Difference between revisions of "1986 AIME Problems/Problem 15"
Aopsuser101 (talk | contribs) (→Solution 2) |
Aopsuser101 (talk | contribs) (→Solution 3) |
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<cmath>\tan (\beta - \alpha) = \frac{2 - 1}{1 + 2} = \frac{1}{3}</cmath>It follows that <math>\cos (\beta - \alpha) = \frac{1}{\sqrt{10}}</math> and <math>\sin (\beta - \alpha) = \frac{3}{\sqrt{10}}</math>. From now on, refer to <math>\theta = \beta - \alpha</math> | <cmath>\tan (\beta - \alpha) = \frac{2 - 1}{1 + 2} = \frac{1}{3}</cmath>It follows that <math>\cos (\beta - \alpha) = \frac{1}{\sqrt{10}}</math> and <math>\sin (\beta - \alpha) = \frac{3}{\sqrt{10}}</math>. From now on, refer to <math>\theta = \beta - \alpha</math> | ||
[asy] | [asy] | ||
− | |||
import graph; size(4cm); | import graph; size(4cm); | ||
real labelscalefactor = 0.5; /* changes label-to-point distance */ | real labelscalefactor = 0.5; /* changes label-to-point distance */ | ||
Line 58: | Line 57: | ||
draw(arc((2,2.6666666666666665),0.6,290.55604521958344,326.3099324740202)--(2,2.6666666666666665)--cycle, linewidth(1)); | draw(arc((2,2.6666666666666665),0.6,290.55604521958344,326.3099324740202)--(2,2.6666666666666665)--cycle, linewidth(1)); | ||
− | + | ||
draw((0,8)--(0,0), linewidth(1) + wrwrwr); | draw((0,8)--(0,0), linewidth(1) + wrwrwr); | ||
draw((0,0)--(6,0), linewidth(1) + wrwrwr); | draw((0,0)--(6,0), linewidth(1) + wrwrwr); | ||
Line 64: | Line 63: | ||
draw((0,8)--(3,0), linewidth(1) + wrwrwr); | draw((0,8)--(3,0), linewidth(1) + wrwrwr); | ||
draw((0,4)--(6,0), linewidth(1) + wrwrwr); | draw((0,4)--(6,0), linewidth(1) + wrwrwr); | ||
− | + | ||
dot((0,0),dotstyle); | dot((0,0),dotstyle); | ||
label("<math>A</math>", (0.08,0.2), NE * labelscalefactor); | label("<math>A</math>", (0.08,0.2), NE * labelscalefactor); | ||
Line 77: | Line 76: | ||
label("theta ", (2.14,2.3), NE * labelscalefactor); | label("theta ", (2.14,2.3), NE * labelscalefactor); | ||
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); | clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); | ||
− | + | ||
[/asy] | [/asy] | ||
Suppose <math>ABC</math> is our desired triangle. Let <math>E</math> be the midpoint of <math>CA</math> and <math>D</math> be the midpoint of <math>AB</math> such that <math>CE = EA = m</math> and <math>AD = DB = n</math>. Let <math>G</math> be the centroid of the triangle (in other words, the intersection of <math>EB</math> and <math>CD</math>). It follows that if <math>CG = 2x</math>, <math>GD = x</math> and if <math>GB = 2y</math>, <math>GE = y</math>. By the Law of Cosines on <math>\triangle GEB</math>, we get that: | Suppose <math>ABC</math> is our desired triangle. Let <math>E</math> be the midpoint of <math>CA</math> and <math>D</math> be the midpoint of <math>AB</math> such that <math>CE = EA = m</math> and <math>AD = DB = n</math>. Let <math>G</math> be the centroid of the triangle (in other words, the intersection of <math>EB</math> and <math>CD</math>). It follows that if <math>CG = 2x</math>, <math>GD = x</math> and if <math>GB = 2y</math>, <math>GE = y</math>. By the Law of Cosines on <math>\triangle GEB</math>, we get that: |
Revision as of 14:18, 18 April 2020
Problem
Let triangle be a right triangle in the xy-plane with a right angle at
. Given that the length of the hypotenuse
is
, and that the medians through
and
lie along the lines
and
respectively, find the area of triangle
.
Solution 1
Translate so the medians are , and
, then model the points
and
.
is the centroid, and is the average of the vertices, so
so
![$3600 = (a - b)^2 + (2b - a)^2$](http://latex.artofproblemsolving.com/1/8/a/18a313e3956fe0bcd1660ecabc9ac19b9aa56b2a.png)
![$3600 = 2a^2 + 5b^2 - 6ab \ \ \ \ (1)$](http://latex.artofproblemsolving.com/e/d/b/edb50bd2bb2dbca072addb484ed9837d12d12621.png)
and
are perpendicular, so the product of their slopes is
, giving
![$\left(\frac {2a + 2b}{2a + b}\right)\left(\frac {a + 4b}{a + 2b}\right) = - 1$](http://latex.artofproblemsolving.com/e/7/2/e72bb70cf41d0dccad96abfa6cd34625f432fe32.png)
![$2a^2 + 5b^2 = - \frac {15}{2}ab \ \ \ \ (2)$](http://latex.artofproblemsolving.com/d/3/9/d39bb21c6df6873302b21d1001209d1fa0a9c0a6.png)
Combining and
, we get
Using the determinant product for area of a triangle (this simplifies nicely, add columns 1 and 2, add rows 2 and 3), the area is , so we get the answer to be
.
Solution 2
The only relevant part about the xy plane here is that the slopes of the medians determine an angle between them that we will use.
This solution uses the tangent subtraction identity . Therefore, the tangent of the acute angle between the medians from A and B will be
.
Let and
. Let the midpoint of AC be M and the midpoint of BC be N. Let the centroid be G. By exterior angles,
. However we know that since
is the acute angle formed by the medians,
. We can express the tangents of the other two angles in terms of
and
.
while
. For simplification, let
.
By the tangent subtraction identity, .
We get the quadratic
, which solves to give
. It does not matter which one is picked, because the two roots multiply to 1, so switching from one root to another is like switching the lengths of AC and BC.
We choose the positive root and we can plug into of course the PYTHAGOREAN THEOREM
and solve for
. We want the area which is
.
Solution 3
We first seek to find the angle between the lines and
.
[asy]
import graph;
size(150);
Label f;
f.p=fontsize(6);
xaxis(-8,8,Ticks(f, 2.0));
yaxis(-8,8,Ticks(f, 2.0));
real f(real x)
{
return (x + 3);
}
real g( real x){
return (2x + 4);
}
draw(graph(f,-8,5),red+linewidth(1));
draw(graph(g,-6,2),blue+linewidth(1));
[/asy]
Let the acute angle the red line makes with the
axis be
and the acute angle the blue line makes with the
axis be
. Then, we know that
and
. Note that the acute angle between the red and blue lines is clearly
. Therefore, we have that:
It follows that
and
. From now on, refer to
[asy]
import graph; size(4cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -5.96, xmax = 24.22, ymin = -10.06, ymax = 14.62; /* image dimensions */
pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451);
draw(arc((2,2.6666666666666665),0.6,290.55604521958344,326.3099324740202)--(2,2.6666666666666665)--cycle, linewidth(1));
draw((0,8)--(0,0), linewidth(1) + wrwrwr); draw((0,0)--(6,0), linewidth(1) + wrwrwr); draw((6,0)--(0,8), linewidth(1) + wrwrwr); draw((0,8)--(3,0), linewidth(1) + wrwrwr); draw((0,4)--(6,0), linewidth(1) + wrwrwr);
dot((0,0),dotstyle);
label("", (0.08,0.2), NE * labelscalefactor);
dot((6,0),dotstyle);
label("
", (6.08,0.2), NE * labelscalefactor);
dot((0,8),dotstyle);
label("
", (0.08,8.2), NE * labelscalefactor);
dot((3,0),linewidth(4pt) + dotstyle);
label("
", (3.08,0.16), NE * labelscalefactor);
dot((0,4),linewidth(4pt) + dotstyle);
label("
", (0.08,4.16), NE * labelscalefactor);
label("theta ", (2.14,2.3), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
[/asy]
Suppose is our desired triangle. Let
be the midpoint of
and
be the midpoint of
such that
and
. Let
be the centroid of the triangle (in other words, the intersection of
and
). It follows that if
,
and if
,
. By the Law of Cosines on
, we get that:
By the Law of Cosines on
, we get that:
Adding yields that:
However, note that
. Therefore,
We also know that by the Law of Cosines on
,
Subtracting this from the
we got earlier yields that:
but recall that
to get that:
Plugging this into the
, we get that:
Aha! How convenient. Recall that
and
. Then, we clearly have that:
But note that the area of our triangle is
. As
, we get a final answer of
.
See also
1986 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Question | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
- AIME Problems and Solutions
- American Invitational Mathematics Examination
- Mathematics competition resources
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.