In triangle the bisector of angle intersects the circumcircle again at , the perpendicular bisector of at , and the perpendicular bisector of at . The midpoint of is and the midpoint of is . Prove that the triangles and have the same area.

Let , , and , and be circumcenter. Note that .

By angle-angle, , thus . by opposite-vertex-angles. Thus, is isosceles, with vertex . Drop the altitude/median from to , let foot be .

is perpendicular to $RC$, and because they are radii of the same circle, thus, . Thus, , ie. bisects .

Clearly, is the mean of and , thus , thus, .

We can express the following areas in terms of and :

The difference in areas and can be expressed as follows:

thus as desired.

Remark: Meh, had to trig bash it.

In the plane, consider a point and a polygon (which is not necessarily convex). Let denote the perimeter of , let be the sum of the distances from the point to the vertices of , and let be the sum of the distances from the point to the sidelines of . Prove that

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1219797#1219797

Solution: Let the segments from X to the vertices be , numbered in increasing order clockwise. Let the segments from X to the sides be , also numbered in increasing order clockwise, and such that is to the side immediately clockwise to . For each side, the foot of partitions it into two segments, that we shall label between and , and between and (where = ).

Thus, , , . Then, we rewrite by difference of squares. (I)

From the Pythagorean Theorem, and .

Thus, and .

Add them, , or

By Cauchy-Schwarz’s Inequality,

The LHS is equivalent to . By (I), , as desired.

Remark: Was amazingly straightfoward, apply pythagorean theorem, sum them, and then noticed that Cauchy kills it. Fun.

Let be a convex pentagon with:

and (angles)

Prove that:

Convex pentagon ABCDE with ^ABC=^ADE and ^AEC=^ADB

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1099822#1099822

Solution:

Let intersection of BD and CE be F.

Since angle AEC = angle ADB, AFDE is cyclic. Thus, angle ADE = angle AFE, and DAE = angle DFE = angle BFC. Since angle ADE + angle DAE + angle AED = 180, from triangle ADE, angle AFB = angle AED. Thus in quadrilateral ABCF, angle ABC + angle AFC = 180, and thus it is cyclic. Thus, angle BFC = angle BAC. Since angle DAE = angle BFC, angle DAE = BAC, and we’re done.

Conjecture: Given concave quad ABCD with concave angle C, let E and F be the points AB and AD such that they are on the extensions of DC and BC. Prove that the circumcircles of AFB, ADE, BCE, DCF, are concurrent. (hint: my solution consisted of angle chasing, like the below proof of The Pivot Theorem)

The Pivot Theorem: Given triangle ABC and points D,E,F on sides a,b,c respectively, prove that the circumcircles of DEC, EFA, FDB are concurrent.

http://www.cut-the-knot.org/proofs/Pivot.shtml

Solution:

Let the intersection of the circumcircles of DEC and EFA be P. From cyclic quads, angle DCP = angle DEC = , angle CPD = angle CDE = , angle EPA = angle EFA , angle APF = angle AEF = . Let angle DFB = x, angle FDB = y, angle DFP = z.

And note that the sum of the angles of triangle DEF is .

We would like to show that BDPF is cyclic, ie. angle DBF and angle DPF are supplementary.

Thus, consider the three straight angles at D,E, and F:

and consider the at P: .

Thus, . Since angle DBF = , it is supplementary with angle DPF, and we’re done.

Given a triangle , and points D on AC and E on BC and F on segment DE, let G be the point on AB such that . Let the intersection of GF with BC be H. Show that .

Solution:

(The labeling on the picture is a little different.)

We construct E on AD with AE/ED = BF/FC by the following.

Extend line DG.

Construct line AD’ parallel to BC, with D’ on DG. Then extend GF to point D” on AD’. From parallel lines, BF/FC = AD”/D”R.

Construct line D”E parallel to DG, with E on AD. From parallel lines, we have constructed E such that AE/ED = AD”/D”R = BF/FC.

Extend EF to T on DG. We want to show that EF/FT = AB/BG. Construct B’ on AG by construction above such that AB’/B’G = EF/FT.

We want to show that B’ = B. Consider the extension of B’F to C’ on DG, and G”’ on AG’. From corresponding angles of parallel lines,
and .
Thus, angle FC’T is a purple angle, and since FCT is also a purple angle, C’ = C. Since both B’ and B are on the intersection of CF and AG, B’ = B. Since AB’/B’G = EF/FT, the same is true of B.

For a given positive integer find, in terms of , the minimum value of for which there is a set of distinct positive integers that has sum greater than but every subset of size has sum at most .

Solution:

WLOG assume that any such set of distinct positive integers is ordered from least to greatest, .

Then the total sum is : we want to minimize the sum so N is minimized.

We also need to satisfy the second condition, that any k element subset has sum at most N/2. It is obvious that the sum of the k greatest elements will be greater than the sum of any other k element subset. It suffices to satisfy .

For a sequence with first term a and 2k+1 elements, define the consecutive-sequence to be the set .

We show that for sufficiently large a, the consecutive-sequence of a satisfies the two conditions in the problem.

Since and ,
then
(1)

,
so the consecutive sequence of satisfies the above conditions for .

Assume is the smallest such first term of a consecutive sequence that satisfies the above conditions, ie. . This has the minimum sum of all sequences with first term greater to or less than .

We show that there is no non-consecutive-sequence with first term smaller than with a smaller sum.

Assume that there is one, , with starting term b. This is not a consecutive-sequence, so there is some such that we can reduce all the terms by 1 to form a new sequence, .

We show that this new sequence also satisfies the conditions of the problem, and thus, we can repeat this process finitely many times to obtain the consecutive-sequence , thus contradicting the fact that is the smallest first term with a consecutive-sequence that satisfies the conditions.

Since satisfies the conditions of the problem, we have from (1)
. (2)

We want to show that this applies to .

If is one of the last k elements, then in we have the sum of all the elements is reduced by some constant C, with the sum of the first k+1 elements unaffected

From (2), we have .

If is one the first k+1 elements, then we have the sum of all the elements is reduced by C, and the sum of the first k+1 elements reduced by 2c<C.

From (2), we have .

So we minimize N when the sequence is . The sum of the last k elements is , so .

Remark: At least I hope the above is right. Anyways, shows that I need to organize my thoughts before writing, since this took way too long to write out.

Prove that for ,

.

- Inequality with Fractions, MOP 1998

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=199005

Solution:

which is true from AM-GM.

Remark: Expand and bash *cough*.

Let be an acute-angled triangle whose side lengths satisfy the inequalities If point is the center of the inscribed circle of triangle and point is the center of the circumscribed circle, prove that line intersects segments and .

- USAMO Challenge #1

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1093419#1093419

Solution:

Consider the lines that pass through the circumcenter O. Extend AO, BO, CO to D,E,F on a,b,c, respectively. We notice that IO passes through sides a and c if and only if I belongs to either regions AOF or COD. Since AO = BO = CO = R, we let , , . We have . Since IA divides angle A into two equal parts, it must be in the region marked by the of angle A, so I is in ABD. Similarly, I is in ACF and ABE. Thus, I is in their intersection, AOF. From above, we have IO passes through a and c.

Remark: Took me a while to figure out the approach of considering where I is in terms of regions defined by O. One could see this from how the circumradii divides the triangle into isosceles triangles, and then we can use the angles and the fact that the inradii bisect the angles.

• Suppose 51 numbers are chosen from 1, 2, 3, …, 99, 100. Show that there are two such that one divides the other.
• Determine the smallest integer n with the property that for any n distinct real numbers , there exist integers i and j such that
• (Poland) Suppose a triangle can be placed inside a square of unit area such that the center of the square is not inside the triangle. Show that a side of the triangle has length less than 1.

From TJ USAMO 2004/10/07, originally from Mathematical Excalibur (the first issue, whatever that was).