Difference between revisions of "2013 Mock AIME I Problems/Problem 13"
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== Solution == | == Solution == | ||
− | <math>\boxed{010}</math>. | + | |
+ | Because <math>[\triangle ABC] = 168</math> and <math>BC=18</math>, we know that the height from <math>A</math> to <math>BC</math> must be <math>18</math>. Thus, because the perpendicular is the shortest segment from a line to a point not on the line, we know that <math>AB \geq 18</math>. Thus, the minimum value of <math>AB</math> is <math>18</math>, when <math>\overline{AB} \perp \overline{BC}</math>. This is technically not possible, because <math>\triangle ABC</math> is acute, but it may be helpful in getting a better sense of the problem. This scenario is shown below: | ||
+ | |||
+ | <asy> | ||
+ | |||
+ | import geometry; | ||
+ | |||
+ | point A = origin; | ||
+ | point B = (0,18); | ||
+ | point C = (18,18); | ||
+ | triangle t = triangle(A,B,C); | ||
+ | |||
+ | point M = midpoint(B--C); | ||
+ | |||
+ | circle c = circumcircle(t); | ||
+ | point O = circumcenter(t); | ||
+ | |||
+ | point G = centroid(t); | ||
+ | |||
+ | // Triangle and Circumcircle | ||
+ | draw(t); | ||
+ | draw(c); | ||
+ | |||
+ | // Median AM | ||
+ | draw(A--M); | ||
+ | |||
+ | // Labelling Points | ||
+ | dot(A); | ||
+ | label("A",A,SW); | ||
+ | dot(B); | ||
+ | label("H=B",B,NW); | ||
+ | dot(C); | ||
+ | label("C",C,NE); | ||
+ | dot(O); | ||
+ | label("O",O,SE); | ||
+ | dot(M); | ||
+ | label("M",M,N); | ||
+ | dot(G); | ||
+ | label("G",G,WSW); | ||
+ | |||
+ | // Length Labels | ||
+ | label("$9$",midpoint(B--M),N); | ||
+ | label("$9$",midpoint(M--C),N); | ||
+ | label("$18$",midpoint(A--B),W); | ||
+ | |||
+ | // Right Angle Mark | ||
+ | markscalefactor = 0.18; | ||
+ | draw(rightanglemark(A,B,C)); | ||
+ | |||
+ | </asy> | ||
+ | |||
+ | Because the orthocenter of <math>\triangle ABC</math> is at <math>B</math>, <math>HM=BM=9</math>. Now, by the [[Pythagorean Theorem]], <math>AM=9\sqrt5</math>. Because the centrod is <math>\tfrac23</math> of the way along the median from the vertex, we know that <math>GM=\tfrac{9\sqrt5}3=3\sqrt5<9</math>. Thus, we have <math>GM<HM</math>, so the given inequality does not hold. Now, let us look at the example where <math>A</math> is collinear with <math>O</math> and <math>M</math>: | ||
+ | |||
+ | <asy> | ||
+ | |||
+ | import geometry; | ||
+ | |||
+ | point A = origin; | ||
+ | point B = (-9,18); | ||
+ | point C = (9,18); | ||
+ | triangle t = triangle(A,B,C); | ||
+ | |||
+ | point M = midpoint(B--C); | ||
+ | |||
+ | circle c = circumcircle(t); | ||
+ | point O = circumcenter(t); | ||
+ | |||
+ | point G = centroid(t); | ||
+ | point H = orthocentercenter(t); | ||
+ | |||
+ | point Cp; | ||
+ | |||
+ | // Triangle and Circumcircle | ||
+ | draw(t); | ||
+ | draw(c); | ||
+ | |||
+ | // Median AM, Segment OB | ||
+ | draw(A--M); | ||
+ | draw(O--B); | ||
+ | |||
+ | // Defining C' | ||
+ | pair[] cp = intersectionpoints(line(O,C),c); | ||
+ | Cp = cp[0]; | ||
+ | |||
+ | // Labelling Points | ||
+ | dot(A); | ||
+ | label("A",A,S); | ||
+ | dot(B); | ||
+ | label("B",B,NW); | ||
+ | dot(C); | ||
+ | label("C",C,NE); | ||
+ | dot(O); | ||
+ | label("O",O,SW); | ||
+ | dot(M); | ||
+ | label("M",M,N); | ||
+ | dot(G); | ||
+ | label("G",G,E); | ||
+ | dot(H); | ||
+ | label("H",H,ENE); | ||
+ | dot(Cp); | ||
+ | label("C$^{\prime}$",Cp,SW); | ||
+ | |||
+ | // Length Labels | ||
+ | label("$9$",midpoint(B--M),N); | ||
+ | label("$9$",midpoint(M--C),N); | ||
+ | |||
+ | // Right Angle Mark | ||
+ | markscalefactor = 0.18; | ||
+ | draw(rightanglemark(A,M,B)); | ||
+ | |||
+ | </asy> | ||
+ | |||
+ | By centroid properties, we know that <math>AG=\tfrac23 AM = \tfrac23 \cdot 18 = 12</math>. Let <math>R</math> be the circumradius of <math>\triangle ABC</math>. Then, by Pythagoras in <math>\triangle OBM</math>, <math>OM=\sqrt{R^2-81}</math>. Because <math>OA=R</math> and <math>AM=18</math>, we have the equation <math>\sqrt{R^2-81}+R=18</math>, which yields <math>AO=R=\tfrac{45}4<12=AG</math>, so <math>G</math> is between <math>O</math> and <math>M</math>. Because <math>\triangle ABC</math> is acute, we know that <math>H</math> is in the interior of <math>\triangle ABC</math>. This fact, combined with the properties of the [[Euler Line]], show that <math>H</math> must be closer to <math>M</math> than <math>G</math> is, so <math>GM>HM</math>, and the inequality is thereby satisfied. | ||
+ | |||
+ | As in the above diagram, let <math>C^{\prime}</math> be the point on the circle such that <math>\overline{CC^{\prime}}</math> is a diameter of the circle. Because <math>\triangle ABC</math> is acute, <math>A</math> and <math>B</math> must be on the opposite sides of <math>\overline{CC^{\prime}}</math> (so that the circumcenter lies inside the triangle). We want <math>A</math> to be as close to <math>C^{\prime}</math> as possible to minimize <math>AB</math>, but, from the first example we explored, we know that when <math>A=C^{\prime}</math>, <math>HM>GM</math>. We would reasonably expect the difference <math>GM-HM</math> to vary continuously as we move <math>A</math> towards <math>C^{\prime}</math>, and this difference is positive in the second example and negative in the first example. Thus, by the [[Intermediate Value Theorem]], there should be a point on the circumcircle between these two locations for <math>A</math> such that this difference is zero, or <math>GM=HM</math>. This point should be as close as we can get to <math>C^{\prime}</math> while still satisfying the inequality. | ||
+ | |||
+ | Now, let <math>GM=HM</math> and <math>D</math> be the foot of the altitude from <math>A</math> to <math>\overline{BC}</math>. Further, let <math>MD=x</math>, so <math>BD=9-x</math>, as shown below: | ||
+ | |||
+ | <asy> | ||
+ | |||
+ | import geometry; | ||
+ | |||
+ | point A = (-5.678648, 1.454964); | ||
+ | point B = (-9,18); | ||
+ | point C = (9,18); | ||
+ | triangle t = triangle(A,B,C); | ||
+ | |||
+ | line a = altitude(t.BC); | ||
+ | point D; | ||
+ | |||
+ | point M = midpoint(B--C); | ||
+ | |||
+ | circle c = circumcircle(t); | ||
+ | point O = circumcenter(t); | ||
+ | |||
+ | point G = centroid(t); | ||
+ | point H = orthocentercenter(t); | ||
+ | |||
+ | // Triangle and Circumcircle | ||
+ | draw(t); | ||
+ | draw(c); | ||
+ | |||
+ | // Median AM, Segment HM | ||
+ | draw(A--M); | ||
+ | draw(H--M); | ||
+ | |||
+ | // Altitude AD | ||
+ | pair[] d = intersectionpoints(a,B--C); | ||
+ | D = d[0]; | ||
+ | draw(A--D); | ||
+ | |||
+ | // Labelling Points | ||
+ | dot(A); | ||
+ | label("A",A,S); | ||
+ | dot(B); | ||
+ | label("B",B,NW); | ||
+ | dot(C); | ||
+ | label("C",C,NE); | ||
+ | dot(O); | ||
+ | label("O",O,S); | ||
+ | dot(M); | ||
+ | label("M",M,N); | ||
+ | dot(G); | ||
+ | label("G",G,WSW); | ||
+ | dot(H); | ||
+ | label("H",H,ESE); | ||
+ | dot(D); | ||
+ | label("D",D,N); | ||
+ | |||
+ | // Length Labels | ||
+ | label("$x$",midpoint(D--M),N); | ||
+ | label("$9$",midpoint(M--C),N); | ||
+ | |||
+ | // Right Angle Mark | ||
+ | markscalefactor = 0.15; | ||
+ | draw(rightanglemark(A,D,B)); | ||
+ | |||
+ | </asy> | ||
+ | |||
+ | By Pythagoras, we know that <math>AM = \sqrt{AD^2+DM^2} = \sqrt{18^2+x^2} = \sqrt{x^2+324}</math>. Thus, by centroid properties, <math>GM = \tfrac13 AM = \tfrac{\sqrt{x^2+324}}3</math>. Now, we desire to find another expression for <math>GM=HM</math>. By using Pythagoras again, we see that <math>AB = \sqrt{(9-x)^2+18^2} = \sqrt{x^2-18x+405}</math> and <math>AC = \sqrt{(9+x)^2+18^2} = \sqrt{x^2+18x+405}</math>. Now, let <math>\measuredangle BAC = \alpha</math>. Also let <math>AC=b</math> and <math>AB=c</math>. By the [[Law of Cosines]] in <math>\triangle ABC</math>, we have the following equation: | ||
+ | \begin{align*} | ||
+ | BC^2 &= AB^2 + AC^2 - 2AB \cdot AC \cos\alpha \ | ||
+ | 18^2 &= (x^2+18x+405)+(x^2-18x+405)-2bc\cos\alpha \ | ||
+ | \frac{324}2 &= x^2+405-bc\cos\alpha \ | ||
+ | bc\cos\alpha &= x^2+405-162 \ | ||
+ | \cos\alpha &= \frac{x^2+243}{bc} | ||
+ | \end{align*} | ||
+ | By rearranging the formula for the area of a triangle <math>\tfrac{abc}{4R}</math> and recalling that, from the problem, <math>[\triangle ABC] = 162</math>, we see that <math>R=\tfrac{abc}{4 \cdot 162}</math>. Because <math>BC=18</math>, this expression equates to <math>\tfrac{18bc}{4 \cdot 162}=\tfrac{bc}{36}</math>. By the formula for the distance from a vertex to the orthocenter and substitution, we know that <math>AH=2R\cos\alpha=2(\tfrac{bc}{36})(\tfrac{x^2+243}{bc})=\tfrac{x^2+243}{18}</math>. Thus, because <math>AD=18</math>, <math>HD=18-\tfrac{x^2+243}{18}=\tfrac{81-x^2}{18}</math>. By Pythagoras in <math>\triangle HDM</math>, <math>HM = \sqrt{\tfrac{(81-x^2)^2}{324}+x^2} = \tfrac{\sqrt{81^2-162x^2+x^4+324x^2}}{18} = \tfrac{\sqrt{x^4+162x^2+81^2}}{18} = \tfrac{x^2+81}{18}</math>. Equating this with our earlier expression for <math>GM</math>, we get the following equation: | ||
+ | \begin{align*} | ||
+ | \frac{x^2+81}{18} &= \frac{\sqrt{x^2+324}}3 \ | ||
+ | \frac{x^2+81}6 &= \sqrt{x^2+324} \ | ||
+ | \frac{x^4+162x^2+81^2}{36} &= x^2+18^2 \ | ||
+ | x^4+162x^2+81^2 &= 36x^2+6^2 \cdot 18^2 \ | ||
+ | x^4+126x^2+81^2-108^2 &= 0 \ | ||
+ | x^4+126x^2-(108-81)(108+81) &= 0 \ | ||
+ | x^4+126x^2-27 \cdot 189 &= 0 \ | ||
+ | x^2 &= \frac{-126 \pm \sqrt{126^2+4\cdot27\cdot189}}2 \ | ||
+ | x^2 &= -63 \pm 36\sqrt7 | ||
+ | \end{align*} | ||
+ | |||
+ | Because <math>x^2>0</math>, <math>x^2=-63+36\sqrt7</math>. Plugging this into <math>GM=HM=\tfrac{x^2+81}{18}</math> yields <math>\tfrac{-63+36\sqrt7+81}{18}=\tfrac{18+36\sqrt7}{18}=1+2\sqrt7</math>. Thus, our answer is <math>1+2+7=\boxed{010}</math>. | ||
== See also == | == See also == |
Latest revision as of 18:16, 31 July 2024
Problem
In acute ,
is the orthocenter,
is the centroid, and
is the midpoint of
. It is obvious that
, but
does not always hold. If
,
, then the value of
which produces the smallest value of
such that
can be expressed in the form
, for
squarefree. Compute
.
Solution
Because and
, we know that the height from
to
must be
. Thus, because the perpendicular is the shortest segment from a line to a point not on the line, we know that
. Thus, the minimum value of
is
, when
. This is technically not possible, because
is acute, but it may be helpful in getting a better sense of the problem. This scenario is shown below:
Because the orthocenter of is at
,
. Now, by the Pythagorean Theorem,
. Because the centrod is
of the way along the median from the vertex, we know that
. Thus, we have
, so the given inequality does not hold. Now, let us look at the example where
is collinear with
and
:
By centroid properties, we know that . Let
be the circumradius of
. Then, by Pythagoras in
,
. Because
and
, we have the equation
, which yields
, so
is between
and
. Because
is acute, we know that
is in the interior of
. This fact, combined with the properties of the Euler Line, show that
must be closer to
than
is, so
, and the inequality is thereby satisfied.
As in the above diagram, let be the point on the circle such that
is a diameter of the circle. Because
is acute,
and
must be on the opposite sides of
(so that the circumcenter lies inside the triangle). We want
to be as close to
as possible to minimize
, but, from the first example we explored, we know that when
,
. We would reasonably expect the difference
to vary continuously as we move
towards
, and this difference is positive in the second example and negative in the first example. Thus, by the Intermediate Value Theorem, there should be a point on the circumcircle between these two locations for
such that this difference is zero, or
. This point should be as close as we can get to
while still satisfying the inequality.
Now, let and
be the foot of the altitude from
to
. Further, let
, so
, as shown below:
By Pythagoras, we know that . Thus, by centroid properties,
. Now, we desire to find another expression for
. By using Pythagoras again, we see that
and
. Now, let
. Also let
and
. By the Law of Cosines in
, we have the following equation:
and recalling that, from the problem,
, we see that
. Because
, this expression equates to
. By the formula for the distance from a vertex to the orthocenter and substitution, we know that
. Thus, because
,
. By Pythagoras in
,
. Equating this with our earlier expression for
, we get the following equation:
Because ,
. Plugging this into
yields
. Thus, our answer is
.