Difference between revisions of "1963 AHSME Problems/Problem 39"
Rockmanex3 (talk | contribs) (Solution to Problem 39) |
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<cmath>a = \tfrac{5}{2}b</cmath> | <cmath>a = \tfrac{5}{2}b</cmath> | ||
Thus, <math>[CAP] = 15b</math>, and since <math>[APE] = 3b</math>, <math>r = \tfrac{CP}{PE} = 5</math>, which is answer choice <math>\boxed{\textbf{(D)}}</math>. | Thus, <math>[CAP] = 15b</math>, and since <math>[APE] = 3b</math>, <math>r = \tfrac{CP}{PE} = 5</math>, which is answer choice <math>\boxed{\textbf{(D)}}</math>. | ||
+ | |||
+ | ==Solution 2 (Mass Geometry)== | ||
+ | Let the mass of point <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math>, and <math>E</math> be <math>mA</math>, <math>mB</math>, <math>mC</math>, <math>mD</math>, and <math>mE</math> respectively. | ||
+ | By mass geometry theorems, we have | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | \frac{CP}{PE} = \frac{mE}{mC} | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | Focusing on the line segment <math>AB</math>, using mass geometry theorems, we have | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | mE = mA + mB | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | and | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | \frac{3}{2} &= \frac{mB}{mA} \ | ||
+ | mA &= \frac{2}{3}mB | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | which leads to <math>mE = \frac{5}{3}mB</math>. | ||
+ | For line segment <math>CB</math>, similarly, we got | ||
+ | <cmath> | ||
+ | mC = \frac{1}{3}mB | ||
+ | </cmath> | ||
+ | Substituting <math>mC</math> and <math>mE</math> back to the equation we obtained at the beginning, we got: | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | \frac{CP}{PE} = \frac{mE}{mC} = \frac{\frac{5}{3}mB}{\frac{1}{3}mB} = 5 | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | which gives us the answer choice <math>\boxed{\textbf{(D)}}</math>. -nullptr07 | ||
==See Also== | ==See Also== |
Latest revision as of 13:12, 29 June 2023
Problem 39
In lines and are drawn so that and . Let where is the intersection point of and . Then equals:
Solution
Draw line , and let , , and , so and . Because and share an altitude, Because and share an altitude, Thus, , and since , , which is answer choice .
Solution 2 (Mass Geometry)
Let the mass of point , , , , and be , , , , and respectively. By mass geometry theorems, we have Focusing on the line segment , using mass geometry theorems, we have and which leads to . For line segment , similarly, we got Substituting and back to the equation we obtained at the beginning, we got: which gives us the answer choice . -nullptr07
See Also
1963 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 38 |
Followed by Problem 40 | |
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All AHSME Problems and Solutions |
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