Difference between revisions of "1964 AHSME Problems/Problem 35"
(→See Also) |
(added another possibly time-consuming but good solution, i guess) |
||
Line 5: | Line 5: | ||
<math>\textbf{(A) }3:11\qquad\textbf{(B) }5:11\qquad\textbf{(C) }1:2\qquad\textbf{(D) }2:3\qquad \textbf{(E) }25:33</math> | <math>\textbf{(A) }3:11\qquad\textbf{(B) }5:11\qquad\textbf{(C) }1:2\qquad\textbf{(D) }2:3\qquad \textbf{(E) }25:33</math> | ||
− | ==Solution== | + | ==Solution 1== |
Using Law of Cosines | Using Law of Cosines | ||
and the fact that the ratio equals cos(a)/[cos(b)cos(c)] | and the fact that the ratio equals cos(a)/[cos(b)cos(c)] | ||
B 5:11 | B 5:11 | ||
+ | ==Solution 2 (coordinates)== | ||
+ | <asy> | ||
+ | draw((0,0)--(15,0)--(6.6,11.2)--(0,0)); | ||
+ | draw((0,0)--(9.6,7.2)); | ||
+ | draw((6.6,0)--(6.6,11.2)); | ||
+ | draw((15,0)--(3267/845,5544/845)); | ||
+ | label("$B$",(15,0),SE); | ||
+ | label("$C$",(6.6,11.2),N); | ||
+ | label("$E$",(6.6,0),S); | ||
+ | label("$15$",(7.5,-0.75),S); | ||
+ | label("$14$",(11,5.75),ENE); | ||
+ | label("$13$",(3,6),WNW); | ||
+ | label("$A$",(0,0),SW); | ||
+ | label("$D$",(9.6,7.2),NE); | ||
+ | label("$H$",(6.6,3.5),E); | ||
+ | </asy> | ||
+ | The reason why we have <math>HD</math> shorter than <math>HA</math> is that all the ratios' left hand side (<math>HD</math>) is less than the ratios' right hand side (<math>HA</math>). | ||
+ | |||
+ | We label point <math>A</math> as the origin and point <math>B</math>, logically, as <math>(15,0)</math>. By Heron's Formula, the area of this triangle is <math>84.</math> Thus the height perpendicular to <math>AB</math> has a length of <math>11.2,</math> and by the Pythagorean Theorem, <math>AE</math> and <math>EB</math> have lengths <math>6.6</math> and <math>8.4,</math> respectively. These lengths tell us that <math>C</math> is at <math>(6.6,11.2)</math>. | ||
+ | |||
+ | The slope of <math>BC</math> is <math>\dfrac{0-11.2}{15-6.6}=-\dfrac{4}{3},</math> and the slope of <math>AD</math> is <math>\dfrac{3}{4}</math> by taking the negative reciprocal of <math>-\dfrac{4}{3}.</math> Therefore, the equation of line <math>AD</math> can best be represented by <math>y=\dfrac{3}{4}x.</math> | ||
+ | |||
+ | We next find the intersection of <math>CE</math> and <math>AD</math>. We automatically know the <math>x</math>-value; it is just <math>6.6</math> because <math>CE</math> is a straight line hitting <math>(6.6,0).</math> Therefore, the <math>y</math>-value is at <math>\dfrac{3}{4}\times 6.6=4.95.</math> Therefore, the intersection between <math>CE</math> and <math>AD</math> is at <math>(6.6,4.95)</math>. | ||
+ | |||
+ | We also need to find the intersection between <math>BC</math> and <math>AD</math>. To do that, we know that the line of <math>AD</math> is represented as <math>y=\dfrac{3}{4}x,</math> and the slope of line <math>BC</math> is <math>-\dfrac{4}{3}.</math> We just need to find line <math>BC</math>'s y-intercept. So far, we have <math>y=-\dfrac{4}{3}x+b,</math> where <math>b</math> is a real y-intercept. We know that <math>B</math> is located at <math>(15,0),</math> so we plug that into the equation and yield <math>b=20.</math> Therefore, the intersection between the two lines is | ||
+ | <cmath> | ||
+ | \dfrac{3}{4}x=-\dfrac{4}{3}x+20, | ||
+ | 9x=-16x+240, | ||
+ | 25x=240, | ||
+ | x=9.6, | ||
+ | y=\dfrac{3}{4}\times 9.6, | ||
+ | y=7.2. | ||
+ | </cmath> | ||
+ | After that, we use the distance formula: <math>HA</math> has a length of <cmath>\sqrt{(6.6-0)^2+(4.95-0)^2}=\sqrt{\dfrac{1089}{25}+\dfrac{9801}{400}}=\sqrt{\dfrac{1089*16+9801}{400}}=\sqrt{\dfrac{27225}{400}}=\sqrt{\dfrac{1089}{16}}=\dfrac{33}{4}=8.25,</cmath> and <math>HD</math> has a length of <cmath>\sqrt{(9.6-6.6)^2+(7.2-4.95)^2}=\sqrt{3^2+(\dfrac{36}{5}-\dfrac{99}{20})^2}=\sqrt{9+\dfrac{81}{16}}=\sqrt{\dfrac{225}{16}}=3.75.</cmath> | ||
+ | Thus, we have that <math>\dfrac{3.75}{8.25}=\dfrac{\frac{15}{4}}{\frac{33}{4}}=\dfrac{15}{33}=\dfrac{5}{11}=\bold{\boxed{B}}.</math> | ||
==See Also== | ==See Also== | ||
{{AHSME 40p box|year=1964|num-b=34|num-a=36}} | {{AHSME 40p box|year=1964|num-b=34|num-a=36}} |
Revision as of 13:19, 21 April 2020
Problem
The sides of a triangle are of lengths , , and . The altitudes of the triangle meet at point . if is the altitude to the side of length , the ratio is:
Solution 1
Using Law of Cosines and the fact that the ratio equals cos(a)/[cos(b)cos(c)] B 5:11
Solution 2 (coordinates)
The reason why we have shorter than is that all the ratios' left hand side () is less than the ratios' right hand side ().
We label point as the origin and point , logically, as . By Heron's Formula, the area of this triangle is Thus the height perpendicular to has a length of and by the Pythagorean Theorem, and have lengths and respectively. These lengths tell us that is at .
The slope of is and the slope of is by taking the negative reciprocal of Therefore, the equation of line can best be represented by
We next find the intersection of and . We automatically know the -value; it is just because is a straight line hitting Therefore, the -value is at Therefore, the intersection between and is at .
We also need to find the intersection between and . To do that, we know that the line of is represented as and the slope of line is We just need to find line 's y-intercept. So far, we have where is a real y-intercept. We know that is located at so we plug that into the equation and yield Therefore, the intersection between the two lines is After that, we use the distance formula: has a length of and has a length of Thus, we have that
See Also
1964 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 34 |
Followed by Problem 36 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.