Difference between revisions of "2018 AMC 12B Problems/Problem 21"
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In <math>\triangle{ABC}</math> with side lengths <math>AB = 13</math>, <math>AC = 12</math>, and <math>BC = 5</math>, let <math>O</math> and <math>I</math> denote the circumcenter and incenter, respectively. A circle with center <math>M</math> is tangent to the legs <math>AC</math> and <math>BC</math> and to the circumcircle of <math>\triangle{ABC}</math>. What is the area of <math>\triangle{MOI}</math>? | In <math>\triangle{ABC}</math> with side lengths <math>AB = 13</math>, <math>AC = 12</math>, and <math>BC = 5</math>, let <math>O</math> and <math>I</math> denote the circumcenter and incenter, respectively. A circle with center <math>M</math> is tangent to the legs <math>AC</math> and <math>BC</math> and to the circumcircle of <math>\triangle{ABC}</math>. What is the area of <math>\triangle{MOI}</math>? | ||
− | <math>\textbf{(A)}\ | + | <math>\textbf{(A)}\ \frac52\qquad\textbf{(B)}\ \frac{11}{4}\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ \frac{13}{4}\qquad\textbf{(E)}\ \frac72</math> |
− | == | + | == Diagram == |
+ | <asy> | ||
+ | /* Made by MRENTHUSIASM */ | ||
+ | size(250); | ||
− | + | pair A, B, C, O, I, M; | |
− | + | C = origin; | |
− | + | A = (12,0); | |
− | + | B = (0,5); | |
+ | C = origin; | ||
+ | O = circumcenter(A,B,C); | ||
+ | I = incenter(A,B,C); | ||
+ | M = (4,4); | ||
+ | fill(M--O--I--cycle,yellow); | ||
+ | draw(A--B--C--cycle^^circumcircle(A,B,C)^^incircle(A,B,C)^^circle(M,4)^^M--O--I--cycle); | ||
+ | dot("$A$",A,1.5*SE,linewidth(4)); | ||
+ | dot("$B$",B,1.5*NW,linewidth(4)); | ||
+ | dot("$C$",C,1.5*SW,linewidth(4)); | ||
+ | dot("$O$",O,1.5*dir((5,12)),linewidth(4)); | ||
+ | dot("$I$",I,1.5*S,linewidth(4)); | ||
+ | dot("$M$",M,1.5*N,linewidth(4)); | ||
+ | </asy> | ||
+ | ~MRENTHUSIASM | ||
− | == Solution | + | == Solution == |
+ | In this solution, let the brackets denote areas. | ||
− | + | We place the diagram in the coordinate plane: Let <math>A=(12,0),B=(0,5),</math> and <math>C=(0,0).</math> | |
+ | |||
+ | Since <math>\triangle ABC</math> is a right triangle with <math>\angle ACB=90^\circ,</math> its circumcenter is the midpoint of <math>\overline{AB},</math> from which <math>O=\left(6,\frac52\right).</math> Note that the circumradius of <math>\triangle ABC</math> is <math>\frac{13}{2}.</math> | ||
+ | |||
+ | Let <math>s</math> denote the semiperimeter of <math>\triangle ABC.</math> The inradius of <math>\triangle ABC</math> is <math>\frac{[ABC]}{s}=\frac{30}{15}=2,</math> from which <math>I=(2,2).</math> | ||
+ | |||
+ | Since <math>\odot M</math> is also tangent to both coordinate axes, its center is at <math>M=(a,a)</math> and its radius is <math>a</math> for some positive number <math>a.</math> Let <math>P</math> be the point of tangency of <math>\odot O</math> and <math>\odot M.</math> As <math>\overline{OP}</math> and <math>\overline{MP}</math> are both perpendicular to the common tangent line at <math>P,</math> we conclude that <math>O,M,</math> and <math>P</math> are collinear. It follows that <math>OM=OP-MP,</math> or <cmath>\sqrt{(a-6)^2+\left(a-\frac52\right)^2}=\frac{13}{2}-a.</cmath> | ||
+ | Solving this equation, we have <math>a=4,</math> from which <math>M=(4,4).</math> | ||
+ | |||
+ | Finally, we apply the Shoelace Theorem to <math>\triangle MOI:</math> <cmath>[MOI]=\frac12\left|\left(4\cdot\frac52+6\cdot2+2\cdot4\right)-\left(4\cdot6+\frac52\cdot2+2\cdot4\right)\right|=\boxed{\textbf{(E)}\ \frac72}.</cmath> | ||
+ | <u><b>Remark</b></u> | ||
+ | |||
+ | Alternatively, we can use <math>\overline{MI}</math> as the base and the distance from <math>O</math> to <math>\overleftrightarrow{MI}</math> as the height for <math>\triangle MOI:</math> | ||
+ | |||
+ | * By the Distance Formula, we have <math>MI=2\sqrt2.</math> | ||
+ | |||
+ | * The equation of <math>\overleftrightarrow{MI}</math> is <math>x-y+0=0,</math> so the distance from <math>O</math> to <math>\overleftrightarrow{MI}</math> is <math>h_O=\frac{\left|1\cdot6+(-1)\cdot\frac52+0\right|}{\sqrt{1^2+(-1)^2}}=\frac74\sqrt2.</math> | ||
+ | |||
+ | Therefore, we get <cmath>[MOI]=\frac12\cdot MI\cdot h_O=\frac72.</cmath> | ||
+ | ~pieater314159 ~MRENTHUSIASM | ||
+ | |||
+ | == Solution 2 (vibe check solution 😎😎😎) == | ||
+ | Let points Q and R be the points of tangency between the incircle and lines <math>AB</math> and <math>BC</math>. Notice that <math>\angle RIB</math> is half of <math>\angle ABC</math>. Let <math>m\angle ABC</math> = θ. Using the half angle tangent formula and keeping the idea that cos(θ) = <math>\frac{5}{13}</math>, we find that tan(θ/2) = sqrt((1-cosθ)/(1+cosθ)) = <math>\sqrt{\frac{4}{9}}</math> = <math>\frac{2}{3}</math>. That's rockin and all, but the real deal starts when we acknowledge that we can find the length of BC in terms of the radius of the incircle, which I'll be calling r. Using the knowledge that <math>\triangle RIB</math> is right, and some trigonometric tomfoolery, we find that <math>BR</math> is 3r/2. We also find that <math>CR</math> is the r, and so we can create the equation <math>BR</math> + <math>CR</math> = <math>BC</math> -> 3r/2 + r = 5 -> 5r/2. We conclude r = 2. Absolutely radical. We now accept the fact that quadrilateral <math>BQIR</math> is a kite, so <math>BR</math> = <math>BQ</math> = 3. We also know that O lies on <math>BA</math> and divides <math>BA</math> in half. Next we collectively determine that <math>MQ</math> = <math>BM</math> - <math>BQ</math> = 13/2 - 3 = 7/2. Fantabular. We also know that <math>IQ</math> has a measure of 2 since it's the radius of circle I (me thinks you would find this relatable (cause u have an iq of 2 lol haha)). | ||
+ | |||
+ | Now we get vibey. We know very little about the placement of <math>M</math>, but that's all about to change. First, we can conclude that <math>\triangle IOQ</math> is part of <math>\triangle MOI</math>. This much is true. We also feel out that <math>M</math> has to be higher up than <math>O</math>. I mean, it just makes sense. Now for the vibey part. We're about to feel out this solution. We find the area of <math>\triangle IOQ</math>. It's 7/2. We glance at the answer choices. 7/2 is the greatest option. erm whart the sigma??? I sincerely doubt that <math>M</math> is below <math>\overline{AB}</math>, but if it were to go past <math>\overline{AB}</math> then the answer would be GREATER than 7/2. This could only mean one thing...*GASP*! This whole time, <math>\overline{QO}</math> was on <math>\overline{AB}</math>!? THEREFORE WE CAN CONCLUDE <math>\triangle IOQ=\triangle MOI</math>. That means the answer must be <math>\boxed{\textbf{(E)}\ \frac72}</math>, and therefore, this question passes the vibe check. Thanks for following fam and I'll catch you in a jiffy! | ||
+ | ~me | ||
==See Also== | ==See Also== | ||
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{{AMC12 box|year=2018|ab=B|num-b=20|num-a=22}} | {{AMC12 box|year=2018|ab=B|num-b=20|num-a=22}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] |
Latest revision as of 04:08, 19 August 2024
Problem
In with side lengths
,
, and
, let
and
denote the circumcenter and incenter, respectively. A circle with center
is tangent to the legs
and
and to the circumcircle of
. What is the area of
?
Diagram
~MRENTHUSIASM
Solution
In this solution, let the brackets denote areas.
We place the diagram in the coordinate plane: Let and
Since is a right triangle with
its circumcenter is the midpoint of
from which
Note that the circumradius of
is
Let denote the semiperimeter of
The inradius of
is
from which
Since is also tangent to both coordinate axes, its center is at
and its radius is
for some positive number
Let
be the point of tangency of
and
As
and
are both perpendicular to the common tangent line at
we conclude that
and
are collinear. It follows that
or
Solving this equation, we have
from which
Finally, we apply the Shoelace Theorem to
Remark
Alternatively, we can use as the base and the distance from
to
as the height for
- By the Distance Formula, we have
- The equation of
is
so the distance from
to
is
Therefore, we get
~pieater314159 ~MRENTHUSIASM
Solution 2 (vibe check solution 😎😎😎)
Let points Q and R be the points of tangency between the incircle and lines and
. Notice that
is half of
. Let
= θ. Using the half angle tangent formula and keeping the idea that cos(θ) =
, we find that tan(θ/2) = sqrt((1-cosθ)/(1+cosθ)) =
=
. That's rockin and all, but the real deal starts when we acknowledge that we can find the length of BC in terms of the radius of the incircle, which I'll be calling r. Using the knowledge that
is right, and some trigonometric tomfoolery, we find that
is 3r/2. We also find that
is the r, and so we can create the equation
+
=
-> 3r/2 + r = 5 -> 5r/2. We conclude r = 2. Absolutely radical. We now accept the fact that quadrilateral
is a kite, so
=
= 3. We also know that O lies on
and divides
in half. Next we collectively determine that
=
-
= 13/2 - 3 = 7/2. Fantabular. We also know that
has a measure of 2 since it's the radius of circle I (me thinks you would find this relatable (cause u have an iq of 2 lol haha)).
Now we get vibey. We know very little about the placement of , but that's all about to change. First, we can conclude that
is part of
. This much is true. We also feel out that
has to be higher up than
. I mean, it just makes sense. Now for the vibey part. We're about to feel out this solution. We find the area of
. It's 7/2. We glance at the answer choices. 7/2 is the greatest option. erm whart the sigma??? I sincerely doubt that
is below
, but if it were to go past
then the answer would be GREATER than 7/2. This could only mean one thing...*GASP*! This whole time,
was on
!? THEREFORE WE CAN CONCLUDE
. That means the answer must be
, and therefore, this question passes the vibe check. Thanks for following fam and I'll catch you in a jiffy!
~me
See Also
2018 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 20 |
Followed by Problem 22 |
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.