Difference between revisions of "2014 AIME II Problems/Problem 14"
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<math>m+n=\boxed{077}</math> | <math>m+n=\boxed{077}</math> | ||
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<cmath>HC=\frac{5\sqrt2+5\sqrt6}{2}</cmath> | <cmath>HC=\frac{5\sqrt2+5\sqrt6}{2}</cmath> | ||
− | <math>MC=\frac{ | + | <math>MC=\frac{BC}{2},</math> |
<cmath>HM=\frac{5\sqrt6}{2}</cmath> | <cmath>HM=\frac{5\sqrt6}{2}</cmath> | ||
<cmath>HN=\frac{5\sqrt6}{4}</cmath> | <cmath>HN=\frac{5\sqrt6}{4}</cmath> | ||
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Draw the <math>45-45-90 \triangle AHC</math>. Now, take the perpendicular bisector of <math>BC</math> to intersect the circumcircle of <math>\triangle ABC</math> and <math>AC</math> at <math>F, L, G</math> as shown, and denote <math>O</math> to be the circumcenter of <math>\triangle ABC</math>. It is not difficult to see by angle chasing that <math>AHBGO</math> is cyclic, namely with diameter <math>AB</math>. Then, by symmetry, <math>EH = HB</math> and as <math>HB, OG</math> are both subtended by equal arcs they are equal. Hence, <math>EH = GO</math>. Now, draw line <math>HL</math> and intersect it at <math>AC</math> at point <math>K</math> in the diagram. It is not hard to use angle chase to arrive at <math>AEOL</math> a parallelogram, and from our length condition derived earlier, <math>AL \parallel HG</math>. From here, it is clear that <math>AK = KG</math>; that is, <math>P</math> is just the intersection of the perpendicular from <math>K</math> down to <math>BC</math> and <math>AD</math>! After this point, note that <math>AP = PF</math>. It is easily derived that the circumradius of <math>\triangle ABC</math> is <math>\frac{10}{\sqrt{2}}</math>. Now, <math>APO</math> is a <math>30-60-90</math> triangle, and from here it is easy to arrive at the final answer of <math>\boxed{077}</math>. ~awang11's sol | Draw the <math>45-45-90 \triangle AHC</math>. Now, take the perpendicular bisector of <math>BC</math> to intersect the circumcircle of <math>\triangle ABC</math> and <math>AC</math> at <math>F, L, G</math> as shown, and denote <math>O</math> to be the circumcenter of <math>\triangle ABC</math>. It is not difficult to see by angle chasing that <math>AHBGO</math> is cyclic, namely with diameter <math>AB</math>. Then, by symmetry, <math>EH = HB</math> and as <math>HB, OG</math> are both subtended by equal arcs they are equal. Hence, <math>EH = GO</math>. Now, draw line <math>HL</math> and intersect it at <math>AC</math> at point <math>K</math> in the diagram. It is not hard to use angle chase to arrive at <math>AEOL</math> a parallelogram, and from our length condition derived earlier, <math>AL \parallel HG</math>. From here, it is clear that <math>AK = KG</math>; that is, <math>P</math> is just the intersection of the perpendicular from <math>K</math> down to <math>BC</math> and <math>AD</math>! After this point, note that <math>AP = PF</math>. It is easily derived that the circumradius of <math>\triangle ABC</math> is <math>\frac{10}{\sqrt{2}}</math>. Now, <math>APO</math> is a <math>30-60-90</math> triangle, and from here it is easy to arrive at the final answer of <math>\boxed{077}</math>. ~awang11's sol | ||
− | =Solution 5== | + | ==Solution 5== |
[[File:2014 AIME II 14.png|500px|right]] | [[File:2014 AIME II 14.png|500px|right]] | ||
Let <math>BO \perp AC, O \in AC.</math> | Let <math>BO \perp AC, O \in AC.</math> | ||
Line 169: | Line 166: | ||
<math>HN = NM, AH||NP||ME \implies AP = PE.</math> | <math>HN = NM, AH||NP||ME \implies AP = PE.</math> | ||
− | In <math>\triangle ABO \angle A = 30^\circ \implies AO = AB \cos 60^\circ = 5 \sqrt{3}.</math> | + | In <math>\triangle ABO \hspace{10mm} \angle A = 30^\circ \implies AO = AB \cos 60^\circ = 5 \sqrt{3}.</math> |
− | In <math>\triangle AEO \angle A = 15^\circ, \angle O = 90^\circ + 45^\circ = 135^\circ \implies</math> | + | In <math>\triangle AEO \hspace{10mm} \angle A = 15^\circ, \angle O = 90^\circ + 45^\circ = 135^\circ \implies</math> |
<cmath>\angle AEO = 30^\circ \implies</cmath> | <cmath>\angle AEO = 30^\circ \implies</cmath> | ||
<cmath>AE = AO \frac {\sin 135^\circ}{\sin 30^\circ} = 5 \sqrt{3} \cdot \sqrt{2} = 5 \sqrt{6} \implies</cmath> | <cmath>AE = AO \frac {\sin 135^\circ}{\sin 30^\circ} = 5 \sqrt{3} \cdot \sqrt{2} = 5 \sqrt{6} \implies</cmath> |
Latest revision as of 19:28, 21 October 2023
Contents
Problem
In , and
. Let
and
be points on the line
such that
,
, and
. Point
is the midpoint of the segment
, and point
is on ray
such that
. Then
, where
and
are relatively prime positive integers. Find
.
Diagram
Solution 1
Let us just drop the perpendicular from to
and label the point of intersection
. We will use this point later in the problem.
As we can see,
is the midpoint of
and
is the midpoint of
is a
triangle, so
.
is
triangle.
and
are parallel lines so
is
triangle also.
Then if we use those informations we get and
and
or
Now we know that , we can find for
which is simpler to find.
We can use point to split it up as
,
We can chase those lengths and we would get
, so
, so
, so
We can also use Law of Sines:
Then using right triangle , we have
So .
And we know that .
Finally if we calculate .
. So our final answer is
.
-Gamjawon
-edited by srisainandan6 to clarify and correct a small mistake
Solution 2
Here's a solution that doesn't need .
As above, get to . As in the figure, let
be the foot of the perpendicular from
to
. Then
is a 45-45-90 triangle, and
is a 30-60-90 triangle. So
and
; also,
,
, and
. But
and
are parallel, both being orthogonal to
. Therefore
, or
, and we're done.
Solution 3
Break our diagram into 2 special right triangle by dropping an altitude from to
we then get that
Since
is a 45-45-90,
We know that
and are 30-60-90.
Thus,
. So our final answer is
.
Solution 4
Draw the
. Now, take the perpendicular bisector of
to intersect the circumcircle of
and
at
as shown, and denote
to be the circumcenter of
. It is not difficult to see by angle chasing that
is cyclic, namely with diameter
. Then, by symmetry,
and as
are both subtended by equal arcs they are equal. Hence,
. Now, draw line
and intersect it at
at point
in the diagram. It is not hard to use angle chase to arrive at
a parallelogram, and from our length condition derived earlier,
. From here, it is clear that
; that is,
is just the intersection of the perpendicular from
down to
and
! After this point, note that
. It is easily derived that the circumradius of
is
. Now,
is a
triangle, and from here it is easy to arrive at the final answer of
. ~awang11's sol
Solution 5
Let
Let
points
are collinear.
In
In
vladimir.shelomovskii@gmail.com, vvsss
Video solution
https://www.youtube.com/watch?v=SvJ0wDJphdU
See also
2014 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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.