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2013 Mock AIME I Problems/Problem 13

Problem

In acute $\triangle ABC$, $H$ is the orthocenter, $G$ is the centroid, and $M$ is the midpoint of $BC$. It is obvious that $AM \ge GM$, but $GM \ge HM$ does not always hold. If $[ABC] = 162$, $BC=18$, then the value of $GM$ which produces the smallest value of $AB$ such that $GM \ge HM$ can be expressed in the form $a+b\sqrt{c}$, for $b$ squarefree. Compute $a+b+c$.

Solution

Because $[\triangle ABC] = 168$ and $BC=18$, we know that the height from $A$ to $BC$ must be $18$. Thus, because the perpendicular is the shortest segment from a line to a point not on the line, we know that $AB \geq 18$. Thus, the minimum value of $AB$ is $18$, when $\overline{AB} \perp \overline{BC}$. This is technically not possible, because $\triangle ABC$ 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 $\triangle ABC$ is at $B$, $HM=BM=9$. Now, by the Pythagorean Theorem, $AM=9\sqrt5$. Because the centrod is $\tfrac23$ of the way along the median from the vertex, we know that $GM=\tfrac{9\sqrt5}3=3\sqrt5<9$. Thus, we have $GM<HM$, so the given inequality does not hold. Now, let us look at the example where $A$ is collinear with $O$ and $M$:

[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 $AG=\tfrac23 AM = \tfrac23 \cdot 18 = 12$. Let $R$ be the circumradius of $\triangle ABC$. Then, by Pythagoras in $\triangle OBM$, $OM=\sqrt{R^2-81}$. Because $OA=R$ and $AM=18$, we have the equation $\sqrt{R^2-81}+R=18$, which yields $AO=R=\tfrac{45}4<12=AG$, so $G$ is between $O$ and $M$. Because $\triangle ABC$ is acute, we know that $H$ is in the interior of $\triangle ABC$. This fact, combined with the properties of the Euler Line, show that $H$ must be closer to $M$ than $G$ is, so $GM>HM$, and the inequality is thereby satisfied.

As in the above diagram, let $C^{\prime}$ be the point on the circle such that $\overline{CC^{\prime}}$ is a diameter of the circle. Because $\triangle ABC$ is acute, $A$ and $B$ must be on the opposite sides of $\overline{CC^{\prime}}$ (so that the circumcenter lies inside the triangle). We want $A$ to be as close to $C^{\prime}$ as possible to minimize $AB$, but, from the first example we explored, we know that when $A=C^{\prime}$, $HM>GM$. We would reasonably expect the difference $GM-HM$ to vary continuously as we move $A$ towards $C^{\prime}$, 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 $A$ such that this difference is zero, or $GM=HM$. This point should be as close as we can get to $C^{\prime}$ while still satisfying the inequality.

Now, let $GM=HM$ and $D$ be the foot of the altitude from $A$ to $\overline{BC}$. Further, let $MD=x$, so $BD=9-x$, 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 $AM = \sqrt{AD^2+DM^2} = \sqrt{18^2+x^2} = \sqrt{x^2+324}$. Thus, by centroid properties, $GM = \tfrac13 AM = \tfrac{\sqrt{x^2+324}}3$. Now, we desire to find another expression for $GM=HM$. By using Pythagoras again, we see that $AB = \sqrt{(9-x)^2+18^2} = \sqrt{x^2-18x+405}$ and $AC = \sqrt{(9+x)^2+18^2} = \sqrt{x^2+18x+405}$. Now, let $\measuredangle BAC = \alpha$. Also let $AC=b$ and $AB=c$. By the Law of Cosines in $\triangle ABC$, 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 $\tfrac{abc}{4R}$ and recalling that, from the problem, $[\triangle ABC] = 162$, we see that $R=\tfrac{abc}{4 \cdot 162}$. Because $BC=18$, this expression equates to $\tfrac{18bc}{4 \cdot 162}=\tfrac{bc}{36}$. By the formula for the distance from a vertex to the orthocenter and substitution, we know that $AH=2R\cos\alpha=2(\tfrac{bc}{36})(\tfrac{x^2+243}{bc})=\tfrac{x^2+243}{18}$. Thus, because $AD=18$, $HD=18-\tfrac{x^2+243}{18}=\tfrac{81-x^2}{18}$. By Pythagoras in $\triangle HDM$, $HM = \sqrt{\tfrac{(81-x^2)^2}{324}+x^2} = \tfrac{\sqrt{81^2-162x^2+x^4+324x^2}}{18} = \tfrac{\sqrt{x^4+162x^2+81^2}}{18} = \tfrac{x^2+81}{18}$. Equating this with our earlier expression for $GM$, 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 $x^2>0$, $x^2=-63+36\sqrt7$. Plugging this into $GM=HM=\tfrac{x^2+81}{18}$ yields $\tfrac{-63+36\sqrt7+81}{18}=\tfrac{18+36\sqrt7}{18}=1+2\sqrt7$. Thus, our answer is $1+2+7=\boxed{010}$.

See also

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