[Izzy]

I went to a math competition today (Seven points away from getting a trophy, but I'm done ranting about that, honest), and I thought the group competition was fairly interesting, so I saved the questions. We were given up to 4 minutes to solve the questions, more points the faster you finish, but I'm not going to time you. You only have pencil and paper. No calculator. Have fun. (Mind, I don't have all the answers, but I'm sure we can figure them out amongst ourselves.)

For those of you that actually want to time yourself to get a proper score, here's how it works. If you answer (correctly..) within the first minute, 16 points. Second minute, 8 points. Third 4. First 1.

In ABC, A + B = 130° and A + C = 110°. How large is A?

The sum of the two exterior angles formed by extending the hypotenuse of a right triangle is how many degrees?

Find the number of degrees in the angle formed by two diagonals drawn from the same vertex of a regular pentagon.

(Ignore this one. The diagram didn't have the numbers, so we were told to skip it.)

The bases of an isosceles trapezoid are AB = 15 inches and DC = 9 inches respectively; each leg is 5 inches. Find the length of a diagonal.

In the triangle ABC, AC = BC and AD bisects angle A. If ADC = 114°, then what is the measure of C?

The perimeter of a rhombus is 80 cm, and one diagonal is 14 cm long. Find the length of the other diagonal.

If the sum of the interior angles of a polygon equals four times the sum of the exterior angles, how many sides does the polygon have?

If the segments of the hypotenuse of a right triangle made by the altitude to the hypotenuse are 3 and 12, find the altitude.

Suppose the medians AA' and BB' of triangle ABC intersect at right angles. If BC = 3 and AC = 4, what is the length of side AB?

The sides of a right triangle are a, a+d, and a+2d, with a and d both positive. Find the ratio of a to d. (This was one of the stupid questions we ended up kicking ourselves about. We did the ratio backwards..)

HIJK and DCBA are congruent parallelograms, with altitude HL. HK = 10 cm, LJ = 8 cm, and the area of ABCD is 112cm^2. What is the length of altitude DM? (Not drawn to scale..)

The coordinates of the vertices of ABC are A(-5,1), B(-2,3), and C(-4,7). The triangle is reflected over the line x=2. What is the sum of the x-coordinates of the images of points A,B, and C?

An isosceles triangle has lengths 6x+40, 4x+104, and 7x-23. What is the greatest possible perimeter?

Two congruent regular polygons are placed side-by-side so they share a common side. The angle formed between the polygons is a right angle. How many sides does each polygon have?

[Guest]

ok this is all I got... my brain hurts now and i'm too hung over to even atempt the other ones.... dont even know if my answers are right for sure.

60

90

90

14 cm

Edited January 31, 2009 by compz

[Guest]

Knowing that these should be timed, I find myself incapable of doing this puzzle until I can get a stop watch. <_< I think my OCD is showing.

[Guest]

A minute a piece to a question right? or the entire thing ?

[Guest]

Actually Compz I think 2 is 270'

Angles go 90:45:45.

If you extend the Hyp. which will inturn produce 2 exterior angles, along the 45' angles.

Since a straight line is 180's. The complimentary of the 45' is 135.

Now it ask for the sum of both so 2(135)=270.

Edited February 1, 2009 by bonanova
Spoiler added.

[Izzy]

Compz got number 1, and Ionno got 2.

[Guest]

1. A = 60°

2. 270°

3. 108°/3 = 36°

5. Height of the trapezoid must be 4. Therefore a diagonal will be the hypothenuse of a 12 by 4 right triangle, which is 4* sqrt(10)

6. A=32°, B=98°, C=50°

7. sqrt(400-49)/2 = 3/2 * sqrt(13)* sqrt(3)

8. 4

9. 144 + 9 + 2*x^2 = 225. => x = 6.

Can't think of a clear solution for n°10;

11. a/d = 2 + 3 (d/a). => a/d = 3

12. HL=10, KL=6. We know that BC*MD=112. So, depending on the way the parallelograms are congruent, we have either (if IJ=BC=10cm) DM=11,2cm ; or (if BC = KJ = 14cm) DM=HL=8cm. The problem doesn't seem to provide enough info to clarify this.

13. 23

14. perimeter is 17x + 121. Greatest value of x is when the equal sides are (6x + 40) and (7x - 23), with x= 63. Result is a perimeter of 1192

15. each interior angle must thus be of (360-90)/2 = 135 degrees. we're talking octagons here. Specifically: n*135° = (n-2)*180° yields n = 8

[Izzy]

See spoiler.

1. A = 60°

Yes

2. 270°

Yes

3. 108°/3 = 36°

Yes

5. Height of the trapezoid must be 4. Therefore a diagonal will be the hypothenuse of a 12 by 4 right triangle, which is 4* sqrt(10)

Yes

6. A=32°, B=98°, C=50°

I don't remember. When we did this one, I misread and solved for ADB.

7. sqrt(400-49)/2 = 3/2 * sqrt(13)* sqrt(3)

Umm.. I'm not sure what you did. I suck at square roots in meh head, so I did it with a calculator just now and got roughly 37. sqrt(400-49) * 2 = 37.469.

8. 4

No.

9. 144 + 9 + 2*x^2 = 225. => x = 6.

Hell if I know. You're probably right.

Can't think of a clear solution for n°10;

Neither could we.

11. a/d = 2 + 3 (d/a). => a/d = 3

Yep. We put 1:3.

12. HL=10, KL=6. We know that BC*MD=112. So, depending on the way the parallelograms are congruent, we have either (if IJ=BC=10cm) DM=11,2cm ; or (if BC = KJ = 14cm) DM=HL=8cm. The problem doesn't seem to provide enough info to clarify this.

HI=KJ=DC=AB=(You have to figure this bit out..). HK=IJ=DA=CB=8.

DM does = 11.2 cm.

13. 23

Yes.

14. perimeter is 17x + 121. Greatest value of x is when the equal sides are (6x + 40) and (7x - 23), with x= 63. Result is a perimeter of 1192

Eh, probably. We didn't get this one either.

15. each interior angle must thus be of (360-90)/2 = 135 degrees. we're talking octagons here. Specifically: n*135° = (n-2)*180° yields n = 8

Yep.

Nice job.

[Guest]

I have not done geometry or been to a math competition in years, but I think that I have this one right.

8 sides.

[Izzy]

I have not done geometry or been to a math competition in years, but I think that I have this one right.

8 sides.

Nope. I think I know where you are going wrong. Remember that the formula for interior angles is 180(n-2).

Edited February 1, 2009 by Izzy

[bonanova]

Timed questions are unrealistic, but kind of fun.

Problem is you can have the right approach and make a mistake in arithmetic.

I got 1-minute answers for most; time ran out on numbers 7 and 10.

In ABC, A + B = 130° and A + C = 110°. How large is A?

A=B+110-A=180; B=70

A=60^o

The sum of the two exterior angles formed by extending the hypotenuse of a right triangle is how many degrees?

180-x + 180-[90-x].

270^o

Find the number of degrees in the angle formed by two diagonals drawn from the same vertex of a regular pentagon.

The five diagonals cycle through 180^o.

Each angle is 1/5 of that: 36^o.

(Ignore this one. The diagram didn't have the numbers, so we were told to skip it.)

The bases of an isosceles trapezoid are AB = 15 inches and DC = 9 inches respectively; each leg is 5 inches. Find the length of a diagonal.

The trapezoid is immediately seen to be a 4x9 rectangle with 3/4/5 right triangles on each side.

The diagonal is the hypotenuse of a 4/12/x triangle, where x=sqrt[160].

In the triangle ABC, AC = BC and AD bisects angle A. If ADC = 114°, then what is the measure of C?

This took 2 minutes.

Let the equal angles be 2x.

Triangle ADB sums to 66 [180-114] + 3x so x=38.

C = 180 - 114 - x = 88^o

The perimeter of a rhombus is 80 cm, and one diagonal is 14 cm long. Find the length of the other diagonal.

Time ran out on this one ... maybe look at it later.

If the sum of the interior angles of a polygon equals four times the sum of the exterior angles, how many sides does the polygon have?

The interior angle is 4 times the exterior angle and their sum is 180.

5x=180 -> x=36^o exterior angle.

The polygon has 360/36 = 10 sides.

If the segments of the hypotenuse of a right triangle made by the altitude to the hypotenuse are 3 and 12, find the altitude.

Let the other sides be x and y.

x² + y² = 15² where x² = a² + 12² and y² = a² + 3²

a=6.

Suppose the medians AA' and BB' of triangle ABC intersect at right angles. If BC = 3 and AC = 4, what is the length of side AB?

Time ran out on this one - maybe later.

The sides of a right triangle are a, a+d, and a+2d, with a and d both positive. Find the ratio of a to d. (This was one of the stupid questions we ended up kicking ourselves about. We did the ratio backwards..)

You have a x/x+1/x=2 right triangle - solve for x.

x=1.5

Typo - it's 3/4/5: x=3.

HIJK and DCBA are congruent parallelograms, with altitude HL. HK = 10 cm, LJ = 8 cm, and the area of ABCD is 112cm^2. What is the length of altitude DM? (Not drawn to scale..)

Left this in algebraic form.

Let the base be 8+KL. Then 112=HL[KL+8] and KL² + HL² = 100. Solve for KL and HL

Let HK and HL mak an angle theta.. Then tan theta = KL/HL. Solve for theta.

Then in ABCD, cos theta = a/[8+KL] where a is the desired altitude. Solve for a.

Too much like work.

Plus it took more than 3 minutes.

The coordinates of the vertices of ABC are A(-5,1), B(-2,3), and C(-4,7). The triangle is reflected over the line x=2. What is the sum of the x-coordinates of the images of points A,B, and C?

Reflecting through x=2 changes their signs and adds 4.

Add them up, reverse the sign and add 12.

-11 -> +11 + 12 = 23.

An isosceles triangle has lengths 6x+40, 4x+104, and 7x-23. What is the greatest possible perimeter?

Let 6x+40= 7x-23. then x is largest: x=63.

Perimeter is 17x + 121 => perimeter = 1192

Two congruent regular polygons are placed side-by-side so they share a common side. The angle formed between the polygons is a right angle. How many sides does each polygon have?

Closed my eyes and put two stop signs together:

8 sides.

[Guest]

I'm in the same boat as bonanova, most of the answers were pretty simple. I ran out of time on 10 as well without really doing anything. So I decided to figure it out with pencil and paper, just to see if it is possible. Turns out it is, the answer is

AB = sqrt(23/3)

I can show the math if you want, but it is really long, quite complex, and would be at least a 3 minute answer if you knew how to do it.

[bonanova]

I'm in the same boat as bonanova, most of the answers were pretty simple. I ran out of time on 10 as well without really doing anything. So I decided to figure it out with pencil and paper, just to see if it is possible. Turns out it is, the answer is

AB = sqrt(23/3)

I can show the math if you want, but it is really long, quite complex, and would be at least a 3 minute answer if you knew how to do it.

Great.

I won't bother then....

BTW - I just noticed #7 said rhombus.

Like, that makes the thing doable ... I was trying a general parallelogram - not easy.

[Guest]

Damn, I hate geometry... Numbers are my game...

[Guest]

n° 8 : I clearly messed up that one. Thought it was 3 times the sum and not 4 ... sounded too easy indeed. Bona got the right thing.

N°7: Divide the rhombus into 4 right triangles. You know that the hypothenuse is 20cm (perimeter/4), and that one of the sides (half a diagonal) is 7 cm. So the length of the other diagonal is sqrt(20^2 - 7^2)*2, as Izzy said (I put /2 instead of *2 -> boo).

That leaves us with n°6 :

I'm quite sure that my answer of: A=32°, B=98°, C=50°. With the bissection, we have triangle ADC with angles of 16°, 114 ° and 50°, and triangle ADB with angles of 16°, 66° and 98°, which both work out to 180°.

Bonanova, I don't understand where you got 3x in "ADB: 66 [180-114] + 3x". This seems to assume that B = 2A, which I can't see why.

[Guest]

And finally

"Suppose the medians AA' and BB' of triangle ABC intersect at right angles. If BC = 3 and AC = 4, what is the length of side AB?"

CBA to post a picture, so take a pencil and paper, and do this:

Draw ABC and the medians AA' and BB'. Let's call their intersection point K.

Now, draw the segment A'B'. According to Thales' theorem, its length (x) will be half of AB's (which will thus be 2x).

Notice how triangles A'B'K and ABK are similar ? The similarity ratio is obviously 2 (see previous statement). Therfore, let's call:

A'K = y => AK = 2y

B'K = z => BK = 2z

Let's take triangle AB'K. We know that AB' = 2 (the whole "median" thing). Therefore, (2y)^2 + z^2 = 4

Let's take triangle BA'K. We know that BA' = 1.5 . Therefore, (2z^2) + y^2 = (3/2)^2

This yields the system of equations:

4y^2 + z^2 = 4

y^2 + 4z^2 = 9/4

Which is easily solved by substitution to yield y^2 = 11/12 and z^2 = 1/3

Now, triangle A'B'K gives us the equation y^2 + z^2 = x^2, where we conclude that x^2 = 15/12 = 5/4.

In conclusion, AB = 2x = 2*sqrt(5/4) = sqrt(5)

... waaaay more than 3 minutes.

[Izzy]

BTW - I just noticed #7 said rhombus.

Haha, that explains it then. I was like "...Why can't he do #7?"

[bonanova]

Bonanova, I don't understand where you got 3x in "ADB: 66 [180-114] + 3x". This seems to assume that B = 2A, which I can't see why.

[a] that angle ADB is 66 [gotten by 180-114] and

that looking at triangle ADB, 3x+66 = 180 so x=38.

Here, 3x is all of angle B and half of angle A.

Now, to finish it correctly:

Looking at triangle ACD, C + 114 + x = 180 so C=28; or,

Looking at triangle ABC, C+4x = 180 so C=28.

[Guest]

I went to a math competition today (Seven points away from getting a trophy, but I'm done ranting about that, honest), and I thought the group competition was fairly interesting, so I saved the questions. We were given up to 4 minutes to solve the questions, more points the faster you finish, but I'm not going to time you. You only have pencil and paper. No calculator. Have fun. (Mind, I don't have all the answers, but I'm sure we can figure them out amongst ourselves.)

For those of you that actually want to time yourself to get a proper score, here's how it works. If you answer (correctly..) within the first minute, 16 points. Second minute, 8 points. Third 4. First 1.

In ABC, A + B = 130° and A + C = 110°. How large is A?

The sum of the two exterior angles formed by extending the hypotenuse of a right triangle is how many degrees?

Find the number of degrees in the angle formed by two diagonals drawn from the same vertex of a regular pentagon.

(Ignore this one. The diagram didn't have the numbers, so we were told to skip it.)

The bases of an isosceles trapezoid are AB = 15 inches and DC = 9 inches respectively; each leg is 5 inches. Find the length of a diagonal.

In the triangle ABC, AC = BC and AD bisects angle A. If ADC = 114°, then what is the measure of C?

The perimeter of a rhombus is 80 cm, and one diagonal is 14 cm long. Find the length of the other diagonal.

If the sum of the interior angles of a polygon equals four times the sum of the exterior angles, how many sides does the polygon have?

If the segments of the hypotenuse of a right triangle made by the altitude to the hypotenuse are 3 and 12, find the altitude.

Suppose the medians AA' and BB' of triangle ABC intersect at right angles. If BC = 3 and AC = 4, what is the length of side AB?

The sides of a right triangle are a, a+d, and a+2d, with a and d both positive. Find the ratio of a to d. (This was one of the stupid questions we ended up kicking ourselves about. We did the ratio backwards..)

HIJK and DCBA are congruent parallelograms, with altitude HL. HK = 10 cm, LJ = 8 cm, and the area of ABCD is 112cm^2. What is the length of altitude DM? (Not drawn to scale..)

The coordinates of the vertices of ABC are A(-5,1), B(-2,3), and C(-4,7). The triangle is reflected over the line x=2. What is the sum of the x-coordinates of the images of points A,B, and C?

An isosceles triangle has lengths 6x+40, 4x+104, and 7x-23. What is the greatest possible perimeter?

Two congruent regular polygons are placed side-by-side so they share a common side. The angle formed between the polygons is a right angle. How many sides does each polygon have?

1. 60

2. 270

3. 36

5. 4 x square root 10

Edited February 2, 2009 by Kevin_3.1415

[Izzy]

Welcome to the Den, Kevin! Please use spoilers.