Difference between revisions of "1997 AHSME Problems/Problem 26"
Talkinaway (talk | contribs) |
m (→Solution) |
||
(4 intermediate revisions by 4 users not shown) | |||
Line 22: | Line 22: | ||
The product of two lengths with a common point brings to mind the [[Power of a Point Theorem]]. | The product of two lengths with a common point brings to mind the [[Power of a Point Theorem]]. | ||
− | Since <math>PA = PB</math>, we can make a circle with radius <math>PA</math> that is centered on <math>P</math>, and both <math>A</math> and <math>B</math> will be on that circle. Since <math>\angle APB = \widehat {AB} = 2 \angle ACB</math>, we can see that point <math>C</math> will also lie on the circle, since the measure of arc <math>\widehat {AB}</math> is twice the | + | Since <math>PA = PB</math>, we can make a circle with radius <math>PA</math> that is centered on <math>P</math>, and both <math>A</math> and <math>B</math> will be on that circle. Since <math>\angle APB = \widehat {AB} = 2 \angle ACB</math>, we can see that point <math>C</math> will also lie on the circle, since the measure of arc <math>\widehat {AB}</math> is twice the measure of inscribed angle <math>\angle ACB</math>, which is true for all inscribed angles. |
<asy> | <asy> | ||
Line 44: | Line 44: | ||
We now extend radius <math>PB = 3</math> to diameter <math>EB = 6</math>. Since <math>EDB</math> is a line, we have <math>ED + DB = EB</math>, which gives <math>ED + 1 = 6</math>, or <math>ED = 5</math>. | We now extend radius <math>PB = 3</math> to diameter <math>EB = 6</math>. Since <math>EDB</math> is a line, we have <math>ED + DB = EB</math>, which gives <math>ED + 1 = 6</math>, or <math>ED = 5</math>. | ||
− | Finally, we apply the | + | Finally, we apply the power of a point theorem to point <math>D</math>. This states that <math>AD \cdot DC = DB \cdot DE</math>. Since <math>DB = 1</math> and <math>DE = 5</math>, the desired product is <math>5</math>, which is <math>\boxed{A}</math>. |
+ | ==Solution 2== | ||
+ | Construct the angle bisector of <math>\angle APD,</math> and let it intersect <math>AD</math> at <math>E.</math> From the angle bisector theorem, we have <math>AE=3a</math> and <math>DE=2a</math> for some <math>a.</math> Then, note that <math>\angle EPD = \angle BCD = x,</math> so <math>EPCB</math> is cyclic. Then, <math>\frac{PD}{ED} = \frac{CD}{BD}</math> or <math>\frac{2}{2x} = \frac{CD}{1}.</math> Thus, <math>AD \cdot DC = 5x \cdot DC = 5,</math> or <math>\boxed{A}.</math> | ||
== See also == | == See also == | ||
{{AHSME box|year=1997|num-b=25|num-a=27}} | {{AHSME box|year=1997|num-b=25|num-a=27}} | ||
+ | {{MAA Notice}} |
Latest revision as of 17:24, 6 November 2022
Contents
Problem
Triangle and point in the same plane are given. Point is equidistant from and , angle is twice angle , and intersects at point . If and , then
Solution
The product of two lengths with a common point brings to mind the Power of a Point Theorem.
Since , we can make a circle with radius that is centered on , and both and will be on that circle. Since , we can see that point will also lie on the circle, since the measure of arc is twice the measure of inscribed angle , which is true for all inscribed angles.
Since is a line, we have , which gives , or .
We now extend radius to diameter . Since is a line, we have , which gives , or .
Finally, we apply the power of a point theorem to point . This states that . Since and , the desired product is , which is .
Solution 2
Construct the angle bisector of and let it intersect at From the angle bisector theorem, we have and for some Then, note that so is cyclic. Then, or Thus, or
See also
1997 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 25 |
Followed by Problem 27 | |
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 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.