use LoC on <math>\triangle ABC \implies cos{\angle C} = \frac{120^2 + 117^2 - 125^2}{2 \cdot 117 \cdot 120} = \frac{41 \cdot 19}{117 \cdot 15}</math>. Our formula now becomes <math>MN^2 = \frac{393120}{181} + \frac{2534}{15 \cdot 117}</math>. After simplifying, we get <math>MN^2 = 3136 \implies MN = \boxed{056}</math>. | + | Project <math>I</math> onto <math>AC</math> and <math>BC</math> as <math>D</math> and <math>E</math>. <math>ID</math> and <math>IE</math> are both in-radii of <math>\triangle ABC</math> so we get right triangles with legs <math>r</math> (the in-radius length) and <math>s - c = 56</math>. Since <math>IC</math> is the hypotenuse for the 4 triangles (<math>\triangle INC, \triangle IMC, \triangle IDC,</math> and <math>\triangle IEC</math>), <math>C, D, M, I, N, E</math> are con-cyclic on a circle we shall denote as <math>\omega</math> which is also the circumcircle of <math>\triangle CMN</math> and <math>\triangle CDE</math>. To find <math>MN</math>, we can use the Law of Cosines on <math>\angle MON \implies MN^2 = 2R^2(1 - \cos{2\angle MCN})</math> where <math>O</math> is the center of <math>\omega</math>. Now, the circumradius <math>R</math> can be found with Pythagorean Theorem with <math>\triangle CDI</math> or <math>\triangle CEI</math>: <math>r^2 + 56^2 = (2R)^2</math>. To find <math>r</math>, we can use the formula <math>rs = [ABC]</math> and by Heron's, <math>[ABC] = \sqrt{181 \cdot 61 \cdot 56 \cdot 64} \implies r = \sqrt{\frac{61 \cdot 56 \cdot 64}{181}} \implies 2R^2 = \frac{393120}{181}</math>. To find <math>\angle MCN</math>, we can find <math>\angle MIN</math> since <math>\angle MCN = 180 - \angle MIN</math>. <math>\angle MIN = \angle MIC + \angle NIC = 180 - \angle BIC + 180 - \angle AIC = 180 - (180 - \frac{\angle A + \angle C}{2}) + 180 - (180 - \frac{\angle B + \angle C}{2}) = \frac{\angle A + \angle B + \angle C}{2} + \frac{\angle C}{2}</math>. Thus, <math>\angle MCN = 180 - \frac{\angle A + \angle B + \angle C}{2} - \frac{\angle C}{2}</math> and since <math>\angle A + \angle B + \angle C = 180</math>, we have <math>\angle A + \angle B + \angle C - \frac{\angle A + \angle B + \angle C}{2} - \frac{\angle C}{2} = \frac{\angle A + \angle B}{2}</math>. Plugging this into our Law of Cosines (LoC) formula gives <math>MN^2 = 2R^2(1 - \cos{\angle A + \angle B}) = 2R^2(1 + \cos{\angle C})</math>. To find <math>\cos{\angle C}</math>, we use LoC on <math>\triangle ABC \implies \cos{\angle C} = \frac{120^2 + 117^2 - 125^2}{2 \cdot 117 \cdot 120} = \frac{41 \cdot 19}{117 \cdot 15}</math>. Our formula now becomes <math>MN^2 = \frac{393120}{181} + \frac{2534}{15 \cdot 117}</math>. After simplifying, we get <math>MN^2 = 3136 \implies MN = \boxed{056}</math>. |
--lucasxia01 | --lucasxia01 | ||
Line 21: | Line 28: | ||
Because <math>\angle CMI = \angle CNI = 90</math>, <math>CMIN</math> is cyclic. | Because <math>\angle CMI = \angle CNI = 90</math>, <math>CMIN</math> is cyclic. | ||
− | Ptolemy on CMIN: | + | Applying Ptolemy's theorem on CMIN: |
− | <math>CN | + | <math>CN \cdot MI+CM \cdot IN=CI \cdot MN</math> |
− | <math>CI^2(\cos \angle ICN \sin \angle ICM + \cos \angle ICM \sin \angle ICN) = CI | + | <math>CI^2(\cos \angle ICN \sin \angle ICM + \cos \angle ICM \sin \angle ICN) = CI \cdot MN</math> |
− | <math>MN = CI \sin \angle MCN</math> by angle addition formula. | + | <math>MN = CI \sin \angle MCN</math> by sine angle addition formula. |
<math>\angle MCN = 180 - \angle MIN = 90 - \angle BCI</math>. | <math>\angle MCN = 180 - \angle MIN = 90 - \angle BCI</math>. | ||
Line 33: | Line 40: | ||
Let <math>H</math> be where the incircle touches <math>BC</math>, then <math>CI \cos \angle BCI = CH = \frac{a+b-c}{2}</math>. | Let <math>H</math> be where the incircle touches <math>BC</math>, then <math>CI \cos \angle BCI = CH = \frac{a+b-c}{2}</math>. | ||
<math>a=120, b=117, c=125</math>, for a final answer of <math>\boxed{056}</math>. | <math>a=120, b=117, c=125</math>, for a final answer of <math>\boxed{056}</math>. | ||
+ | |||
+ | Note: This is similar to Solution 2 after the first four lines | ||
+ | |||
+ | ==Solution 5 (Trig Bash)== | ||
+ | Applying [[Ptolemy's Theorem]] on the cyclic quadrilateral <math>MINC</math>, we find that | ||
+ | |||
+ | <math>MI\cdot CN + IN\cdot MC = MN\cdot IC</math>. | ||
+ | |||
+ | <math>\angle CIN=\frac{\alpha+\gamma}{2}</math> and <math>\angle MIC=\frac{\beta+\gamma}{2}</math> by the Exterior Angle Theorem, so from properties of sine and cosine, we can find that | ||
+ | |||
+ | <math>MI=IC\cdot\cos\left(\frac{\beta+\gamma}{2}\right),</math> <math>MC=IC\cdot\sin\left(\frac{\beta+\gamma}{2}\right),</math> <math>IN=IC\cdot\cos\left(\frac{\alpha+\gamma}{2}\right),</math> <math>CN=IC\cdot\sin\left(\frac{\alpha+\gamma}{2}\right).</math> | ||
+ | |||
+ | Plugging in the values and simplifying results in <math>MN = IC\cdot\sin\left(\frac{\alpha+\beta+2\gamma}{2}\right)</math> by the angle-addition identity <math>\sin(A+B)=\sin(A)\cos(B)+\sin(B)\cos(A)</math>. | ||
+ | |||
+ | Before we continue, we would like to simplify the value in the sine function. We see that <math>\frac{\alpha+\beta+2\gamma}{2}=\frac{\gamma}{2}+\frac{\alpha+\beta+\gamma}{2}=\frac{\gamma}{2}+90</math>. Using the fact that <math>\cos(A)=\sin(90-A)</math> results in | ||
+ | |||
+ | <math>\sin\left(\frac{\alpha+\beta+2\gamma}{2}\right)=\sin\left(90+\frac{\gamma}{2}\right)=\sin\left(90-(-\frac{\gamma}{2})\right)=\cos\left(-\frac{\gamma}{2}\right)=\cos\left(\frac{\gamma}{2}\right).</math> | ||
+ | |||
+ | How do we simplify <math>IC</math>? Well, we can perform the [[Law of Sines]] on triangle <math>AIC</math>. This results in: | ||
+ | |||
+ | <math>\frac{AC}{\sin(\angle AIC)}=\frac{IC}{\sin\left(\frac{\alpha}{2}\right)}</math> | ||
+ | |||
+ | The value of <math>\angle AIC</math> is <math>\frac{\alpha+2\beta+\gamma}{2}</math> by the Exterior Angle Theorem on <math>\triangle ABI</math>, so the value of <math>\sin(\angle AIC)</math> is equivalent to the value of <math>\cos\left(\frac{\beta}{2}\right)</math> by a similar argument as above. Then rearranging yields <math>IC = b\cdot\frac{\sin\left(\frac{\alpha}{2}\right)}{\cos\left(\frac{\beta}{2}\right)}</math>. | ||
+ | |||
+ | Going back to the previous formula <math>MN = IC\cdot\sin\left(\frac{\alpha+\beta+2\gamma}{2}\right)</math> and substituting values yields: | ||
+ | |||
+ | <math>MN = b\cdot\frac{\sin\left(\frac{\alpha}{2}\right)\cos\left(\frac{\gamma}{2}\right)}{\cos\left(\frac{\beta}{2}\right)}</math>. | ||
+ | |||
+ | Finally, using the formulae <math>\sin\left(\frac{\alpha}{2}\right)=\sqrt{\frac{(s-b)(s-c)}{bc}}</math> and <math>\cos\left(\frac{\alpha}{2}\right)=\sqrt{\frac{s(s-a)}{bc}}</math> (where <math>s</math> is half the perimeter of the triangle), we reach our final value: | ||
+ | |||
+ | <math>MN = b\cdot\frac{\sqrt{\frac{(s-b)(s-c)}{bc}}\cdot\sqrt{\frac{s(s-c)}{ab}}}{\sqrt{\frac{s(s-b)}{ac}}}</math> | ||
+ | |||
+ | <math>=\frac{\sqrt{s(s-b)(s-c)^2}}{\sqrt{s(s-b)}}</math> | ||
+ | |||
+ | <math>=\sqrt{(s-c)^2}</math> | ||
+ | |||
+ | <math>=s-c</math> | ||
+ | |||
+ | <math>=181-125</math> | ||
+ | |||
+ | <math>=\boxed{056}.</math> | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=yIUBhWiJ4Dk | ||
+ | ~Mathematical Dexterity | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=vkniYGN45F4 | ||
+ | |||
+ | ~Shreyas S | ||
+ | |||
+ | Alternate Solution: https://www.youtube.com/watch?v=L2OzYI0OJsc&t=12s | ||
== See also == | == See also == | ||
{{AIME box|year=2011|n=I|num-b=3|num-a=5}} | {{AIME box|year=2011|n=I|num-b=3|num-a=5}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 12:08, 15 July 2024
Contents
Problem
In triangle ,
,
and
. The angle bisector of angle
intersects
at point
, and the angle bisector of angle
intersects
at point
. Let
and
be the feet of the perpendiculars from
to
and
, respectively. Find
.
Solution 1
Extend and
such that they intersect line
at points
and
, respectively.
Since
is the angle bisector of angle
and
is perpendicular to
,
must be an isoceles triangle, so
, and
is the midpoint of
. For the same reason,
, and
is the midpoint of
.
Hence
. Since
so
.
Solution 2
Let be the incenter of
. Since
lies on
and
,
and
, so
. This means that
is a cyclic quadrilateral. By the Law of Sines,
, where
is the radius of the circumcircle of
. Since
, we have that
. Letting
be the point of contact of the incircle of
with side
, we have
. Thus,
.
Solution 3 (Bash)
Project onto
and
as
and
.
and
are both in-radii of
so we get right triangles with legs
(the in-radius length) and
. Since
is the hypotenuse for the 4 triangles (
and
),
are con-cyclic on a circle we shall denote as
which is also the circumcircle of
and
. To find
, we can use the Law of Cosines on
where
is the center of
. Now, the circumradius
can be found with Pythagorean Theorem with
or
:
. To find
, we can use the formula
and by Heron's,
. To find
, we can find
since
.
. Thus,
and since
, we have
. Plugging this into our Law of Cosines (LoC) formula gives
. To find
, we use LoC on
. Our formula now becomes
. After simplifying, we get
.
--lucasxia01
Solution 4
Because ,
is cyclic.
Applying Ptolemy's theorem on CMIN:
by sine angle addition formula.
.
Let be where the incircle touches
, then
.
, for a final answer of
.
Note: This is similar to Solution 2 after the first four lines
Solution 5 (Trig Bash)
Applying Ptolemy's Theorem on the cyclic quadrilateral , we find that
.
and
by the Exterior Angle Theorem, so from properties of sine and cosine, we can find that
Plugging in the values and simplifying results in by the angle-addition identity
.
Before we continue, we would like to simplify the value in the sine function. We see that . Using the fact that
results in
How do we simplify ? Well, we can perform the Law of Sines on triangle
. This results in:
The value of is
by the Exterior Angle Theorem on
, so the value of
is equivalent to the value of
by a similar argument as above. Then rearranging yields
.
Going back to the previous formula and substituting values yields:
.
Finally, using the formulae and
(where
is half the perimeter of the triangle), we reach our final value:
Video Solution
https://www.youtube.com/watch?v=yIUBhWiJ4Dk ~Mathematical Dexterity
Video Solution
https://www.youtube.com/watch?v=vkniYGN45F4
~Shreyas S
Alternate Solution: https://www.youtube.com/watch?v=L2OzYI0OJsc&t=12s
See also
2011 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.