Art of Problem Solving
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2013 Mock AIME I Problems/Problem 12: Difference between revisions

Sugar rush (talk | contribs)
596 edits
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 <..."
 
Thepowerful456 (talk | contribs)
editor
642 edits
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 09:42, 31 July 2024

Problem

In acute triangle $ABC$, the orthocenter $H$ lies on the line connecting the midpoint of segment $AB$ to the midpoint of segment $BC$. If $AC=24$, and the altitude from $B$ has length $14$, find $AB\cdot BC$.

Solution (easy coordinate bash)

Toss on the coordinate plane with $A=(0,0)$, $B=(t,14)$, and $C=(24,0)$, where $t$ is a real number and $0<t<24$.

Then, the line connecting the midpoints of $AB$ and $BC$ runs from $\left(\frac{t}{2}, 7\right)$ to $\left(\frac{24+t}{2}, 7\right)$, or more simply the line $y=7$.

The orthocenter of $\triangle ABC$ will be at the intersection of the altitudes from $A$ and $B$.

The slope of the altitude from $A$ is the negative reciprocal of the slope of $\overline{BC}$. The slope of $\overline{BC}$ is $-\frac{14}{24-t}$, and its negative reciprocal is $\frac{24-t}{14}$. Since the altitude from $A$ passes through the origin, its equation is $y=\frac{24-t}{14}x$.

The altitude from $B$ is the vertical line running through $B=(t, 14)$ which has equation $x=t$.

Thus the lines $x=t$ and $y=\frac{24-t}{14}x$ meet on the line $y=7$. Substituting the first equation into the second, $\frac{t(24-t)}{14}=7$.

Multiplying both sides by $14$, we have $t(24-t)=98$.

This rearranges to the quadratic $t^{2}-24t+98=0$, and completing the square by adding $46$ to each side gives us $(t-12)^{2}=46$. Thus $t=12\pm\sqrt{46}$.

The cases where $t=12-\sqrt{46}$ and $t=12+\sqrt{46}$ are similar; they merely correspond to two triangles that can each be obtained by reflecting the other across the perpendicular bisector of $\overline{AC}$, so we consider the case where $t=12-\sqrt{46}$.

So $A=(0,0),B=(12-\sqrt{46},14),C=(24,0)$.

Thus \[AB\cdot BC=\sqrt{14^{2}+\left(12-\sqrt{46}\right)^{2}}\sqrt{14^{2}+\left(12+\sqrt{46}\right)^{2}}=\sqrt{\left(386-24\sqrt{46}\right)\left(386+24\sqrt{46}\right)}=\sqrt{386^{2}-46\cdot 24^{2}}=\boxed{350}.\]

The cases where $t=12-\sqrt{46}$ and $t=12+\sqrt{46}$ are shown below, labeled $\triangle ABC$ and $\triangle AB^{\prime}C$, 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 $AB\cdot BC$ is the same for both triangles.

[asy] real t = 12 - sqrt(46); pair A = (0, 0); pair B = (t, 14); pair C = (24, 0); draw(A--B--C--cycle); draw((A+B)/2--(B+C)/2, dashed); draw(B--(t, 0), blue); draw(A--(8.572, 11.55), red); draw(C--(2.927, 7.854), green); label("$A$", A, W); label("$B$", B, N); label("$C$", C, E); real t = 12 + sqrt(46); pair A = (0, 0); pair B = (t, 14); pair C = (24, 0); draw(A--B--C--cycle); draw((A+B)/2--(B+C)/2, dashed); draw(B--(t, 0), blue); draw(A--(21.073, 7.854), red); draw(C--(15.428, 11.55), green); label("$A$", A, W); label("$B^{\prime}$", B, N); label("$C$", C, E); [/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 $M$ be the midpoint of $\overline{AB}$ and $N$ be the midpoint of $\overline{BC}$. Further let $D$ be the foot of the altitude from $A$, $E$ from $B$, and $F$ from $C$, as in the diagram.

Because $\overline{MN}$ is a midpoint connector of $\triangle ABC$ amd $H$ is on $\overline{MN}$ and $\overline{BE}$, we know that $H$ is the midpoint of altitude $BE$. Thus, because, from the problem, $BE=14$, $BH=HE=7$. Now we see that $\overline{MH}$ is a midpoint connector of $\triangle BAE$, so $MH=\tfrac12 AE$.

Now, let $\measuredangle CHE = \theta$. We know that $\measuredangle BHF = \measuredangle CHE = \theta$, because they are vertical angles. Because $\triangle BHF$ is right (by the definition of an altitude), we know that $\measuredangle MBH = 90^{\circ}-\theta$. $\triangle BHM$ is also right, so $\measuredangle BMH = 90^{\circ} -(90^{\circ}-\theta)=\theta$.

From $\triangle HEC$, we know that $\tan\theta = \tfrac{CE}7$. From $\triangle BHM$, we know that $\tan\theta = \tfrac7{AE/2} = \tfrac{14}{AE}$. Equating these two expressions for $\tan\theta$, we see that $AE \cdot CE = 7 \cdot 14 = 98$. From the problem, we know that $AE+CE=AC=24$.

Now, we can proceed as in Solution 1 by using the quadratic formula to solve for $AE$ and the Pythagorean Theorem to find $AB$ and $BC$. We do this to obtain our answer $AB \cdot BC = \boxed{350}$.

See also

Retrieved from "https://wiki.aopstest.com/wiki/index.php?title=2013_Mock_AIME_I_Problems/Problem_12&oldid=224048"