Difference between revisions of "2013 Mock AIME I Problems/Problem 12"
Sugar rush (talk | contribs) (Created page with "==Problem== In acute triangle <math>ABC</math>, the orthocenter <math>H</math> lies on the line connecting the midpoint of segment <math>AB</math> to the midpoint of segment <...") |
(added second solution) |
||
(2 intermediate revisions by 2 users not shown) | |||
Line 53: | Line 53: | ||
label("$C$", C, E); | label("$C$", C, E); | ||
</asy> | </asy> | ||
+ | |||
+ | == Solution 2 (no coordinates) == | ||
+ | |||
+ | <asy> | ||
+ | |||
+ | import geometry; | ||
+ | |||
+ | point A = origin; | ||
+ | point B = (12-sqrt(46),14); | ||
+ | point C = (24,0); | ||
+ | point M = midpoint(A--B); | ||
+ | point N = midpoint(B--C); | ||
+ | |||
+ | triangle t = triangle(A,B,C); | ||
+ | point H = orthocentercenter(t); | ||
+ | |||
+ | point D, E, F; | ||
+ | |||
+ | line a = altitude(t.BC); | ||
+ | line b = altitude(t.AC); | ||
+ | line c = altitude(t.AB); | ||
+ | |||
+ | // Triangle ABC and Segment MN | ||
+ | draw(t); | ||
+ | draw(M--N); | ||
+ | |||
+ | // Altitudes | ||
+ | pair[] e = intersectionpoints(b,A--C); | ||
+ | E = e[0]; | ||
+ | draw(B--E); | ||
+ | |||
+ | pair[] d = intersectionpoints(a,B--C); | ||
+ | D = d[0]; | ||
+ | draw(A--D); | ||
+ | |||
+ | pair[] f = intersectionpoints(c,A--B); | ||
+ | F = f[0]; | ||
+ | draw(C--F); | ||
+ | |||
+ | // Labeling Points | ||
+ | dot(A); | ||
+ | label("A",A,SW); | ||
+ | dot(B); | ||
+ | label("B",B,NW); | ||
+ | dot(C); | ||
+ | label("C",C,SE); | ||
+ | dot(M); | ||
+ | label("M",M,WSW); | ||
+ | dot(N); | ||
+ | label("N",N,ENE); | ||
+ | dot(H); | ||
+ | label("H",H,SSE); | ||
+ | dot(D); | ||
+ | label("D",D,NE); | ||
+ | dot(E); | ||
+ | label("E",E,S); | ||
+ | dot(F); | ||
+ | label("F",F,NW); | ||
+ | |||
+ | // Right angle marks | ||
+ | markscalefactor = 0.15; | ||
+ | draw(rightanglemark(B,H,M)); | ||
+ | draw(rightanglemark(B,D,C)); | ||
+ | |||
+ | </asy> | ||
+ | |||
+ | Let <math>M</math> be the midpoint of <math>\overline{AB}</math> and <math>N</math> be the midpoint of <math>\overline{BC}</math>. Further let <math>D</math> be the foot of the altitude from <math>A</math>, <math>E</math> from <math>B</math>, and <math>F</math> from <math>C</math>, as in the diagram. | ||
+ | |||
+ | Because <math>\overline{MN}</math> is a midpoint connector of <math>\triangle ABC</math> amd <math>H</math> is on <math>\overline{MN}</math> and <math>\overline{BE}</math>, we know that <math>H</math> is the midpoint of altitude <math>BE</math>. Thus, because, from the problem, <math>BE=14</math>, <math>BH=HE=7</math>. Now we see that <math>\overline{MH}</math> is a midpoint connector of <math>\triangle BAE</math>, so <math>MH=\tfrac12 AE</math>. | ||
+ | |||
+ | Now, let <math>\measuredangle CHE = \theta</math>. We know that <math>\measuredangle BHF = \measuredangle CHE = \theta</math>, because they are vertical angles. Because <math>\triangle BHF</math> is right (by the definition of an altitude), we know that <math>\measuredangle MBH = 90^{\circ}-\theta</math>. <math>\triangle BHM</math> is also right, so <math>\measuredangle BMH = 90^{\circ} -(90^{\circ}-\theta)=\theta</math>. | ||
+ | |||
+ | From <math>\triangle HEC</math>, we know that <math>\tan\theta = \tfrac{CE}7</math>. From <math>\triangle BHM</math>, we know that <math>\tan\theta = \tfrac7{AE/2} = \tfrac{14}{AE}</math>. Equating these two expressions for <math>\tan\theta</math>, we see that <math>AE \cdot CE = 7 \cdot 14 = 98</math>. From the problem, we know that <math>AE+CE=AC=24</math>. | ||
+ | |||
+ | Now, we can proceed as in Solution 1 by using the quadratic formula to solve for <math>AE</math> and the [[Pythagorean Theorem]] to find <math>AB</math> and <math>BC</math>. We do this to obtain our answer <math>AB \cdot BC = \boxed{350}</math>. | ||
+ | |||
+ | ==See also== | ||
+ | * [[2013 Mock AIME I Problems]] | ||
+ | * [[2013 Mock AIME I Problems/Problem 11|Preceded by Problem 11]] | ||
+ | * [[2013 Mock AIME I Problems/Problem 13|Followed by Problem 13]] | ||
+ | [[Category:Intermediate Geometry Problems]] |
Latest revision as of 10:42, 31 July 2024
Problem
In acute triangle , the orthocenter
lies on the line connecting the midpoint of segment
to the midpoint of segment
. If
, and the altitude from
has length
, find
.
Solution (easy coordinate bash)
Toss on the coordinate plane with ,
, and
, where
is a real number and
.
Then, the line connecting the midpoints of and
runs from
to
, or more simply the line
.
The orthocenter of will be at the intersection of the altitudes from
and
.
The slope of the altitude from is the negative reciprocal of the slope of
. The slope of
is
, and its negative reciprocal is
. Since the altitude from
passes through the origin, its equation is
.
The altitude from is the vertical line running through
which has equation
.
Thus the lines and
meet on the line
. Substituting the first equation into the second,
.
Multiplying both sides by , we have
.
This rearranges to the quadratic , and completing the square by adding
to each side gives us
. Thus
.
The cases where and
are similar; they merely correspond to two triangles that can each be obtained by reflecting the other across the perpendicular bisector of
, so we consider the case where
.
So .
Thus
The cases where and
are shown below, labeled
and
, respectively, where the dotted line is a midline in both triangles. As you can see, the orthocenter falls perfectly on that line for both triangles, and the value of
is the same for both triangles.
Solution 2 (no coordinates)
Let be the midpoint of
and
be the midpoint of
. Further let
be the foot of the altitude from
,
from
, and
from
, as in the diagram.
Because is a midpoint connector of
amd
is on
and
, we know that
is the midpoint of altitude
. Thus, because, from the problem,
,
. Now we see that
is a midpoint connector of
, so
.
Now, let . We know that
, because they are vertical angles. Because
is right (by the definition of an altitude), we know that
.
is also right, so
.
From , we know that
. From
, we know that
. Equating these two expressions for
, we see that
. From the problem, we know that
.
Now, we can proceed as in Solution 1 by using the quadratic formula to solve for and the Pythagorean Theorem to find
and
. We do this to obtain our answer
.