Difference between revisions of "2008 AMC 10B Problems/Problem 24"
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<math>I</math> is the intersection of the two diagonals. The diagonals each form two isosceles triangles, <math>\triangle BCD</math> and <math>\triangle ABC</math>. | <math>I</math> is the intersection of the two diagonals. The diagonals each form two isosceles triangles, <math>\triangle BCD</math> and <math>\triangle ABC</math>. | ||
− | Using this, we find: <math>\angle DBC = \angle CDB = 5</math> and <math>\angle BAC = \angle BCA = 55</math>. Expanding on this, we can fill in a couple more angles. | + | Using this, we find: <math>\angle DBC = \angle CDB = 5^\circ</math> and <math>\angle BAC = \angle BCA = 55^\circ</math>. Expanding on this, we can fill in a couple more angles. |
− | <math>\angle ABD = 70 - 5 = 65</math>, <math>\angle ACD = 170 - 55 = 115</math>, <math>\angle BIA = \angle CID = 180 - (65 + 55) = 60</math>, <math>\angle BIC = | + | <math>\angle ABD = 70^\circ - 5^\circ = 65^\circ</math>, <math>\angle ACD = 170^\circ - 55^\circ = 115^\circ</math>, <math>\angle BIA = \angle CID = 180^\circ - (65^\circ + 55^\circ) = 60^\circ</math>, <math>\angle BIC = |
− | \angle AID = 180 - 60 = 120</math>. | + | \angle AID = 180^\circ - 60^\circ = 120^\circ</math>. |
− | We can rewrite <math>\angle CAD</math> and <math>\angle BDA</math> in terms of <math>x</math>. <math>\angle CAD = x - 55</math> and <math>\angle BDA = 180 - (120 + x - 55) = 115 - x</math>. | + | We can rewrite <math>\angle CAD</math> and <math>\angle BDA</math> in terms of <math>x</math>. <math>\angle CAD = x - 55^\circ</math> and <math>\angle BDA = 180^\circ - (120^\circ + x - 55^\circ) = 115^\circ - x</math>. |
Let us relabel <math>AB = BC = CD = a</math> and <math>AD = b</math>. | Let us relabel <math>AB = BC = CD = a</math> and <math>AD = b</math>. | ||
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By Rule of Sines on <math>\triangle ACD</math> and <math>\triangle ABD</math> respectively, <math>\frac{\sin(\angle CAD)}{a} = \frac{\sin(\angle ACD)}{b}</math>, and <math>\frac{\sin(\angle ABD)}{b} = \frac{\sin(\angle BDA)}{a}</math> | By Rule of Sines on <math>\triangle ACD</math> and <math>\triangle ABD</math> respectively, <math>\frac{\sin(\angle CAD)}{a} = \frac{\sin(\angle ACD)}{b}</math>, and <math>\frac{\sin(\angle ABD)}{b} = \frac{\sin(\angle BDA)}{a}</math> | ||
− | In a more convenient form, <math>\frac{\sin(x-55)}{a} = \frac{\sin(115)}{b} \implies \frac{a}{b} = \frac{\sin(x-55)}{\sin(115)}</math> | + | In a more convenient form, <math>\frac{\sin(x-55^\circ)}{a} = \frac{\sin(115^\circ)}{b} \implies \frac{a}{b} = \frac{\sin(x-55^\circ)}{\sin(115^\circ)}</math> |
− | and <math>\frac{\sin(65)}{b} = \frac{\sin(115-x)}{a} \implies \frac{a}{b} = \frac{\sin(115-x)}{\sin(65)}</math> | + | and <math>\frac{\sin(65^\circ)}{b} = \frac{\sin(115^\circ-x)}{a} \implies \frac{a}{b} = \frac{\sin(115^\circ-x)}{\sin(65^\circ)}</math> |
− | <math>\implies \frac{\sin(115-x)}{\sin(65)} = \frac{\sin(x-55)}{\sin(115)}</math> | + | <math>\implies \frac{\sin(115^\circ-x)}{\sin(65^\circ)} = \frac{\sin(x-55^\circ)}{\sin(115^\circ)}</math> |
− | Now, by identity <math>\sin(\theta) = \sin(180-\theta)</math>, <math>\sin(65) = \sin(115)</math> | + | Now, by identity <math>\sin(\theta) = \sin(180^\circ-\theta)</math>, <math>\sin(65^\circ) = \sin(115^\circ)</math> |
− | Therefore, <math>\sin(115-x) = \sin(x-55).</math> This equation is only satisfied by option | + | Therefore, <math>\sin(115^\circ-x) = \sin(x-55^\circ).</math> This equation is only satisfied by option $C.\boxed{(85^\circ)}$ |
Note: I'm pretty bad at Asymptote, if anyone could edit this and fill in the angles into the diagram, that would be pretty cool. | Note: I'm pretty bad at Asymptote, if anyone could edit this and fill in the angles into the diagram, that would be pretty cool. |
Revision as of 16:03, 20 August 2022
Problem
Quadrilateral has , angle and angle . What is the measure of angle ?
Solution 1
This solution requires the use of cyclic quadrilateral properties but could be a bit time-consuming during the contest.
To start off, draw a diagram like in solution one and label the points. Now draw and and call their intersection point . Note that triangle is an isosceles triangle so angles and are each degrees. Since equals , angle equals degrees, thus making angle equal to degrees. We can also find out that angle equals degrees.
Extend and and let their intersection be . Since angle plus angle equals degrees, quadrilateral is a cyclic quadrilateral.
Next, draw a line from point to point . Since angle and angle point to the same arc, angle is equal to degrees. Since is an isosceles triangle (based on angle properties) and is also an isosceles triangle, we can find that is also an isosceles triangle. Thus, each of the other angles is degrees. Finally, we have angle equals degrees.
~Minor edits by BakedPotato66
Solution 2
First, connect the diagonal , then, draw line such that it is congruent to and is parallel to . Because triangle is isosceles and angle is , the angles and are both . Because angle is , we get angle is . Next, noticing parallel lines and and transversal , we see that angle is also , and subtracting off angle gives that angle is .
Now, because we drew , triangle is equilateral. We can also conclude that meaning that triangle is isosceles, and angles and are equal.
Finally, we can set up our equation. Denote angle as . Then, because is a parallelogram, the angle is also . Then, is . Again because is a parallelogram, angle is . Subtracting angle gives that angle equals . Because angle equals angle , we get , solving into .
Side note: this solution was inspired by some basic angle chasing and finding some 60 degree angles, which made me want to create equilateral triangles.
~Someonenumber011
Solution 3(Using Trig.)
Let the unknown be .
First, we draw diagonal and . is the intersection of the two diagonals. The diagonals each form two isosceles triangles, and .
Using this, we find: and . Expanding on this, we can fill in a couple more angles. , , , .
We can rewrite and in terms of . and .
Let us relabel and .
By Rule of Sines on and respectively, , and
In a more convenient form,
and
Now, by identity ,
Therefore, This equation is only satisfied by option $C.\boxed{(85^\circ)}$
Note: I'm pretty bad at Asymptote, if anyone could edit this and fill in the angles into the diagram, that would be pretty cool.
~Raghu9372
See also
2008 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.