Difference between revisions of "Power of a Point Theorem/Introductory Problem 4"
| (2 intermediate revisions by 2 users not shown) | |||
| Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
| − | ([[ARML]]) Chords <math> AB </math> and <math> CD </math> of a given circle are perpendicular to each other and intersect at a right angle. Given that <math> BE = 16, DE = 4, </math> and <math> AD = 5 </math>, find <math> CE </math>. | + | ([[ARML]]) Chords <math> AB </math> and <math> CD </math> of a given circle are perpendicular to each other and intersect at a right angle at <math>E</math> . Given that <math> BE = 16, DE = 4, </math> and <math> AD = 5 </math>, find <math> CE </math>. |
== Solution == | == Solution == | ||
| − | <math> ADE </math> is a right triangle with hypotenuse 5 and leg 4. Thus, by the [[Pythagorean Theorem]], <math> AE = 3 </math> (or by just knowing your [[Pythagorean Triple]]s). Applying the Power of a Point Theorem gives <math> AE\cdot BE = CE\cdot DE </math> or <math> 3\cdot 16 = x\cdot 4 </math>. Solving gives <math> x = 12 </math>. | + | <asy> |
| + | import olympiad; | ||
| + | pair E = (0,0); | ||
| + | pair A = (3,0); | ||
| + | pair D = (0,4); | ||
| + | pair B = (-16,0); | ||
| + | pair C = (0,-12); | ||
| + | dot("E", E, SW); | ||
| + | dot("A", A, dir(0)); | ||
| + | dot("D", D, N); | ||
| + | dot("B", B, W); | ||
| + | dot("C", C, SE); | ||
| + | draw(A--D--E--cycle); | ||
| + | draw(B--E); | ||
| + | draw(C--E); | ||
| + | label("5", A--D, NE); | ||
| + | label("3", E--A, S); | ||
| + | label("4", D--E, W); | ||
| + | label("16", B--E, N); | ||
| + | label("x", C--E, W); | ||
| + | markscalefactor=0.1pt; | ||
| + | draw(rightanglemark(D, E, A)); | ||
| + | draw(circumcircle(B, D, A)); | ||
| + | </asy> | ||
| + | <math> ADE </math> is a right triangle with hypotenuse 5 and leg 4. Thus, by the [[Pythagorean Theorem]], <math> AE = 3 </math> (or by just knowing your [[Pythagorean Triple]]s). Applying the Power of a Point Theorem gives <math> AE\cdot BE = CE\cdot DE </math>, or <math> 3\cdot 16 = x\cdot 4 </math>. Solving gives <math> x = 12 </math>. | ||
''Back to the [[Power of a Point Theorem]].'' | ''Back to the [[Power of a Point Theorem]].'' | ||
Latest revision as of 02:14, 26 November 2020
Problem
(ARML) Chords 
and 
of a given circle are perpendicular to each other and intersect at a right angle at 
. Given that 
and 
, find 
.
Solution
![[asy] import olympiad; pair E = (0,0); pair A = (3,0); pair D = (0,4); pair B = (-16,0); pair C = (0,-12); dot("E", E, SW); dot("A", A, dir(0)); dot("D", D, N); dot("B", B, W); dot("C", C, SE); draw(A--D--E--cycle); draw(B--E); draw(C--E); label("5", A--D, NE); label("3", E--A, S); label("4", D--E, W); label("16", B--E, N); label("x", C--E, W); markscalefactor=0.1pt; draw(rightanglemark(D, E, A)); draw(circumcircle(B, D, A)); [/asy]](http://latex.artofproblemsolving.com/f/a/f/fafd9b1f8fb2407c8e9079fb6e40589e4415ce5e.png)
is a right triangle with hypotenuse 5 and leg 4. Thus, by the Pythagorean Theorem, 
(or by just knowing your Pythagorean Triples). Applying the Power of a Point Theorem gives 
, or 
. Solving gives 
.
Back to the Power of a Point Theorem.
