Dimentional Change Questions III: Similar Shapes

There are numerous similar triangle questions on Mathcounts.

Here are the basics:

If two triangles are similar, their corresponding angles are congruent and their corresponding sides will have the same ratio or proportion.

Δ ABC and ΔDEF are similar. \(\frac{AB}{DE}\) = \(\frac{AC}{DF}\) = \(\frac{BC}{EF}\)= their height ratio = their perimeter ratio.







Once you know the linear ratio, you can just square the linear ratio to get the area ratio and cube the linear ratio to get the volume ratio. 

Practice Similarity of Triangles here.  Read the notes as well as work on the practice problems.  There is instant feedback online. 

Other practice sheets on Similar Triangles                                                        

Many students have trouble solving this problem when the two similar triangles are superimposed. 

Just make sure you are comparing smaller triangular base with larger triangular base and smaller triangular side with corresponding larger triangular side, etc... In this case:

\(\frac{BC}{DE}\)= \(\frac{AB}{AD}\) = \(\frac{AC}{AE}\)




Questions to ponder (Solutions below)

#1: Find the area ratio of Δ ABC to trapezoid BCDE to DEGF to FGIH. You can easily get those ratios using similar triangle properties. All the points are equally spaced and line \(\overline{BC}\)// \(\overline{DE}\) // \(\overline{FG}\) // \(\overline{HI}\). 

#2: Find the volume of the cone ABC to Frustum BCDE to DEGF to FGIH. Again, you can use the similar cone, dimensional change property to easily get those ratios.Same conditions as the previous question.

 

Answer key: 

#1:

 #2:

Mathcounts : Geometry -- Medians ; Squares in Isosceles Right Triangle, Similar Triangles, Triangles share the same vertex

Check out Mathcounts here, the best competition math program for middle school students.
Download this year's Mathcounts handbook here.

This one appears at 91 Mathcounts National Target #4. It's interesting.

Q: triangle ABC is equilateral and G is the midpoint. BI is the same length as BC and the question asks about the area of quadrilateral polygon GDBC.
We'll also find the area of triangle ADG and triagle DBI here. 

Solution I: The side length is 2 and it's an equilateral triangle so the area of triangle ABC is 
\(\dfrac {\sqrt {3}} {4}\times 2^{2}=\sqrt {3}\).
Using similar triangles GCF and ACE, you get \(\overline {GF}\)= \(\dfrac {\sqrt {3}} {2}\)
From there, you get area of  \(\Delta\)GIC = \(4*\dfrac {\sqrt {3}} {2}*\dfrac {1} {2}=\sqrt {3}\)
 \(\Delta\)FCH is a 30-60-90 degree right triangle so \(\overline {DH}=\sqrt {3}*\overline {BH}\)
Using similar triangles IDH and IGF, you get \(\dfrac {2+\overline {BH}} {3.5}=\dfrac {\sqrt {3}\overline {BH}} {\dfrac {\sqrt {3}} {2}}\) = 2\(\overline {BH}\);
2 +\(\overline {BH}\)= 7\(\overline {BH}\); \(\overline {BH}\)= \(\dfrac {1} {3}\)
\(\overline {DH}=\dfrac {\sqrt {3}} {3}\) = the height
 Area of \(\Delta\)DBI = 2 * \(\dfrac {\sqrt {3}} {3}\)*\(\dfrac {1} {2}\) = \(\dfrac {\sqrt {3}} {3}\)
Area of quadrilateral polygon GDBC =  \(\sqrt {3}-\dfrac {\sqrt {3}} {3}\)= \(\dfrac {2\sqrt {3}} {3}\)
The area of both  \(\Delta\)ABC and \(\Delta\)GIC is \(\sqrt {3}\) and both share the quadrilateral polygon GDBC (It's like a Venn diagram).
Thus, the area of both triangles are the same \(\dfrac {\sqrt {3}} {3}\).

Solution II:

There are some harder AMC-10 questions using the same technique.
Using triangles share the same vertex,  you get the two same area "a"s and "b"s because the length of the base is the same.
From the area we've found at solution I, we have two equations:
2a + b = \(\sqrt {3}\)
2b + a = \(\sqrt {3}\)
It's obvious a = b so you can find the area of the quadrilateral being the \(\dfrac {2} {3}\)of the equilateral triangle so the answer is \(\dfrac {2\sqrt {3}} {3}\).




Vinjai's solution III:

Draw the two extra lines and you can see that the three lines are medians, which break the largest triangle into 6 equal parts. [You can proof this using triangles share the same vertices : notes later.]
Quadrilateral polygon FECB is  \(\dfrac {2} {3}\)of the equilateral triangle so the answer is \(\dfrac {2\sqrt {3}} {3}\).







This is similar to 2005 Mathcounts National Target #4 questions. AMC might have similar ones.
Q: Both triangles are congruent and isosceles right triangles and each has a square embedded in it. If the area of the square on the left is 12 square unites, what is the area of the square on the right? 
Solution I: 

Solution II:

Triangular Numbers & Word Problems: Chapter Level

Triangular Numbers :  From Math is Fun. 1, 3, 6, 10, 15, 21, 28, 36, 45...

Interesting Triangular Number Patterns: From  Nrich

Another pattern: The sum of two consecutive triangular numbers is a square number.

What are triangular numbers? Let's exam the first 4 triangular numbers:

The 1st number is  "1".
The 2nd number is "3" (1 + 2)
The 3rd number is "6" (1 + 2 + 3)
The 4th number is "10" (1 + 2 + 3 + 4)
.
.
The nth number is \(\frac{n(n+1)}{2}\)

It's the same as finding out the sum of the first "n" natural numbers.

Let's look at this question based on the song "On the Twelve Day of Christmas"
(You can listen to this on Youtube,)

On the Twelve Day of Christmas
On the first day of Christmas
my true love gave to me
a Partridge in a Pear Tree

On the second day of Christmas,
My true love gave to me,
Two Turtle Doves,
And a Partridge in a Pear Tree.

On the third day of Christmas,
My true love gave to me,
Three French Hens,
Two Turtle Doves,
And a Partridge in a Pear Tree

On the fourth day of Christmas,
My true love gave to me,
Four Calling Birds,
Three French Hens,
Two Turtle Doves,
And a Partridge in a Pear Tree.


The question is "How many gifts were given out on the Day of Christmas?"

Solution I"
1st Day:       1
2nd Day:      1 + 2
3rd Day       1 + 2 + 3
4th Day        1 + 2 + 3 + 4
.
.
12th Day      1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12

Altogether, you'll have 12 * 1 + 11 * 2 + 10 * 3 + 9 * 4 + 8 * 5 + 7 * 6 + 6 * 7 + 5 * 8 + 4 * 9 + 3 * 10 + 2 * 11 + 1 * 12 = 12 + 22 + 30 + 36 + 40 + 42 + 42 + 40 + 36 + 30 + 22 + 12 = 364

Solution II: The sum of the "n" triangular number is a tetrahedral number.
To get the sum, you use \(\frac{n(n+1)(n+2)}{6}\)

n = 12 and \(\frac{12 (13)(14)}{6}= 364 \)

Here is a proof without words.

Applicable questions: (Answers and solutions below)

#1 Some numbers are both triangular as well as square numbers. What is the sum of the first three positive numbers that are both triangular numbers and square numbers?

#2 What is the 10th triangular number? What is the sum of the first 10 triangular numbers?

#3 What is the 20th triangular number?

#4 One chord can divide a circle into at most 2 regions, Two chords can divide a circle at most into 4 regions. Three chords can divide a circle into at most seven regions. What is the maximum number of regions that a circle can be divided into by 50 chords? 

#5: Following the pattern, how many triangles are there in the 15th image? 

Answers: 
#1 1262  The first 3 positive square triangular numbers are: 1, 36 (n = 8) and 1225 (n = 49).

#2 55; 220    a. \(\frac{10*11}{2}=55\)   b. \(\frac{10*11*12}{6}=220\)

#3 210   \(\frac{20*21}{2}=210\)  

#4 1276 
1 chord :    2 regions or \(\boxed{1}\) + 1
2 chords:   4 regions or  \(\boxed{1}\)  + 1 + 2
3 chords:   7 regions or \(\boxed{1}\)  + 1 + 2 + 3
.
.
50 chords:\(\boxed{1}\)  + 1 + 2 + 3 + ...+ 50 = 1 +  \(\frac{50*51}{2}\) =1276

#5: 120
The first image has just one triangle, The second three triangles. The third 6 triangles total. 
It follows the triangular number pattern. The 15th triangular is  \(\frac{15*16}{2}\) =120 

2015 Mathcounts State Prep : Inscribed Cricle Radius and Similar Triangles

Question : \(\Delta\) ABC is an equilateral triangle. Circle "O" is the inscribed circle and it's radius is 15. 

What is the length of the radius of the smaller circle p which is tangent to circle "O" and the two sides?








Here is the link to the basics of inscribed circle radius as well as circumscribed circle radius of an equilateral triangle.

Solution I :
The length of the radius of an inscribed circle of an equilateral triangle is \(\dfrac {1} {3}\) of the height so you know AO is \(\dfrac {2} {3}\) of the height or 30 (the height is 15 + 30 = 45 unit long)

\(\Delta\) AEP is similar to \(\Delta\) AFO \(\rightarrow\) \(\dfrac {r} {15}=\dfrac {AP} {30}\)
so \(\overline {AP}\) = 2r.
\(\overline {AP}+\overline {PO}=30\)  \(\rightarrow\)2r + r + 15 = 30   \(\rightarrow\)  3r = 15 so r = 5 
or \(\dfrac {1} {3}\) of the larger radius 

Solution II:
\(\Delta\) APE is a 30-60-90 right triangle, so \(\overline {AP}\) = 2r
\(\overline {PO}\) = r + 15
\(\overline {AP}+\overline {PO}\)  \(\rightarrow\)  2r + r + 15 = 30  \(\rightarrow\)  3r  =  15 so  r  = 5 
or \(\dfrac {1} {3}\) of the larger radius 

This is an AMC-10 question.

\(\Delta\) ABC is an isosceles triangle.

The radius of the smaller circle is 1 and the radius of the larger circle is 2,

A: what is the length of \(\overline {AP}\) ?

B. what is the area of \(\Delta\) ABC?







Solution for question A: 
\(\Delta\) AEP is similar to \(\Delta\) AFO \(\rightarrow\) \(\dfrac {1} {2}=\dfrac {AP} {AP +3}\)
 2\(\overline {AP}\) =  \(\overline {AP}\) + 3 \(\rightarrow\) AP = 3

Using Pythagorean theorem, you can get \(\overline {AE}\) = \(2\sqrt {2}\)
\(\Delta\) AEP is similar to \(\Delta\) ADC [This part is tricky. Make sure you see that !!]
 \(\rightarrow\) \(\dfrac {1} {\overline {DC}}=\dfrac {AE} {AD}\) = \(\dfrac {2\sqrt {2}} {8}\)
\(\overline {DC}\) = \(2\sqrt {2}\)  and \(\overline {BC}\) = 2 * \(2\sqrt {2}\) = \(4\sqrt {2}\)
The area of  \(\Delta\) ABC =  \(\dfrac{1}{2}\)*\(4\sqrt {2}\) * 8 =  \(16\sqrt {2}\)

Question: If you know the length of x and y, and the whole length of \(\overline {AB}\),

A: what is the ratio of a to b and 

B: what is the length of z.







Solution for question A:
\(\Delta\)ABC and \(\Delta\)AFE are similar so \(\dfrac {z} {x}=\dfrac {b} {a+b}\). -- equation 1
Cross multiply and you have z ( a + b ) = bx

\(\Delta\)BAD and \(\Delta\)BFE are similar so \(\dfrac {z} {y}=\dfrac {a} {a+b}\). -- equation 2
 Cross multiply and you have z ( a + b ) = ay

bx = ay so \(\dfrac {x} {y}=\dfrac {a} {b}\)  same ratio

Solution for question B: 
Continue with the previous two equations, if you add equation 1 and equation 2, you have:
\(\dfrac {z} {x}+\dfrac {z} {y}=\dfrac {b} {a+b}+\dfrac {a} {a+b}\)
\(\dfrac {zy+zx } {xy}=1\) \(\rightarrow\) z = \(\dfrac {xy} {x+y}\)

Applicable question: 

\(\overline {CD}=15\) and you know \(\overline {DB}:\overline {BC}=20:30=2:3\) 
 so  \(\overline {DB}=6\) and \(\overline {BC}=9\) 

\(\overline {AB}=\dfrac {20\times 30} {\left( 20+30\right) }\) = 12

Testimonials for my services. So far, all through words of mouth locally or chance meets online and it's great.

Testimonials from my students/parents  :) 

Dear Mrs Lin,

I just did not get the opportunity earlier to thank you for all your help. _____ was able to make to the National MathCounts largely because of your excellent guidance and coaching. I do not know how to thank you. He had a great once-in-a-lifetime experience there and he really loved the competition as well as meeting other people.

If you can please provide me your mailing address, he wants to send you a gift as a token of thanks for your guidance and tutoring. 

Sincerely,

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In school, his teacher focuses more on details like, put all the steps, write neatly, don't disturb the class by asking unnecessary questions, don't ask for more work, behave properly etc....so he is not too happy with math in school.

Thank you so much for making such a big difference in his life. He  enjoys doing the homework you are assigning him and has not complained at all. He said "Mrs Lin is so smart and I love her classes...wish she lived next to our house...so I can go to her and have live classes' [disclaimer : I'm not ; my students are much smarter than I and I learn along with them and it's exhilarating ]   Thank you so much!!

One thing he said was it would be helpful if he can have targeted practice worksheets on the tips covered during the class, after that class, so he can practice those shortcuts/tips more.

Hello Mrs. Lin,

This is ______ 's mom. We greatly appreciate your help in working with
______, keeping him motivated and providing him wth constant
encouragement. We are fortunate to have great mentors like you who

are so selfless in their services to our younger generation.

We have a request, can you please share your address. ______ wanted
to send you a card to express his thanks.

BTW, _______ has his chapter level competition tomorrow.

Warm Regards

Mrs. Lin,

Good Afternoon.

Me and my son, ___ and I, have used your blog pages and got a great insight into several of the techniques that you use to solve the problems. I am glad to inform you that _____ was placed 14th in the State Mathcounts for ____. Your blog information has helped us a lot in this preparation and we really want to appreciate all that you do in sharing the information. 

_____ is completing Geometry this year and he hopes to have a exciting next year for AMC 8 and Mathcounts. We hope to learn a lot more from your blog pages.

Regards,

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I just finished The One World School House by Salman Khan and I have been thinking of you and the online community you created. Even though _____ and ____ have taken web based courses before, what makes your class different is your inspiration and enthusiasm - you really care about them as unique individuals and you sincerely expect the best from them, more than they (and I sometimes) think they can achieve. Thank you for encouraging them to become a more responsible and self driven learner.

Dear Mrs.Lin,

             I went to NSF tests today and got first place in the Math Bee III. The problems didn't seem that bad although I'm waiting to see my score online. I will most likely be going to nationals which will be held in Ohio this year so that's convenient. Lastly I would like to thank you for all you have done for me. I believe I've grown much more as a student under you and really appreciate everything you have shown me and taught me. I wouldn't be succeeding now if it wasn't for you. :)

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So far,in January state, in algebra 2 he came individual 2nd and one regional February he came 1st (competing with same state high school students while as an 8th grader) 

He could not participate in one regional bec of conflict with MathCounts chapter. We also want to share with you that he got admission into a private school for 9th gr with 100% scholarship. [more than 50 students in that high school are national merit semifinalists, so highly competitive]
Thank you,

An elementary whiz kid, national winners at Math Kangaroo and Math Olympiad. 

_____ will be off to summer sleep away camp where he is not allowed to have any electronics starting _____.  

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He will be back online with you at _______ . 

There are many more but I'll take my time to update/upload these infor. sheets. 

To be continued ... 

I'm quite busy these days with resuming our Math Circle + many other projects (my students don't just excel at math, but many other areas equally fun and challenging + most Asian students have much bigger problems with critical reading/not to mention writing, taking initiatives and being strong leaders, and those are my other projects. 

Work-Life balance is utmost important. Less is more.