Week 5

10/2/2015 1003 Precalculus Art of Problem Solving

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HOMEWORK WEEK 5

Samer Seraj

TuesdayApr 14, 2015 Sep 8, 20157:309:00 PM ET, 4:306:00 PM PT

Overview Homework Message Board Report

Week: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22

You have completed 8 of 8 challenge problems.Your 2 writing problem responses have been graded.Past Due May 20.

Lesson 5: Sections 4.2, 4.3Lesson 6: Sections 4.1, 4.4Lesson 5 Transcript: p Tue, May 12

Challenge Problems 3

Problem 1 – Correct! – Score: 7 / 7 (4529)

Problem:Points , , and are on a circle such that , , and . Find the area of thecircle.

Solution:Let be the radius of the circle. From the Law of Sines, we have

Therefore, we have , so the area of the circle is .

Your Response(s):$ 32\pi


Problem:Two rays with common endpoint form a angle. Point lies on one ray, point on the other ray, and

. What is the maximum possible length of ?

Solution:

Precalculus (1003)

Homework: Week 5

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We begin with the diagram below.

By the Law of Sines, we have

so, with equality if and only

if . Therefore, the maximum possible length of is .

Incidentally, we notice that is a 306090 right triangle. The diagram below is more accurate:

Your Response(s):$ 2


Problem:In triangle , we have , and . Calculate .

Solution:From the Law of Cosines, we have , or

Plugging in the known values for , , and yields

Your Response(s):% \frac157168$ 59/84



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Problem:If Jane walks North for 3 miles, turns to the right, and then walks another 4 miles, how many miles will Jane befrom her starting point? Give your answer as a decimal rounded to the nearest hundredth. (You may use a calculatorto compute the approximation.)

Solution:We'll start with a diagram. Jane starts at , walks to , turns to the right, and then walks to point .

Since , applying the Law of Cosines to triangle gives

Therefore, we have .

Your Response(s):$ 6.48


Problem:In , we have and . Side and themedian from to have the same length. What is ?

Solution:



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Let be the midpoint of , let , and let . Then ,and Applying the Law of Cosines to andto yields, respectively,

and

Adding, we obtain , so and .

Alternate answer: Applying the Law of Cosines to and yields, respectively,

and

From the first equation, Substituting this into the second equation gives

The remainder of the solution continues as above.

Your Response(s):$ sqrt(10)


Problem: is an equilateral triangle with side length 4. is the midpoint of , and is a diagonal of square

. Find the area of the region common to both and .

Solution:Solution 1: We start with a diagram:



Let be the intersection of and , and let be the intersection of and . The desired area is . Since is a square, we have , so

From 306090 triangle , we have , so we have

Next, since and , triangle gives us . Applying the Law of Sines to , we have

so

We have

so



Finally, we have

Solution 2: We use the same notation as in Solution 1. Let be the projection of onto , and let .

Triangle is , so . Triangle is , so

. Hence, .

As in Solution 1, , so



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Then the area of triangle is

so the area of quadrilateral is

Your Response(s):% 6+6sqrt(6)$ 6+6sqrt(3)

Problem 7 Part (a) – Correct! – Score: 7 / 7 (4541)

Problem:In , we have , , and . Find the sum of all possible values of .

Solution:Applying the Law of Cosines to gives

so letting gives

so . Factoring gives , so or . Both are valid

solutions, so the sum of all possible values of is .

Your Response(s):$ 16

Problem 7 Part (b) – Correct! – Score: 7 / 7 (4542)

Problem:In , we have , , and . Find the sum of all possible values of .

Read closely; this is not the same as the previous part!

Solution:Once again, applying the Law of Cosines to gives

so letting gives



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so . Applying the quadratic formula gives . However, is

negative, and cannot be negative. Therefore, the only possible value of is .

Your Response(s):$ 7+sqrt(109)

Problem 8 – Graded – Technical: 5 / 7 – Style: 0.8 / 1 (8823)

Problem:In most geometry courses, we learn that there's no such thing as "SSA Congruence". That is, if we have triangles

and such that , , and , then we cannot deduce that and are congruent.

However, there are a few special cases in which SSA "works". That is, suppose is a triangle. Let , , and . For some values of , , and , we can uniquely determine the third side, .

(a) Use the Law of Cosines to derive a quadratic equation in .

(b) Use the quadratic polynomial you found in part (a) in order to find conditions on and which guarantee thatthe side is uniquely determined.

Solution:(a) Applying the Law of Cosines gives . We write our equation as aquadratic in :

(b) There is a unique value of if and only the quadratic derived in (a) has one distinct positive root. (It has at leastone root, since we assumed was a triangle.) Next, we apply the quadratic formula to get:

Now, we consider two cases:

Case 1: One distinct root: This occurs if and only if , which gives .

Case 2: Two roots, one nonpositive: This occurs if and only if , so



. Right away, we note that this holds whenever , since

in this case. (Note that implies .) Next, we consider , in which case we have

. Rearranging this inequality gives . Since and are positive, gives us .

We conclude that we have two roots, one nonpositive, whenever .

Combining our two cases, SSA "works" if or if and . Otherwise, either no triangle is

possible, or there are two noncongruent possibilities.

Your Response:a) For this part, we have to use the Law of Cosines to derive a quadratic equation in AC.

First, let's write the Law of Cosines for this specific triangle:

The side lengths are already defined for us in the problem, so let's substitute x for AB, y for BC, and theta for angleBAC:

Let's rewrite it:

b) To find specific conditions in which and guarantee that the side is uniquely determined, we shouldhave an equation that equals .One way to do this is to complete the square. Another way is to use the quadratic formula.

Now, we are going to attempt to complete the square, so that we can get an equation that is equal to AC.

First, move the value of to the other side of the equal sign.

Now, we must complete the square by adding to both sides, half the value of , squared.



Simplify:

Now, we have our value of AC in the form of and .

But what restrictions can we make so that the value of AC is only one value?

First of all, AC is a measurement of distance, so it cannot be negative. It must be positive.

What restrictions can we make so that AC is only one positive value?

Their are 5 possible situations:Both values of AC are negative,one value is negative, the other is positive,one value is negative, the other is 0one value is positive, the other is 0, andboth values of AC are positive.

We are looking for either the second or the fourth possibilities.

In order for this to be possible, we must have the value of the second term be larger than or equal to the value of thefirst term. If this is so, one value will be positive, and the other will be negative or 0.

So the following restriction must be made:

If we simplify:

The restriction is:

or



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Technical Score: 5 / 7Style Score: 0.8 / 1Comments:Good work! You did a good job fully explaining your steps and (almost) reaching the correct conclusion. You arecorrect that there is a unique solution when , but there is one more case that you missed: exactly one positivesolution for . This is the case where the discriminant is equal to . On that note, if you are solving a quadraticequation and it looks complicated, it is fine to just use the quadratic formula instead of completing the square.

You should be careful when squaring both sides of an inequality (or an equality!), as it is entirely possible that you areintroducing extraneous solutions when doing so. For example, , but squaring both sides gives us an

incorrect result. We can take case of this in this instance by noting that is negative when , which

implies directly that , so it's not a case we need to worry too much about, but you should still explicitly handlethis in your solution.

Regarding style, each sentence should end with a period, even if the sentence ends with an equation. I noticed thatyou did not put a period at the end of certain equations even though they started off as sentences.

You have thanked the grader!

Problem 9 – Graded – Technical: 7 / 7 – Style: 0.9 / 1 (4548)

Problem:Show that in any triangle, we have

where is the circumradius of the triangle.

Solution:From the Law of Sines, , , and . Substituting these values, thelefthand side becomes

and the righthand side becomes

Thus, the given equation becomes



The numerators in both fractions are equal to , so this equation is equivalent to

We offer two solutions from here, followed by a third solution that does not use this setup.

Solution 1: Grind through trig identities.We will prove this equation by manipulating the lefthand side. By the double angle identity,

so

By the sumtoproduct identity,

Since , . Therefore,

Again by the sumtoproduct identity,

But

so



and

so

Therefore,

as desired.

Solution 2: Whizzy cool geometry.We use doubleangle identities to write

Now, consider the case in which is acute, and let be its circumcenter:

Since is inscribed in minor arc , we have . Similarly, we have and . Next, we note that



However, we also can use the Law of Sines once again to note

Combining our two expressions for gives the desired

The righttriangle case is straightforward: if (so and ), then , and

as desired.

Finally, if angle is obtuse, then we can use essentially the same area argument as above, but with the wrinkle together with the fact that , so

. The rest of the details are the same.

Solution 3: More geometry, less trig.We use the same setup as in "whizzy cool geometry," but we calculate the areas slightly differently, and we don't usethe trigonometry from the beginning. Suppose that is acute. We note that is an isoscelestriangle with a base of length , apex angle , and legs of length . We find its area by dropping an altitude from to the midpoint of side .

Since has the same measure as and , we can use right triangle to conclude that



. Therefore .

Similarly, we have and . This allows us to conclude that

Therefore

We can derive the same result in the case when is not acute, as in whizzy cool geometry.

On the other hand, from the law of sines, we have

Putting this together with the result from the previous paragraph, we find that

On the other hand, the law of sines and the formula for the area of a triangle allow us to derive theidentity

We can rewrite this as

Picking up where we left off, we find that



Hint(s):

as desired.

Turn the problem into a trig problem: get rid of the side lengths.Remember that , so .

Your Response:Let's try to manipulate the left side of the equation to get the right side.

First, let's replace the values of and with other values.

Using the Extended Law of Sines:

Since the same will be true for and , we can replace those values of the equation:

The numerator of our fraction looks quite similar to the numerator of the fraction we are trying to get.

Now, we must work on the denominator.

Let's try replacing the values of and in the equation with other values.We could use the Extended Law of Sines again:



Let's substitute this value and the corresponding values for the variables b and c into the denominator:

We have seen the values of before in the double angle formulas. So, substitute again.

We can then simplify this equation using sum to product formulas and other identities:

We can use the identities

,,

and

to simplify the denominator further:



Use the Extended Law of Sines again:

Finally, we can plug this simplified denominator into the original equation and simplify.

The equation is proven!

*Note: Detailed work is attached .

File: Week 5 Problem 9.pdf [Download]

Technical Score: 7 / 7Style Score: 0.9 / 1Comments:Good job on a correct proof! One thing that you should try to do in your proofs, however, is explain every step that youtake. Right now, you structure your proof very well, spacing it out vertically, but there are some steps that you skipover. You include these in the pdf that you've attached, but it is difficult to understand the reasoning there, because thedetailed work is missing the explanation and the motivation behind the steps.

For instance, when you simplify

to

you use the sumtoproduct identity, but you don't mention this identity before, and you don't show the manipulationswith the angle measures that are necessary to show this step. As a result, you lose your reader, who has to try andprove this on his/her own. Remember, the best proofs are the ones that fully explain every single step taken.

Regarding style, each sentence should end with a period, even if the sentence ends with an equation.

You have thanked the grader!

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