# Stumped by a mathematics question - Area of Triangle **URL:** \ **Category:** Factual Questions\ **Created:** [November 4, 2007, 9:05am UTC](https://boards.straightdope.com/t/stumped-by-a-mathematics-question-area-of-triangle/425175 "2007-11-04T09:05:10Z")\ **Posts on this page:** 9\ **Page:** 2
**Author:** ![pmwgreen](https://avatars.discourse-cdn.com/v4/letter/p/edb3f5/32.png) [@pmwgreen](https://boards.straightdope.com/u/pmwgreen)\ **Post date:** [November 5, 2007, 8:06pm UTC](https://boards.straightdope.com/t/stumped-by-a-mathematics-question-area-of-triangle/425175/21 "2007-11-05T20:06:31Z")
[QUOTE=Bytegeist] We are given that AE = ED, and that AED is a straight line. This makes the height of triangle BDE (the distance of point E from line BC) exactly half the height of ABC (the distance of A from the same line). [/QUOTE] Or consider AED the base line and then the two triangles have the same height and bases in the ratio 2 to 1. ---
**Author:** ![Tyrrell\_McAllister](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/tyrrell_mcallister/32/16772_2.png) [@Tyrrell\_McAllister](https://boards.straightdope.com/u/Tyrrell_McAllister)\ **Post date:** [November 5, 2007, 8:45pm UTC](https://boards.straightdope.com/t/stumped-by-a-mathematics-question-area-of-triangle/425175/22 "2007-11-05T20:45:51Z")
One way to make the problem easier is to translate point _A_ rightwards and parallel to line _BC_ until it lies directly above the point _D_. If you simultaneously translate _F_ so that line _BF_ continues to bisect _AD_, then you will not change the areas of any of the regions. (In other words, treat _D_ as the origin in **R** [sup]2[/sup] and let _BC_ lie on the _x_-axis. Then, for some real number _r_, multiplication by the matrix [1 _r_] [0 1] preserves all areas (because its determinant is 1) and maps the point _A_ to a point directly above the origin.) So, we may as well assume that _AD_ is perpendicular to _BC_. It is then easy to see that triangles _ABE_ and _BDE_ have the same area _A_[sub]1[/sub]. Set _x_[sub]0[/sub] to be the length of _DC_ and _y_[sub]0[/sub] to be the length of _AD_. Then Area of _ABC_ = 5 cm[sup]2[/sup] = 3/2 _x_[sub]0[/sub] _y_[sub]0[/sub], and Area of _ABD_ = 2 _A_[sub]1[/sub] = _x_[sub]0[/sub] _y_[sub]0[/sub], so you can solve for _A_[sub]1[/sub] to find that _A_[sub]1[/sub] = 5/3 cm[sup]2[/sup], which is part (a). To solve for part (b), you can write down the equations for the lines containing _AC_ and _BE_ in terms of _x_[sub]0[/sub] and _y_[sub]0[/sub] to get the coordinates of their intersection at _F_. You find that _F_ = (2/5 _x_[sub]0[/sub], 3/5 _y_[sub]0[/sub]). Hence, triangle _AEF_ has “base” 1/2 _y_[sub]0[/sub] and “height” 2/5 _x_[sub]0[/sub]. Thus, its area _A_[sub]2[/sub] is 1/10 _x_[sub]0[/sub] _y_[sub]0[/sub]. But we saw earlier that _x_[sub]0[/sub] _y_[sub]0[/sub] = 2 _A_[sub]1[/sub] = 10/3 cm[sup]2[/sup], so this gives us _A_[sub]2[/sub] = 1/3 cm[sup]2[/sup], solving part (b). ---
**Author:** ![Hari\_Seldon](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/hari_seldon/32/5173_2.png) [@Hari\_Seldon](https://boards.straightdope.com/u/Hari_Seldon)\ **Post date:** [November 6, 2007, 1:19am UTC](https://boards.straightdope.com/t/stumped-by-a-mathematics-question-area-of-triangle/425175/23 "2007-11-06T01:19:26Z")
[QUOTE=Malacandra] To elaborate, draw PQ parallel to BC such that it passes through E. If any line that runs from A to any point on BC is bisected by PQ, all such lines are (and we are told that AD, which is one such, is indeed so bisected). In particular, the height of ABC, being a perpendicular from BC to A, is bisected. Thus, although we know the height of neither, we do know that the height of BED is half the height of ABC. [/QUOTE] I stand corrected. But that is what I meant by some clever argument. I haven’t read the rest of it, but I guess it is correct. The question is underdetermined in the sense that there are many dissimilar triangles that satisfy the conditions, but these answers are always the same. I made one of them, not entirely at random and sure enough the areas were 5/3 and 1/3. It was done so that the first was 5/3 and the second had quadratic irrationalities and I used a high precision program to calculate the area of the small triangle. So I was getting convinced the answer was right. ---
**Author:** ![garysoh](https://avatars.discourse-cdn.com/v4/letter/g/6a8cbe/32.png) [@garysoh](https://boards.straightdope.com/u/garysoh)\ **Post date:** [November 6, 2007, 1:31am UTC](https://boards.straightdope.com/t/stumped-by-a-mathematics-question-area-of-triangle/425175/24 "2007-11-06T01:31:33Z")
I think we got it! Thanks for your help! I can’t believe I was stumped by triangles. ---
**Author:** ![Malacandra](https://avatars.discourse-cdn.com/v4/letter/m/45deac/32.png) [@Malacandra](https://boards.straightdope.com/u/Malacandra)\ **Post date:** [November 6, 2007, 10:47am UTC](https://boards.straightdope.com/t/stumped-by-a-mathematics-question-area-of-triangle/425175/25 "2007-11-06T10:47:11Z")
[QUOTE=Hari Seldon] The question is underdetermined in the sense that there are many dissimilar triangles that satisfy the conditions, but these answers are always the same. [/QUOTE] Absolutely. We’re in complete ignorance about the lengths of the sides, and the size of the angles, and none of the formulae above do anything to address this (nor can). But it turns out that we need surprisingly little information to work out the areas of the small triangles. (Given the area, we can specify some constraints on the sides by working backwards through the semiperimeter formula (Heron’s formula), but that doesn’t get us very close to the actual size of the sides.) My hat’s off to **CJJ** \* - I reckon myself a pretty fair trigonometrician, but I was a little stuck until I read his excellent solution. ---
**Author:** ![CJJ](https://avatars.discourse-cdn.com/v4/letter/c/ecc23a/32.png) [@CJJ](https://boards.straightdope.com/u/CJJ)\ **Post date:** [November 6, 2007, 3:12pm UTC](https://boards.straightdope.com/t/stumped-by-a-mathematics-question-area-of-triangle/425175/26 "2007-11-06T15:12:03Z")
[QUOTE=Malacandra] My hat’s off to **CJJ** \* - I reckon myself a pretty fair trigonometrician, but I was a little stuck until I read his excellent solution. [/QUOTE] Thx…I too was stumped for a little while. Abandoning attempts to determine congruent lengths/angles was a key step for me (I spent quite a bit of time trying to pointlessly prove the angle at B was bisected by BF, convinced somehow that the two shaded triangles were similar). Once I started manipulating everything in terms of area and the (1/2)bh equation, the problem unravelled (drawing the line FD was also a breakthru). ---
**Author:** ![wolf\_meister](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/wolf_meister/32/15202_2.png) [@wolf\_meister](https://boards.straightdope.com/u/wolf_meister)\ **Post date:** [November 6, 2007, 3:59pm UTC](https://boards.straightdope.com/t/stumped-by-a-mathematics-question-area-of-triangle/425175/27 "2007-11-06T15:59:47Z")
**garysoh** I think _ **we** _ got it? LOL **Omphaloskeptic** , **CJJ** \*, **Tyrrell McAllister** and many others were the major contributors don’t you think? By the way **Malacandra** , I was going along a similar path. My great “discovery” was constructing a rectangle around triangle ABC such that its width would be AC and its height would be ‘h’ - the height of triangle ABC. Therefore, the rectangle’s area would be 2 x ABC = 10. Then I went on to Heron’s formula, the _1/2 b x h_ formula and realized I was _not_ going to find the solution that way and gave up. 🙂 ---
**Author:** ![Malacandra](https://avatars.discourse-cdn.com/v4/letter/m/45deac/32.png) [@Malacandra](https://boards.straightdope.com/u/Malacandra)\ **Post date:** [November 6, 2007, 4:05pm UTC](https://boards.straightdope.com/t/stumped-by-a-mathematics-question-area-of-triangle/425175/28 "2007-11-06T16:05:05Z")
**CJJ** \*, I think a nice way to close in on the finish is this: BED = 5/3 BEA = 5/3 AEF = x DEF = x FDC = y BFC = 3y Then we get a couple of easy-to-solve simultaneous equations for x and y: 5/3 + 5/3 + 2x + y = 5 5/3 + x + y = 3y reducing down to 2x + y = 5/3 2y - x = 5/3 giving y = 1 and x = 1/3. > [@CJJ\*](#): > > (drawing the line FD was also a breakthru). Yup! 🙂 ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [November 6, 2007, 6:43pm UTC](https://boards.straightdope.com/t/stumped-by-a-mathematics-question-area-of-triangle/425175/29 "2007-11-06T18:43:37Z")
I guess this problem’s been solved to death now. But since I find it aesthetically preferable to solve geometric problems “geometrically”, rather than with algebra and solutions to simultaneous linear equations, here’s my small contribution: The following is a solution for part two that doesn’t require getting your hands dirty with even the simple algebra of **CJJ** \*'s/ **Malacandra** ’s. You don’t even have to assume that the area of a large region is decomposable into the sum of the areas of its subregions. Everything is established directly in terms of distance ratios. First, we establish the ratio AF/AC, by way of establishing AF/FC. [In the usual overloading of notation, I’ll use YZ to indicate the length of the line segment YZ; in my own idiosyncratic notation, I’ll also use X\_YZ to mean the distance of the point X from the line containing YZ.]. Now, note that AF/AC = F\_AB/C\_AB and FC/AC = F\_BC/A\_BC. Therefore, AF/FC = (F\_AB/F\_BC)/(C\_AB/A\_BC). Furthermore, note that E\_AB = F\_AB \* BE/BF while E\_BC = F\_BC \* BE/BF. Therefore, E\_AB/E\_BC = F\_AB/F\_BC. Furthermore, E\_AB = 1/2 D\_AB and D\_AB = 2/3 C\_AB. Finally, E\_BC = 1/2 A\_BC. Putting this all together, we find that F\_AB/F\_BC = (1/3 C\_AB)/(1/2 A\_BC) = 2/3 C\_AB/A\_BC. That is to say, (F\_AB/F\_BC)/(C\_AB/A\_BC) = 2/3, telling us that AF/FC = 2/3, from which we can naturally conclude that AF/AC = 2/5. Having done the hard work, the rest is easy. Consider the triangle ADC; it has 1/3 the area of ABC [because DC is 1/3 of BC; you can view this as shrinking the base by a factor of 1/3 while holding the height the same or as shrinking the height by a factor of 1/3 while holding the base the same, depending on whether you originally took the perspective of BC or AC as “the base”]. And, in turn, AEC has 1/2 the area of ADC [because AE is 1/2 of AD; same reasoning]. Finally, AEF has 2/5 the area of AEC. Thus, since ABC has area 5 cm^2, we see that AEF has area 2/5 \* 1/2 \* 1/3 \* 5 = 1/3 cm^2. Ta-da! (I got this solution by starting with **Omphaloskeptic** ’s barycentric approach and reworking it into a more directly geometric intuition, though the resemblances to his solution may now be tortured and lost). [Previous page](https://boards.straightdope.com/t/stumped-by-a-mathematics-question-area-of-triangle/425175.md?page=1)