Classic problems (mainly geometry)
by math154, Nov 15, 2009, 4:10 AM
If you care...
Inheritance problem: Excellence in Mathematics Contest 2009.18
20-80-80 triangle: Excellence in Mathematics Contest 2009.20
Let
be isosceles with vertex 
such that 
, with point 
on 
and point 
on 
(unless stated otherwise).
Problem 1: Find
if 
and 
.
Solution 1
Solution 2
Problem 2: Find if 
and 
.
Solution 1
Solution 2
Problem 3: Find
if 
and 
.
Solution
Problem 4: Find
if is on and 
.
Solution 1
Solution 2
Problem 5: Prove that
given that 
, is on , 
is on , and 
are all on a circle with center 
.
Solution
OK so at this contest the test writer(s) decided to put two classic problems in... some inheritance problem and the original 20-80-80 triangle problem. The first one was acceptable but not that great for a multiple choice test, but the second one definitely shouldn't have been there because obviously most people who got it right probably didn't know how to solve it. If the answer had been "none of the above" then maybe that would be better. The main problem is time though, because this problem is not meant to be speedsolved.
Inheritance problem: Excellence in Mathematics Contest 2009.18
The wealthy mother died leaving an estate of 
gold coins, where is an integer greater than 
. She has daughters and they inherit the coins as follows. The eldest receives 
coin plus 
of those remaining, where 
is an integer. The second eldest receives 
coins plus of those remaining. The third eldest receives 
coins plus of those remaining. This pattern continues for the rest of the daughters, except that the youngest receives all the coins that were not inherited by her sisters. If each of the daughters received the same number of coins, then 
is
![\[ \textbf{(A)}6k\quad\textbf{(B)}k^2 + k\quad\textbf{(C)}k^2 + 1\quad\textbf{(D)}(k + 1)^2\quad\textbf{(E)}\text{None of these}\]](//latex.artofproblemsolving.com/6/9/2/6929740d80e528edb308721d7e0ba46a02019211.png)
Natural interpretation, in my opinion
Other interpretation
gold coins, where is an integer greater than 
. She has 
daughters and they inherit the coins as follows. The eldest receives 
coin plus 
of those remaining, where 
is an integer. The second eldest receives 
coins plus of those remaining. The third eldest receives 
coins plus of those remaining. This pattern continues for the rest of the daughters, except that the youngest receives all the coins that were not inherited by her sisters. If each of the daughters received the same number of coins, then 
is![\[ \textbf{(A)}6k\quad\textbf{(B)}k^2 + k\quad\textbf{(C)}k^2 + 1\quad\textbf{(D)}(k + 1)^2\quad\textbf{(E)}\text{None of these}\]](http://latex.artofproblemsolving.com/6/9/2/6929740d80e528edb308721d7e0ba46a02019211.png)
Natural interpretation, in my opinion
First, note that each daughter gets 
coins. Here, we assume that the 
eldest daughter first gets 
coins and then of the remaining coins (not including the coins). Because 
daughters each get coins before the daughter goes, we find that for 
,
![\[ \frac {G}{D} = n + \frac {G - (n - 1)\frac {G}{D} - n}k \\
\\
\implies G((k - 1) - D) = n(D(k - 1) - G).\]](//latex.artofproblemsolving.com/c/1/7/c176b8909a80b398ef9494d550b9f2830596bd31.png)
The wording implies that there are at least two daughters. If there are only two daughters, then
and the only valid value of is , so
![\[ G(k - 3) = 2(k - 1) - G\implies(G - 2)(k - 2) = 2,\]](//latex.artofproblemsolving.com/e/c/d/ecd330a5ee9fc547706097f85c36dc275c724697.png)
but clearly then
.
However, if there are at least three daughters, then if
is nonzero, then can only take on one value, while it must take on at least 
, which is impossible. Hence
![\[ D(k - 1) - G = G((k - 1) - D) = 0,\]](//latex.artofproblemsolving.com/0/c/3/0c368ed83bafaf9daa2bc753e2e3123602c484d4.png)
so because
, 
and 
, and hence
![\[ D + G = (k - 1) + (k - 1)^2 = k^2 - k\implies\boxed{\textbf{(E)}\text{None of these}}.\]](//latex.artofproblemsolving.com/a/1/3/a13f2089b0d056a20fa148cd404c3578773aad2f.png)
coins. Here, we assume that the 
eldest daughter first gets 
coins and then of the remaining coins (not including the coins). Because 
daughters each get coins before the daughter goes, we find that for 
,![\[ \frac {G}{D} = n + \frac {G - (n - 1)\frac {G}{D} - n}k \\
\\
\implies G((k - 1) - D) = n(D(k - 1) - G).\]](http://latex.artofproblemsolving.com/c/1/7/c176b8909a80b398ef9494d550b9f2830596bd31.png)
The wording implies that there are at least two daughters. If there are only two daughters, then
and the only valid value of is , so![\[ G(k - 3) = 2(k - 1) - G\implies(G - 2)(k - 2) = 2,\]](http://latex.artofproblemsolving.com/e/c/d/ecd330a5ee9fc547706097f85c36dc275c724697.png)
but clearly then
.However, if there are at least three daughters, then if
is nonzero, then can only take on one value, while it must take on at least 
, which is impossible. Hence![\[ D(k - 1) - G = G((k - 1) - D) = 0,\]](http://latex.artofproblemsolving.com/0/c/3/0c368ed83bafaf9daa2bc753e2e3123602c484d4.png)
so because
, 
and 
, and hence![\[ D + G = (k - 1) + (k - 1)^2 = k^2 - k\implies\boxed{\textbf{(E)}\text{None of these}}.\]](http://latex.artofproblemsolving.com/a/1/3/a13f2089b0d056a20fa148cd404c3578773aad2f.png)
Other interpretation
Again, each daughter gets coins. Here, we assume that the eldest daughter first gets of the remaining coins before receiving more coins. Because daughters each get coins before the daughter goes, we find that for ,
![\[ \frac {G}{D} = n + \frac {G - (n - 1)\frac {G}{D}}k \\
\\
\implies G((k - 1) - D) = i(Dk - G) = 0\]](//latex.artofproblemsolving.com/c/c/d/ccdf9f3868b2d064f3a8c93f748f9a949a47a02d.png)
The wording implies that there are at least two daughters. If there are only two daughters, then and the only valid value of is , so
![\[ G(k - 3) = 2k - G\implies(G - 2)(k - 2) = 4,\]](//latex.artofproblemsolving.com/b/7/e/b7eee7aaa070eae146c8b94befc920b8e26ac1e2.png)
but clearly then.
However, if there are at least three daughters, then if
is nonzero, then can only take on one value, while it must take on at least , which is impossible. Hence
![\[ Dk - G = G((k - 1) - D) = 0,\]](//latex.artofproblemsolving.com/2/6/a/26a4ea078c91bd42adcab5c223421f22123afb90.png)
so because, and 
, and hence
![\[ D + G = (k - 1) + (k^2 - k) = k^2 - 1\implies\boxed{\textbf{(E)}\text{None of these}}.\]](//latex.artofproblemsolving.com/7/e/0/7e07eef3b126c58294f2549fdb4d83bce8838a17.png)
coins. Here, we assume that the eldest daughter first gets of the remaining coins before receiving more coins. Because daughters each get coins before the daughter goes, we find that for ,![\[ \frac {G}{D} = n + \frac {G - (n - 1)\frac {G}{D}}k \\
\\
\implies G((k - 1) - D) = i(Dk - G) = 0\]](http://latex.artofproblemsolving.com/c/c/d/ccdf9f3868b2d064f3a8c93f748f9a949a47a02d.png)
The wording implies that there are at least two daughters. If there are only two daughters, then
and the only valid value of is , so![\[ G(k - 3) = 2k - G\implies(G - 2)(k - 2) = 4,\]](http://latex.artofproblemsolving.com/b/7/e/b7eee7aaa070eae146c8b94befc920b8e26ac1e2.png)
but clearly then
.However, if there are at least three daughters, then if
is nonzero, then can only take on one value, while it must take on at least , which is impossible. Hence![\[ Dk - G = G((k - 1) - D) = 0,\]](http://latex.artofproblemsolving.com/2/6/a/26a4ea078c91bd42adcab5c223421f22123afb90.png)
so because
, and 
, and hence![\[ D + G = (k - 1) + (k^2 - k) = k^2 - 1\implies\boxed{\textbf{(E)}\text{None of these}}.\]](http://latex.artofproblemsolving.com/7/e/0/7e07eef3b126c58294f2549fdb4d83bce8838a17.png)
20-80-80 triangle: Excellence in Mathematics Contest 2009.20
Triangle 
is isosceles (). Point is on side , such that . Point is on side , such that . The triangle is shown in the figure but not to scale. If 
then 
, in degrees, is
![\[ \textbf{(A)}60\qquad\textbf{(B)}45\qquad\textbf{(C)}50\qquad\textbf{(D)}40\qquad\textbf{(E)}\text{None of these}\]](//latex.artofproblemsolving.com/c/7/7/c770a2e0caddd5b42082336e2cd3be826afcd15a.png)
Solution
is isosceles (
). Point is on side 
, such that 
. Point 
is on side 
, such that 
. The triangle is shown in the figure but not to scale. If 
then 
, in degrees, is![\[ \textbf{(A)}60\qquad\textbf{(B)}45\qquad\textbf{(C)}50\qquad\textbf{(D)}40\qquad\textbf{(E)}\text{None of these}\]](http://latex.artofproblemsolving.com/c/7/7/c770a2e0caddd5b42082336e2cd3be826afcd15a.png)
Solution
We prove below in problem 2 that 
, so it's easy to see that
![\[ \angle AED = 50^\circ\implies\boxed{\textbf{(C)}50}.\]](//latex.artofproblemsolving.com/a/8/d/a8dbb3942f865355ee07f6d34fd63401e2ab95e4.png)
, so it's easy to see that![\[ \angle AED = 50^\circ\implies\boxed{\textbf{(C)}50}.\]](http://latex.artofproblemsolving.com/a/8/d/a8dbb3942f865355ee07f6d34fd63401e2ab95e4.png)
Let
be isosceles with vertex 
such that , with point on and point on (unless stated otherwise).Problem 1: Find
if 
and 
.Solution 1
Originally discussed here.
Equilateral triangles are awesome, so construct
on such that 
(
is parallel to 
), and let the intersection of line 
with 
be point . Note that 
is equilateral. Furthermore, is equidistant from and by symmetry, so 
is precisely the angle bisector of 
. This means that 
. Because 
(since 
), 
, and 
, we have that 
, so 
. But because 
is equilateral, 
. Recall that is parallel to , so now
![\[ \angle EE'D = 80^\circ\implies\angle DEE' = \angle EDE' = 50^\circ,\]](//latex.artofproblemsolving.com/2/f/b/2fb3e3c2d3115cbd9959659ccae703ca2f72d538.png)
and since
![\[ 30^\circ = \angle BDE' = \angle EDE' - \angle BDE = 50^\circ - \angle BDE,\]](//latex.artofproblemsolving.com/9/7/b/97b3b204bfad35d87ef61d870f8b61fdedf883d8.png)
we finally arrive at
.
Equilateral triangles are awesome, so construct
on such that 
(
is parallel to 
), and let the intersection of line 
with 
be point 
. Note that 
is equilateral. Furthermore, is equidistant from and by symmetry, so 
is precisely the angle bisector of 
. This means that 
. Because 
(since 
), 
, and 
, we have that 
, so 
. But because 
is equilateral, 
. Recall that is parallel to , so now![\[ \angle EE'D = 80^\circ\implies\angle DEE' = \angle EDE' = 50^\circ,\]](http://latex.artofproblemsolving.com/2/f/b/2fb3e3c2d3115cbd9959659ccae703ca2f72d538.png)
and since
![\[ 30^\circ = \angle BDE' = \angle EDE' - \angle BDE = 50^\circ - \angle BDE,\]](http://latex.artofproblemsolving.com/9/7/b/97b3b204bfad35d87ef61d870f8b61fdedf883d8.png)
we finally arrive at
.Solution 2
Found on the first post here, solution 1.
As in solution 1, construct on such that ( is parallel to ), and let the intersection of line with be point . From problem 4 (solution 1), 
, and because we do have 
, . Then from equilateral and this, we may let 
. Realize also that 
, 
, and 
, so
![\[ DE' = AE' - AD = AE' - \alpha = AE - \alpha = EC - \alpha \\
= EC - FC = EF = EE',\]](//latex.artofproblemsolving.com/b/c/3/bc3bc9eefd500dcefa3d71756c5b09fb2a0f008e.png)
so
and because that is parallel to ,
and since
we finally arrive at.
As in solution 1, construct
on such that ( is parallel to ), and let the intersection of line with be point . From problem 4 (solution 1), 
, and because we do have 
, 
. Then from equilateral and this, we may let 
. Realize also that 
, 
, and 
, so![\[ DE' = AE' - AD = AE' - \alpha = AE - \alpha = EC - \alpha \\
= EC - FC = EF = EE',\]](http://latex.artofproblemsolving.com/b/c/3/bc3bc9eefd500dcefa3d71756c5b09fb2a0f008e.png)
so
and because that is parallel to ,and since
we finally arrive at
.Problem 2: Find
if and .Solution 1
Equilateral triangles are good. Construct 
on such that 
and let the intersection of 
and 
be 
(
is equilateral). Note that 
so 
, and because 
is equilateral 
. Draw 
; we see that 
because 
, and 
because is equilateral. Because 
, 
. Basic angle chasing tells us that 
. But because 
is parallel to , 
, and we also know that 
, so
![\[ \angle BDD' = \frac12\angle BDD' = \frac12(180^\circ - 80^\circ - 40^\circ) = 30^\circ.\]](//latex.artofproblemsolving.com/2/8/1/28113c870cb46fd7e65d91f03d7493f6116f5af9.png)
on such that 
and let the intersection of 
and 
be 
(
is equilateral). Note that 
so 
, and because 
is equilateral 
. Draw 
; we see that 
because 
, and 
because is equilateral. Because 
, 
. Basic angle chasing tells us that 
. But because 
is parallel to , 
, and we also know that 
, so![\[ \angle BDD' = \frac12\angle BDD' = \frac12(180^\circ - 80^\circ - 40^\circ) = 30^\circ.\]](http://latex.artofproblemsolving.com/2/8/1/28113c870cb46fd7e65d91f03d7493f6116f5af9.png)
Solution 2
Construct on so that 
. First realize that by problem 1, 
, so 
. By problem 4, 
. But because 
is isosceles, , so 
. Note also that because 
, 
so 
and hence 
.
on so that 
. First realize that by problem 1, 
, so 
. By problem 4, 
. But because 
is isosceles, , so 
. Note also that because 
, 
so 
and hence 
.Problem 3: Find
if 
and 
.Solution
This is pretty much a trivial setup compared to the others. Let be the intersection of and 
. Because 
and 
so is perpendicular to , we conclude that 
and because 
, the Pythagorean Theorem gives us 
, so 
by SSS congruency and 
.
be the intersection of and 
. Because 
and 
so is perpendicular to , we conclude that 
and because 
, the Pythagorean Theorem gives us 
, so 
by SSS congruency and 
.Problem 4: Find
if is on and .Solution 1
Previously posted here.
Reflect
across to get 
, and then across 
to get 
. Essentially we have placed two triangles congruent to next to it so that 
is equilateral (since it is isosceles with vertex angle 
). With some simple angle chasing, we find that 
, 
, and 
. But because 
, 
, and 
, SAS congruence tells us that 
, and hence 
.
Reflect
across to get 
, and then 
across 
to get 
. Essentially we have placed two triangles congruent to next to it so that 
is equilateral (since it is isosceles with vertex angle 
). With some simple angle chasing, we find that 
, 
, and 
. But because 
, 
, and 
, SAS congruence tells us that 
, and hence 
.Solution 2
Originally discussed here.
Recall solution 1 to problem 1 (just swap and , and and in that solution to make this easier to see). It's clear that 
, so when 
, 
. But if 
, then it's clear that 
will be shorter or greater than .
Recall solution 1 to problem 1 (just swap
and , and and in that solution to make this easier to see). It's clear that 
, so when 
, 
. But if 
, then it's clear that 
will be shorter or greater than .Problem 5: Prove that
given that 
, is on , is on , and 
are all on a circle with center .Solution
Let the intersection of circle and line 
be , different from . Note that 
because they are all radii of circle , so because and 
, and simple angle chasing (remember that ) yields 
. Because 
is an inscribed angle, arc 
and hence 
. By more simple angle chasing, 
, so 
. But because 
and subtend the same arc of 
, 
, as desired.
and line 
be , different from . Note that 
because they are all radii of circle , so because and 
, and simple angle chasing (remember that ) yields 
. Because 
is an inscribed angle, arc 
and hence 
. By more simple angle chasing, 
, so 
. But because 
and subtend the same arc of 
, 
, as desired.This post has been edited 1 time. Last edited by math154, Aug 5, 2010, 8:57 PM
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